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K2
Writing Team Stan Jones
Ken Couchman Keith Slinn
Frank Morgan
Shakespeare Head Press Sydney ·Auckland
First published 1983 Reprinted 1984 (twice)
Reprinted 1985 Reprinted 1986 Reprinted 1987 Reprinted 1988
by Shakespeare Head Press The Education Division of Golden Press Pty. Ltd.
Incorporated in N.S.W. 46 Egerton Street,
Silverwater, N.S.W. 2141, Australia
© 1983 S.B. Jones & K.E. Couchman
This book is copyright. No part of it may be reproduced or transmitted without the written permission of the publisher.
ISBN 0 85558 773 3
Typeset in Hong Kong by Asco Trade Typesetting Ltd.
Printed in Singapore by Singapore National Printers Ltd
This book completes the 2 Unit course in senior mathematics. For flexibility of programming, Chapter 13 on Probability and Chapter 14 on Sequences and Series overlap with Book 1.
As in Book 1, Practice Papers are placed at intervals throughout the text and a set of 150 Revision Exercises are given at the back of the book for additional practice.
Some proofs in the calculus have again been given in the appendix where students can study them as their mathematical maturity allows.
S.B.J. and K.E.C.
·--·-- ~-.--~----I
\ J~r\..1'- I·~j\HiE CLASS C01·..:u1(!\, -~~
u ~ _de\ --~.J 'vc/
I -
I I j
i
TI-IIS BOOK IS THE PROPERTY OF ' SHOALHAVEN HIGH SCHOOL
-·------·- --
Equally likely outcomes- set notation- range of probability - ways of counting n(S) and n (E) - bias - probability trees - complementary events- non mutually exclusive events.
The general term - the arithmetic sequence -sigma notation - geometric sequences -compound interest - superannuation - time payments- limiting sums.
Graphs of quadratic functions - quadratic inequalities - the discriminant of a quadratic equation - sum and product of the roots -maximum and minimum - the sign of the quadratic function - quadratic identities -equations reducible to quadratics.
The second derivative - points of inflexion -turning points- curve sketching- the primitive function - application of the primitive function.
Examples of locus - the circle - the parabola - general locus proofs.
Area under a curve - the definite integral -the relation between the integral and the primitive function - the indefinite integral -calculating areas under curves - numerical methods of integration - trapezoidal rule -Simpson's rule-volumes of solids of revolution.
The graph and derivative of y = ax - the exponential function y = ex - differentiation of exponential functions - integration - the logarithm function - differentiation and integration -the derivative of y = ax -derivative of log8 x- a value for e.
Radian measure of angles - length of circular arc - area of a sector - graphs of the trigonometric functions - graphical solution of equations - differentiation of the trigonometric functions - integration of the trigonometric functions.
The derivative applied to velocity and acceleration - the use of primitives with motion in a straight line - exponential growth and decay.
I
The results given by this probability machine are related to Pascal's Triangle.
Pr ilit
CHAPTER 13
Probability is the study of unpredictable events and originated in the mid-17th century inspired by the enquiry of gamblers seeking information to help them win at cards and dice. Today probability has many important commercial and scientific applications. Life assurance companies, for example, have tables of life expectancy to help them calculate their premiums, businessmen carry out surveys to help predict the probable market size for a given item. These are attempts to make fairly precise statements about the chance or probability of an event happening.
An important idea in the study of probability is the idea of randomness. Suppose five different name cards are placed in a hat and one name is drawn at random; what do we mean? We mean that each name is equally likely to be drawn from the hat and that it is not possible to predict which name would be drawn.
1. Suppose one card is selected at random from this set of five cards. What colour card would most probably be selected? Is the card selected more likely to be a King or a ten?
2. A box contains 5 black balls and 4 red balls. You are blindfolded and asked to draw out one ball from the box. Is this a random choice? What colour ball would you most probably draw?
3. The numbers 1 to 9 are written on separate cards and the cards are turned face down on a table. One card is chosen at random. Is it more likely to be more than 7 or less than 7? Is it more likely to be odd or even? Why?
4. Suppose a bag contains blue marbles and red marbles. You are asked to take one marble at random from the bag and you are told that it will probably be blue. What does that suggest to you about the marbles in the bag?
When the occurrence of events is equally likely we can state the expected probability of an event occurring by considering the possible outcomes.
Tossing a Coin Consider the tossing of a coin. There are two possible outcomes, heads or tails, and if it is a fair coin then each of the two outcomes is equally likely to occur. There is one chance in two of tossing a head. We say the probability of tossing a head is!. What is the probability of tossing a tail?
2
Rolling a Die Each ofthe numbers 1 to 6 is equally likely to be on top. There is one chance in 6 ofthrowing the number 4. We say the probability of throwing a 4 is i. What is the probability of throwing a 6?
Drawing a Marble from a Bag
Possible Outcomes
A bag contains 3 red and 2 black marbles. If one marble is drawn at random it is equally likely that any one of the five marbles will be chosen. Consider the probability that the marble chosen is red. Now of the 5 possible outcomes, 3 of them are favourable for the choice of a red marble. We say the probability of drawing a red marble is~. What is the probability of drawing a black marble?
In general the probability that an event will occur can be given as P(Event) where:
P(E t) _ number of favourable outcomes
ven - . . number of posstble outcomes
Further: (i) if there are a favourable outcomes for an event and b unfavourable outcomes and all outcomes are equally likely then:
the probability of the event occurring = ____!!____b a+
the probability of the event not occurring = ___!!__b a+
Note that the sum of these two probabilities is 1 since ____!!____b + ___!!__b = a + bb = 1. a+ a+ a+
(ii) if an experiment has n equally likely outcomes E1 , E2 , E3 , ..• , E,. then P(E1) + P(E2 ) + P(E3) + · · · + P(E,.) = 1.
1. Toss a coin 50 times and keep a record of the number of heads and tails. (a) Based on your sample is the probability of tossing a
head close to ! ? (b) Compare your results with other groups. (c) Now combine the results of ten groups. Based on
this larger sample, is the probability of tossing a head close to the expected probability?
2. Roll a die 60 times and record the results using a tally (a) Based on your sample, what is the probability of
throwing a 3? (b) What would be the theoretical probability of
throwing a 3? (c) Combine the results of ten groups. What is the
ratio of the number of times 3 appeared to the total number of tosses? Is this ratio close to the expected probability of i?
3
TAILS
1
2
3
4
5
6
7f!./ tfi.J I I
7fl/. I
11/.1.
7tf/.. II
Ill
7111. Ill
7'fl.l..
3. (a) From a well shuffled pack of 52 playing cards draw a card and record the result with a tally mark for the events listed in the table. Replace the card and reshuffle the pack. Do this 50 times.
(b) Combine your results with other groups and find the experimental probability of each of the outcomes. Compare this with the theoretical probability of each outcome.
BlACI< CARD
DIAMOND
J, a. 1<. A OF ANY SUIT
7IIJ. 7f/.l. /Ill
7IIJ. Ill
7fll. 7111. I
I. A bag contains 3 black marbles and I white marble. If one marble is drawn out of the bag without looking what is the probability that it is black?
2. A coin is chosen at random from 3 two-cent and 4 five-cent pieces. What is the probability that it is a two-cent piece?
3. From a pack of cards the four aces are turned face down on a table and thoroughly mixed. A girl selects one card. What are her chances of selecting: (a) the ace of clubs? (b) a red ace?
4. Five pupils: Susan, Robert, Janet, Ian and Gloria write their names on separate cards and put them into a hat. One name is drawn from the hat. What is the probability that: (a) the name is Janet? (b) the name is a girl's name? (c) the name is a boy's name?
5. A die-cube has 2 faces red, I face white and the other faces blue. What is the probability if the die is rolled on a table that: (a) a red face would be uppermost? (b) a blue face would be uppermost?
6. If the probability of an event is-!-, about how many times would you expect it to occur in 80 trials?
7. A small deck of cards contains 10 blue and 10 yellow cards. A blue card is drawn from the pack and not returned. If a random draw is then made, which colour is most likely to be selected?
The gambling game Roulette illustrates simple probability very well. In the game a wheel is set spinning and a white marble is thrown in the opposite direction. The marble comes to rest in one of the slots which have numbers on them.
4
0
/ 1 2 3
Passe 4 5 6 Manque high low
19-36 7 8 9 1-18
10 11 12
13 14 15 Pair Impair all 16 17 18 all the the
even 19 20 21 odd numbers numbers
22 23 24
25 26 27
Noir 28 29 30 Rouge (black) (red)
31 32 33
34 35 36
' r /)'II
~ Colonnes
(Column of 12 numbers)
The slots are numbered 1-36 of which half are red and half are black. They are so chosen that some red numbers are odd and some are even, some are low numbers (1-18) and some are high numbers (19-36). With only 36 numbers the probability of an odd number is! or 50%. In this case you would have just as much chance of winning as the casino and there would be no profit for the operator. For this reason a zero (0) slot is included on the roulette wheel and the probability is altered slightly in favour of the casino. All winning bets are paid as if there are 36 slots. You can bet on the zero but it does not count for bets on odd or even, red or black and if the marble falls in the zero slot all bets on these are lost.
People indicate their bet by placing tokens on appropriate places on the table. The token can be placed on one number, or between a pair of numbers or on a corner where four numbers meet, indicating the choice of any of the four. French words abound in roulette because of its French origin. Thus a token placed at the end of a row of three numbers is called transversale and at the end of a group of six numbers is called sixaine and indicates bets on these numbers. Other words are shown above on the diagram and you will meet some in the exercises. The call for players to place bets is Faites vos jeux and when the croupier calls Rien ne va plus there are no more bets allowed.
Example: Carre refers to a group of 4 numbers. What is the probability that one of these numbers will win?
Solution: P(group of 4) = 3
47 •
5
These questions refer to a 37 slot French roulette wheel.
1. What is the probability for a black number (nair)?
2. What is the probability for any even number (pair)?
3. A person choosing colonnes wins if any of the twelve numbers in that column occurs. What is the probability for this?
4. What is the probability for success of a single number (en plein)?
5. Sixaine refers to a group of six numbers. What is the probability that the marble will end against one of them?
6. What is the chance of success with: (a) cheval (a pair of numbers)? (b) transversale (a group of 3 numbers)? (c) douzaines (a group of 12 numbers)?
7. Ken put his tokens on the individual numbers 7, 19,20 and 33. What chance has he of success?
8. Mary chose all the odd numbers and, as well, the numbers 6 and 14. What is her probability for success?
9. What is the chance that one of the numbers from 1 to 8 will win?
10. Jenny covered all the low numbers (1 to 18) and also from 31 to 36. What chance did she have of success?
We can apply the notation of sets to the statement of probability in favour of an event. Suppose we wish to find the probability of throwing a number greater than 4 with one roll of an ordinary die. Let S be the set of possible outcomes called the sample space, then S = {1, 2, 3, 4, 5, 6}. Let E be the subset of outcomes favourable to the required event, then E = {5, 6}. If P (Event) is the probability of the given event occurring then:
P(Event) = n(E) = number of elements favourable to event n(S) number of elements in the sample space
Thus P (number greater than 4) = % _ _1 - 3
Example (i):
Sample Space
Event Subset
From a normal pack of 52 cards, one card is selected at random. What is the probability that it is: (a) an ace? (b) a club?
6
Solution: (a) Now if E is the set of favourable elements,
n(E) = 4, since there are 4 aces in the pack and n(S) = 52, since there are 52 cards in the pack. Since every element in the sample space is equally likely to be selected
P ( ) _ n(E)
ace - n(S)
= s4z
_ _1_ - 13
Thus the probability of drawing an ace from a deck of 52 cards is / 3 •
(b) Now if E is the set of favourable outcomes n(E) = 13, since there are 13 clubs in the pack.
P (club) = n(E) n(S)
_ _u - 52 _ _!_ -4
Thus the probability of drawing a club is!.
Example (ii): A four-digit number is to be formed from the digits 1, 3, 5 and 8 which are printed on cards. What is the probability that the number will: (a) start with the digit 3? (b) be odd? (c) be greater than 5000?
Solution: (a) We are in this part concerned only with the first
digit in the number. It is possible to fill this place with any one of the four given digits. Thus the sample space,
s = {1, 3, 5, 8} Now only the digit 3 will give the required event.
.'. E = {3}
P (number starting with 3) = n(E)) n(S
=! Thus the probability that the number starts with the digit 3 is!.
7
3 1 8
(b) In this part we are concerned only with the last digit (the units digit). It is possible to fill this place with any one of the given digits .
... s = {1, 3, 5, 8} For the number to be odd the last digit must be odd, thus E = {1, 3, 5}.
P (odd) = ~~~~ _i\_ -4
Thus the probability that the number is odd is l (c) The number will be greater than 5000 provided it
starts with 5 or 8. Again considering the first place s = {1, 3, 5, 8} E = {5, 8}
. n(E) .. P (greater than 5000) = n(S)
-1. -4 -1. - 2
:. The probability that the number is greater than 5000 is t·
If the probability in favour of event A is {0 and the probability in favour of event B is~' then event A is more likely to occur than event B. We could say that there is a high probability that event A will occur and a low probability that event B will occur.
The lowest probability that can ever be quoted is zero, and this when an event can never occur, e.g., selecting a 5 cent coin from a box containing only 10 cent coins.
The highest probability that can ever be quoted in favour of an event is 1 and this only when the event is certain, e.g., drawing a red marble from a box containing only red marbles.
Thus the range of probability for an event is from 0 to 1 and to quote any number outside this range is meaningless.
1. If a die is rolled once, what is the probability of throwing: (a) 6? (b) an even number? (c) a number less than 3?
1
5
1
1 2
0
5 8 3
1 8
Probability
...,.__ CERTAIN
...,.__ LIKELY
...,.__ EVEN CHANCE
...,.__ UNLIKELY
...,.__.. IMPOSSIBLE
2. The probability that a car driver will be stranded by mechanical failure in 15 000 km of driving is 1
30 • What is the probability that the driver will have trouble free motoring for the 15 000 km?
3. At a school assembly 62% of the students said they were in favour of a new design for the school badge. If the headmaster asked a student at random about the proposed design, what is the probability that this student was: (a) in favour of the new design? (b) not in favour?
4. In a raffle 40 tickets are sold and there is one prize. What is the probability that someone buying five tickets wins the prize?
5. Give the probability of each of the following selections occurring at a single random draw from a pack of 52 playing cards: (a) the ace of clubs (b) a seven of any suit (c) a black Jack
(d) a court card (Jack, Queen, King, Ace) (e) a red card (f) either the king of spades or the queen of diamonds.
6. A shelf contains 9 different books, arranged in any order between a wall and a book-end. Three are yellow, 5 are blue, and one is green. What is the probability that: (a) a yellow book is in the middle position? (b) a blue book is against the wall? (c) the green book is beside the book-end?
7. The numbers I to 20 are written on twenty cards. If a card is chosen at random what is the probability that : (a) the number is a multiple of 5? (b) the number is even? (c) the number is odd?
(d) the number will contain the digit 7? (e) the number will contain the digit 2?. (f) the number will contain the digit I?
8. A bag contains 3 blue, 5 red, 4 green and 2 white balls. Find the chance of drawing at random one ball which is:
9.
(a) red (b) either blue or green (c) not white.
The organisers of a local procession which is held every year on the first Saturday in June have available a record of the weather over the past fifty years on the procession day.
Based on the record given in the table what is the probability that this year's procession will be: (a) held on a sunny day with no cloud? (b) held on a rainy day?
4 / 10 / 15 / / 21 /
10. A number is formed by using all five digits 2, 3, 4, 5, 6. What is the probability that the number: (a) starts with 4? (d) is greater than 30 000? (b) is even? (e) is divisible by 3? (c) is odd?
11. Joel and Debbie are each drawing one card at random from a pack of ten cards numbered 1 to 10. Joel draws a "6" which is replaced and the pack shuffled. What is the probability that Debbie will draw a higher card?
12. From a set of 18 discs numbered from 1 to 18, one is drawn at random. What is the probability that the number on the disc is: (a) a multiple of 5 (b) a multiple of 3 (c) a multiple of 4
13. A bag contains 5 white and 8 green marbles.
(d) a multiple of either 4 or 5 (e) a multiple of either 3 or 5 (f) a multiple of either 3 or 4.
(a) What is the probability of drawing at random, a white marble from the bag? (b) If a white marble is drawn first and not replaced, what is the probability of picking up a
white marble at the second draw? (c) Suppose white marbles have been selected at the first and second draws and not replaced,
what is the probability of picking a green marble at the third draw?
9
In some problems it is helpful to draw a diagram or use some other systematic procedure in order to determine the elements in the sample space.
TWO DICE ROLLED SIMULTANEOUSLY
The diagram shows the sample space (i.e. possible outcome) when two dice are rolled together. We have shown one die white and one die red to indicate, for example, that there are two distinct
sample points whose sum is 3, i.e., 2 on the white and 1 on the red, and vice versa.
Example:
36 SAMPLE POINTS
What is the probability of throwing a total of five in one roll of a pair of dice?
Solution: We can see that all the sample points whose sum is five lie in one line parallel to a diagonal.
If we write the number on the white die first, the sample points can be written as ordered pairs. Now E = { (1, 4), (2, 3), (3, 2), (4, 1)} thus n(E) = 4, and from the diagram n(S) = 36.
. . n(E) .. P(f1ve) = n(S)
= 3~ _ _1 - 9
Hence the probability of throwing a five total with a pair of dice is ! . 10
1. A pair of dice is thrown. Use the diagram to count the number of sample points favourable to the events listed below and hence find the probability of the event in each case. (a) a total of 7 (b) a double six (c) any double (d) a total of 10 (e) either a score of 7 or a score of 11 (f) a score greater than 6 (g) an even score (h) an odd score also greater than 4 (i) at least one six on the uppermost face of a die (j) at least one three on the uppermost face of a die.
2. Which would give the better chance; rolling one die or rolling two dice, if you wanted to throw a score of, (a) 6? (b) 2?
3. A game is played so that a player can start only if he throws a double or a total of six, using two dice. What is the chance of a player's starting on the first throw?
Craps is a game of chance played by rolling two dice. The total made by the two dice is what matters. This is a game played in all big casinos.
WIN LINE IN CRAPS
This is an even money wager and there are three rules: 1. If, when the two dice are rolled, the total is 7 or 11, the player wins. 2. If a total of 2, 3 or 12 shows, the player loses immediately. 3. If any other total such as 5, shows; this is called the player's point. The player keeps on rolling the
dice. He wins if this total (his point) appears before a 7 and he loses if a 7 appears first.
1. Complete the following table showing the number of ways various totals can be obtained with two dice.
TOTAL OF 2 DICE 2 3 4 5 6 7 8 9 10 11 12 All outcomes
Number of ways 36
2. What is the chance of losing on the first throw, that is by scoring a total of 2 or 3 or 12?
3. If on the first throw a total of 5 appears is it more likely to gain another 5 before a 7?
4. Is it more likely to win or lose on the first throw?
5. What is the probability that the game will end one way or the other with the first throw?
11
6. What is the probability that the game will not end on the first throw?
7. If the first throw is 10 what is the probability that the next throw will be a 10? (Hint: The dice can't remember what their first throw was! This question is simply asking what the chance is of throwing a 10.)
8. If none of the numbers 7, 11, 2, 3 or 12 occurs first up what is the next best total to get? Mention two if they are equally desirable.
9. Write all the totals in the order they would be preferred. For example a 7 or an 11 are the most desirable totals. What is the remaining order?
A tree diagram can be used to find the elements in the sample space.
Example (i): The numbers 2, 3, 4 and 5 are written on separate cards. One card is drawn at random to give the tens digit of a two digit number and then one of the remaining cards is also drawn at random to give the units digit. What is the probability that the number formed will be divisible by 6?
Solution: From the tree diagram we can see that the set of two digit numbers in the sample space contains 4 x 3 = 12 elements.
The numbers in the sample space which are divisible by 6 have been ticked.
Thus the probability that the number formed is divisible by 6 is given by:
P(number divisible by 6) = ~~;? 3
12 1 4
3 23
2 ~ 4 24
5 25
2 32
3 4 34
5 35
2 42
4 3 43
5 45
2 52
5 3 53
4 54
The main value of the tree diagram is that it lists every possible element in the sample space. In this case every two digit number that could be formed is found by following all possible branches on the tree diagram.
Another value of the tree diagram is that it illustrates an important principle of counting possibilities known as the product rule.
If a selection can be made in r different ways and if a second selection can be made ins different ways then the two selections can be made in sucession in r x s different ways.
Example (ii): A coin is tossed three times and the results noted. What is the probability of the event that two heads and a tail will be thrown, if the order is not considered?
12
Solution: Draw a tree diagram to obtain the sample space.
H H
T H
H T
T
H H
T T
H T
T
The three sample points ticked give the required event .
.'. P(two heads and one tail) = :~~~ _.J. - 8
.·. The probability of throwing two heads and a tail is i.
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
Note: If three coins are tossed together the sample space is identical with that in the tree diagram as the headings 1st toss, 2nd toss, 3rd toss, can be replaced by 1st coin, 2nd coin and 3rd coin.
Example (iii): A simple poker machine has only two wheels. On one of them there are the numbers 1, 2, 3 and 4. On the other one the letters A, A, B and C. (a) How many arrangements can show on the dial (including repetitions)? (b) In how many ways can the machine stop with the first wheel on a 3 and the second wheel showing
anA? (c) What is the probability that the arrangement 3-A will appear?
Solution: We draw a tree diagram:
A 1A
A 1A 8 18
c 1C 2A (a) Altogether 16 arrangements are
2A possible (not all different). 2
28 (b) 3-A occurs twice. --c 2C (c) P(3-A) = n(E)
3A n(S) A 3A
_ _.1_ - 16
3 38 _1_ - 8
3C
~/A 4A 4A
4 48
~,c 4C
13
Where necessary draw a tree diagram or use some systematic procedure for finding the sample space in each of the following problems.
1. A set of three cards numbered 1, 2 and 3 are placed in a hat. (a) How many different two-digit numbers would it be possible to form by selecting two cards
in succession and placing them in order? (b) Use a tree diagram to list all the possible numbers. (c) What is the probability that any two-digit number formed in this way is even?
2. Two white balls and one black ball are placed in a bag. Two balls are selected at random. Find the probability of choosing:
0 00 (a) 2 white balls. 0 (b) 1 white and 1 black ball. • oe
0 00 0
• oe 0 eo • 0 eo
3. A coin is tossed three times. Use a tree diagram to find the probability in favour of the event: (a) one head and two tails in any order (b) three heads or three tails (c) at least one tail.
4. A tennis team of four players A, B, C and D select a captain and vice-captain. (a) List all the possibilities. (b) What is the probability that player A will be either captain or vice-captain?
5. Four Jacks are taken from a pack of cards and placed face down. Two of these cards are chosen at random. Use the tree diagram given to help find the probability of choosing: (a) 2 black Jacks. (b) at least 1 black Jack.
SPADE
CLUB
HEART
CLUB
HEART
DIAMOND
SPADE
HEART
DIAMOND
SPADE
CLUB
DIAMOND
SPADE
DIAMOND CLUB
6. A coin is tossed 4 times. What is the probability of 3 heads and a tail in any order?
7. Two coins are tossed together. What is the probability of: (a) 2tails? (b) 1 head and 1 tail?
14
HEART
8. If you toss a coin and throw a die, what is the probability of getting: (a) a head and a 4? (b) a tail and an odd number? (c) a head and a number less than 3?
9. There are two roads m and n between towns A and B. There are three roads x, y and z between towns Band C. If the path is chosen at random from A to C, what is the probability that a traveller will journey over roads m andy?
A
n
10. A two-digit number is to be formed from the set of digits {3, 4, 5, 6}.
m
(a) If any digit may be repeated how many two-digit numbers can be formed? (b) What is the probability that the number formed is divisible by 5? (c) What is the probability that the number formed has both digits the same?
11. Use the tree diagram given to find the sample space for the composition of a family of 3 children. For such a family what is the probability of: (a) 3 girls? (b) 2 boys and 1 girl? (c) the eldest child being a girl? (d) the youngest child being a boy?
B
G
/
X
z
B
/ B
G
B
G G
B B
G
B
G G
12. From a set of 4 cards, consisting of the Kings from a normal pack, 3 are chosen at random. Find the probability that the two red Kings would be included in the 3 cards chosen.
13. Cards in a set are numbered 1, 2, 3, 4 and 5. If two cards are drawn at random: (a) write down all the possible two-digit numbers that can be formed (b) what is the probability that the number formed is odd?
14. Three yellow cards are numbered 1, 2 and 3. Three red cards are also numbered 1, 2 and 3. -If a yellow and a red card are selected at random, what is the probability that the cards selected:
(a) are the yellow 2 and the red 3? (b) give a total of 5? (c) are the yellow 3 and the red 3? (d) give a total of 2? (e) have a total greater than 3?
15. A drawer contains 2 black and 2 red ball-point pens, and nothing else. If the pens are identical in size and shape, what is the probability if two are chosen in the dark, that they will both be red?
16. Three books are on a shelf. One book has a red cover and the other two have blue covers. If two books are selected by chance what is the probability that the red book is included in the selection?
17. Five golf balls are in a pouch. Four have red numbers and one has a blue number printed on it. If two are selected at random, what is the probability that the one with the blue number is in the pair selected?
15
18. (a) A poker machine has 4 wheels and 20 symbols on each. How many arrangements are possible? (b) Suppose it is possible for a certain winning combination to occur in 4 ways, what is the prob
ability of this occurring?
19. (a) A poker machine with 3 wheels has 4 Aces on the first wheel, 2 Kings on the second wheel and 3 Jacks on the third. In how many ways can the machine stop to show Ace, King, Jack?
(b) If each of the three wheels has 10 symbols, then how many total positions are possible and what is the probability for Ace, King, Jack?
At times outcomes of experiments are not equally likely. Sometimes this is due to bias. If the nature of the bias is known then allowance can be made for it in the sample space.
Example: A die is biased so that to throw a six is twice as likely as any other number. Find the probability of throwing: (a) a three (b) a six.
Solution: The six is given extra weight in the sample space by counting it twice.
Thus S = {1, 2, 3, 4, 5, 6, 6}
(a) P (throwing a three) = :~~j _ _i
(b) P (throwing a six)
- 7
n(E) n(S)
_l_ - 7
1. A die is biased so that to throw a 2 is twice as likely as any other number. Find the probability of throwing: (a) a six (b) a two.
2. The 38 slots on an American roulette wheel are marked 1-36, 0, 00. If the wheel is biased so that 00 is twice as likely as any other slot to pick up the ball, find the probability of a player winning if he bets on number 7.
3. A coin is so weighted that over many trials the ratio of heads to tails will be 5 to 3. Find the probability, at a single trial, of throwing a tail.
In some questions on probability, basic tree diagrams become very large and unwieldy. It is then a useful economy to use a probability tree.
Examples for study
Example (i): A bag contains 3 white and 4 brown balls. Two balls are drawn at random, the first not being replaced before the second is drawn.
16
(a) The probability of White at the first draw is ~. Now given the condition that the first draw resulted in White, the probability of White at the second draw is ~' because there are now six balls in the bag of which two are White.
The probability of getting White at each draw can be calculated by multiplying the probabilities along the coloured branch.
P(White, White) = ~ x ~ = fz = ~
2 6 w
This rule can be verified by noting that these are 3 ways of picking a white ball at the first draw and 2 ways at the second draw. Thus by the product rule the sequence White, White can be chosen in 3 x 2 = 6 ways. There are 7 possible outcomes of the first draw and 6 for the second draw, giving 7 x 6 = 42 possible outcomes for the two draws in sequence.
This gives a probability for White, White of ,f2 which corresponds with the result obtained by multiplying the probabilities along the branch giving White, White.
The product rule for a probability tree states that the probability of an event is the product of the probabilities along the branch which yields the event.
(b) Now the probability of each outcome can be easily calculated by using the product rule above.
P(White, Brown) = ~ X ~ = g = t S · l P(White, White) = l x ~ = ,f2 = ~ t P(Brown, White) = 4 x ~ = !~ = t urn ts P(Brown, Brown)= 4 X~= g = t
Notice if the sum of the probabilities along the separate branches is 1, we have a check on our work.
(c) If the required event is drawing a white ball and a brown ball in any order, we note that this can be done in two ways (White, Brown) or (Brown, White). Thus to find the probability of this event the two probabilities concerned are added.
P(White and Brown) = P(White, Brown) + P(Brown, White)
=t+t -± - 7
The addition rule for a probability tree states that the probability of a set of mutually exclusive events is the sum of the probabilities of corresponding branches.
Example (ii):
If a coin is tossed and a die thrown, find the probability that a head and a number greater than 2 results.
Solution: The coloured branch gives the required event. Thus the probability of the event is the product of the probabilities along this branch.
P (head and > 2) = ! x ~ = ti _.!_ - 3
17
Example (iii): There are nine cards. The numeral 4 is printed on three of the cards, the numeral 5 on two of them and the numeral 7 on four of them. One card is drawn at random and not replaced, giving the tens digit of a number. Then another card is drawn to give the units digit. Find the probability that the number formed: (a) is 44 (b) contains a repeated digit.
Solution: Using the probability tree diagram. (a) P(44) = ~ X i
_ _§_ - 72 _ _L - 12
(b) The branches giving repeated digits are shown in colour. The probabilities along each branch must be found by multiplication and then P(44), P(55) and P(77) must be added.
P(repeated digits) = P(44) + P(55) + P(77) = (~ X i) + (~ X i) + (~ X ~) = 7
62 + /2 + n
= ~~ = ls
1. In Example (iii) above, find the probability that the number formed, (a) is 47 (c) has a digit sum of 11 (b) is 74 (d) contains the digit 5 at least once.
2. A box contains 3 red and 2 black marbles. Two are drawn at random, the first not being replaced before the second draw. Find the probability that: (a) 2 red are drawn (b) a black is drawn followed by a red (c) the selection contains a red and a black.
3. (a) A die is thrown three times. Find the prob-ability of:
(i) three sixes (ii) no sixes
(iii) two sixes (iv) at least two sixes.
(b) Would the probabilities above be altered if three dice were thrown together?
4. To win a golf tournament a professional is faced with two putts, one on the 17th green which he has an even chance to sink, and one on the 18th green that he would sink 3 times out of 4. What is his chance of: (a) winning the tournament? (b) missing both putts? (c) sinking the putt on the 17th green and failing on the 18th green?
18
5. A certain canteen plans its lunch menu in 6 day units. On two days potato pie comprises the first course and on 3 days fruit salad is served as dessert. What is the probability that a given meal chosen at random includes: (a) neither? (b) both?
6. The probability that a certain missile will blow up its target is~. If two missiles are launched in quick succession, what is the probability that the target will escape?
7. What is the probability of throwing six on a die four times in succession?
8. The probability that a car driving on sealed roads will have a puncture in 25 000 km is l What is the probability that this car does not have a puncture during the first 100 000 km? A new set of tyres is fitted every 25 000 km.
9. If i of the boys in a certain class carry a briefcase, and i are out of uniform, find the probability that a boy selected by lot, would be in uniform and carrying a briefcase.
10. Of7 equally matched toy racing cars, 2 are white, 4 are red, and one is green. Find the probability that if two races are run : (a) both will be won by a red car (b) both will be won by a car of the same colour (c) the first race will be won by a red car and the second by a white car (d) that one race will be won by a green car and one by a red car.
11. The probability that a tail-end batsman will lose his wicket on any ball is 0·1. The probability that he survives the first ball of an over is therefore 0·9. The probability that he survives the first and second balls is given by the product rule as 0·9 x 0·9. What is the probability that he survives: (a) the first 3 balls? (b) a complete 6 ball over?
12. What is the probability of drawing the ace of hearts on two successive occasions from a pack of 52 playing cards with replacement and shuffling between each draw?
13. A television set makes a whistling noise 5 times out of 6 when it is switched on. When the technician comes to repair the set what is the probability that it does not make the whistling sound when switched on: (a) once? (b) twice? (c) three times?
14. When an egg is broken from the shell into a frying pan the probability that the yoke breaks is said to be 1 in 10. If three eggs are fried, what is the probability that no yokes will be broken?
15. On a three-wheel poker machine these are 10 symbols on each wheel. There are 5 lemons on the first wheel, 3 on the second and 2 on the last. (a) What is the probability that 3 lemons will show on the centre line? (b) What is the probability that the two outside wheels will show a lemon with something dif
ferent on the middle wheel?
In a given experiment if E is the subset of outcomes favourable to an event then E, the complement of E, represents the subset of outcomes not favourable to the event and P(E) is the probability that the complementary event occurs. The event E is often referred to as not E.
E and E together form the total sample space thus P(E) + P(E) = 1
or P(E) = 1 - P(E) This result can be useful in finding probabilities where the words at least are used.
Example (i): What is the probability that in a family of five children there is at least one boy.
19
Solution: The probability of no boys = t x t x t x t x t
_ __!_ - 32
The complement of no boys is at least one boy. Hence the probability of at least one boy = 1 - l 2
_ _li - 32
Example (ii): In three throws of a die, what is the probability of at least one six?
Solution: The probability of no six in three throws = i x i x i
=iU The complement of no six is at least one six. Hence the probability of at least one six = 1 - iU
- 91 - 216'
1. If a coin is tossed three times what is the probability of getting at least one head?
2. What is the probability that in a family of four children there is at least one girl?
3. Two dice are thrown. What is the probability of throwing at least one six?
4. (a) In four throws of a die what is the probability of getting at least one six? (b) If five dice are thrown, what is the probability of throwing at least one six? Hint: The
calculations when throwing five dice can be done in the same way as for throwing one die five times.
5. The four Aces from a pack of cards are laid face down. One is selected at random and replaced. In two such selections what is the probability of drawing the Ace of spades at least once?
6. There are two boxes A and B. Box A contains 3 white balls and 2 black balls. Box B contains 4 white balls and 3 black balls. One ball is taken at random from each box. What is the probability that at least one of the balls is white?
7. A die is thrown and a coin is tossed, what is the probability (a) that the die shows six and the coin shows a head? (b) of at least one of the events in part (a) occurring?
8. In a mixture of brown and white pebbles in a gravel, brown and white pebbles occur in the ratio of 2: 3. Find the probability that if three pebbles are chosen from the mixture, at least one is white.
9. On a three wheel poker machine there are 20 symbols on each wheel. There are 5 cherries on the first wheel, 3 on the second and 2 on the third. What is the probability that: (a) 3 cherries will show across the centre line? (b) at least one cherry will show on the centre line?
Consider the following example. From a pack of cards numbered 1 to 10, one card is drawn at random. What is the probability that the number on the card is : (a) less than 5 or divisible by 7? (b) less than 5 or divisible by 2?
20
Solution: (a) Let A be the event of drawing a number less
than 5 and B be the event of drawing a number divisible by 7. The events A and B are mutually exclusive; we cannot select a number less than 5 and a number divisible by 7 at the same time.
We require A or B and denote this by A u B (called the sum of A and B). Now from the diagram.
P(A u B) = P(A) + P(B)
= 140 + lo
_.!_ - 2
(b) Let A be the event of drawing a number less than 5 and B be the event of drawing anumber divisible by 2.
The events A and B are not mutually exclusive; drawing a number less than 5 does not exclude the possibility of drawing a number divisible by 2.
Since A and B are not mutually exclusive events, A u B contains the points which are common to A and B; these are the points An B.
We want P(A u B), but if we add P(A) and P(B) the points of P(A n B) will have their probabilities counted twice. Thus P(A u B) = P(A) + P(B) - P(A n B)
= 1~ + 15o - /o
= 2o
s
s
6
8
10 5 9
A B
6 7
3 10
9 5
Note: In probability theory A n B is called the intersection or product of the events A and B. It is written as AB and denotes the event that both A and B occur.
1. If A and B are two events in a random experiment P(A u B) = P(A) + P(B) - P(AB)
2. In the special case where A and B are mutually exclusive events P(AB) = 0. Hence P(A u B) = P(A) + P(B).
Example: One card is selected at random from a pack of 52 playing cards. What is the probability that it is either a red card or a ten?
Solution: A is the event of a red card drawn; P(A) = ;~ B is the event of a ten drawn; P(B) = 5i A and B are not mutually exclusive events as the event AB is a red ten drawn; P(AB) = l 2
P(A u B) = P(A) + P(B) - P(AB)
= ~~ + 542 - l2
=~ = 23
Hence the probability that the card is a red card or a ten is 23 .
21
Note: In solving problems the student can work from the above formula or work from a counting diagram such as a Venn diagram or a simple tree diagram. The important thing is to recognise a situation where the events are not mutually exclusive being careful not to count twice those elements in the intersection.
In the following exercise some events are mutually exclusive and some are not.
1. From a set of cards numbered 1 to 12, one is selected at random. What is the probability that the number on the card is : (a) less than 5 or divisible by 6? (b) less than 5 or divisible by 4?
2. A card is selected at random from a pack of 52 playing cards. What is the probability that it is: (a) a black ten or an Ace? (c) a club or an Ace? (b) a black card or an Ace? (d) a club or a court card?
3. If two dice are thrown what is the probability of: (a) a double or a total of 8? (b) a double or a total of 9?
4. A bag contains 3 white marbles, 4 yellow marbles and 2 red marbles. A marble is drawn at random, what is the probability that it is either white or yellow?
5. From the integers 1 to 50 an integer is chosen at random. What is the probability that it is: (a) divisible by 4 and 6? (b) divisible by 4 or 6?
6. Two dice are thrown. What is the probability of obtaining a total which is either even or greater than 7?
7. From a set of 30 discs numbered 1 to 30, one disc is selected at random. What is the probability that the number is: (a) a multiple of 5? (b) amultipleof7? (c) a multiple of 3?
(d) a multiple of either 5 or 7? (e) a multiple of either 5 or 3? (f) a multiple of either 3 or 7?
8. (a) If A and Bare mutually exclusive events and P(A) = 152 and P(B) = 1
32 , find P(A u B).
(b) If A and Bare not mutually exclusive events and P(A) = { 0 , P(B) = 240 and P(AB) = l 0 ,
find P(A u B).
9. From the integers 1 to 11, one is chosen at random. What is the probability that it is less than 9 or divisible by 4?
10. From the numbers 4, 5, 6, 7, 8, 9; what is the probability of selecting at random an odd number or a number less than 7?
22
A computer can be programmed to give the monthly instalments on loans which are subject to reducible interest. The calculations involve geometric series.
Photograph courtesy of /.B.M.
CHAPTER 14
If a set of numbers is written down in order according to some rule or pattern, the set of numbers is called a sequence and each number is called a term of the sequence. In some sequences the pattern can readily be seen and this allows us to predict subsequent terms in the sequence.
For example consider the sequence 1, 3, 6, 10, 15, 21, 28, .... The difference between successive terms increases by 1 each time, so the sequence may be continued indefinitely. The next three terms in the sequence are 36, 45 and 55.
Discover the pattern in each of the following sequences and use this knowledge to write down the next three terms of the sequence.
1. 3, 7, 11, 15, 19, -, _,- 7. 31, 24, 17, 10, _, _,-
2. 3, 4, 6, 9, 13, 18, -, _,- 8. -17, -8, 1, 10, _, _,-
3. 19, 16, 13, 10, 7,_,_,_ 9. 1, -2,4, -8, 16, -32,_,_,_
4. 3,6, 12,24,48,_,_,_ 10. 1, 8, 27, 64, _, -,-
5. 1, 4, 9, 16, 25, _, _,- 11. 1, 1, 2, 3, 5, 8, 13, _, -,-
6. 2, 6, 18, 54, 162, _, _,- 12. 4, 3, 7, 6, 10, 9, 13, 12, _, -,-
It is possible to indicate a sequence by giving its general term. The symbol T,, is used to represent the general term or nth term of a sequence.
For example if T,, = 3n + 2, then the first term is found by substituting 1 for n in this expression and so on for the other terms: Thus T1 = 3 x 1 + 2 = 5
T2 =3x2+2=8 T3=3X 3+2=11 T4 = 3 X 4 + 2 = 14
giving the sequence 5, 8, 11, 14, .... We could say this is the sequence whose general term is 311 + 2.
1. Find the first four terms of the following sequences whose nth terms are given.
(a) T,, = 211 (d) T,, = 4n - 2 (g) 11
T,, = 11 + 2 (j) T,, = 2"-1
(b) T,, = 3(11 - 1) (e) T,, = 11 + 6 (h) T11 = 113 (k) T = 11 + 1
n 11
2
(c) T,,=11-4 (f) 1
T,, = 211 (i) T,, = 12 - 3n (1) T,, = 3"
2. Write down the first six terms of the sequence given T,, = ~(n + 1). Do you recognise this
sequence?
24
3. If n is an integer, (a) is 2n odd or even? (b) is 2n + 5 odd or even? (c) What is the simplest general statement of an odd number?
4. What is the value of ( -l)n when (a) n is even? (b) n is odd?
5. Find the first three terms of the following sequences whose nth terms are given. (a) T,, = ( -1)n(2n + 3) (c) T,, = ( -l)n+13n
(b) T,, = ( -l)n! (d) T,, = ( -1)nn2•
n
6. The nth term of a sequence is given by T,, = 5n - 23. (a) How many terms of the sequence are negative? (b) Is 32 a term of the sequence? If so, which term? (c) What is the difference between successive terms of this sequence?
7. Which term of the sequence T,, = 3n + 7 is 34?
8. If T,, = 35 - 3n (a) How many terms are positive? (b) Is 16 a term of the sequence? (c) What is the first term? How could you use the first term to find the second?
An arithmetic sequence or arithmetic progression is a sequence in which each term after the first is formed by adding a constant number to the preceding term.
The sequence 6, 10, 14, 18, 22, 26, ... for example, progresses by adding the constant number 4. This sequence is an arithmetic sequence and the constant number added is called the common difference of the sequence.
Examples of arithmetic sequences are : 8, 10, 12, 14, 16, . . . common difference 2 15, 12, 9, 6, 3, . . . common difference -3 4, 51, 6, 71, 9, . . . common difference 11.
We note from the above that the common difference may be positive or negative.
Example: Find the common difference of the arithmetic progression 17, 15, 13, 11, ... and write the next two terms of the sequence.
Solution: Any term subtracted from the term that follows it gives the common difference d .
. ·. Common difference = d = 15 - 17 = -2
. ·. The next two terms are 9 and 7.
1. Find the common difference in each of the following arithmetic sequences and then write the next two terms. (a) 1, 3, 5, 7, 9, .. . (f) 5, 12, 19, 26, .. . (b) 18, 16, 14, 12, .. . (g) -5, -1, 3, 7, .. . (c) 12, 91, 7, 41, .. . (h) 4, 4!, 51, 6!, .. . (d) 60, 70, 80, 90, .. . (i) J3, JU, J27, .. . (e) X, X + 3, X + 6, X + 9, ... (j) 4x, 7x, lOx, ...
25
2. Determine whether or not the following numbers form arithmetic sequences. (a) 15, 23, 31 (c) -5, -1,4 (b) 2!, 3i, 5 (d) 3·7, 4·1, 4·5
3. Write down the first three terms of an arithmetic sequence given that (a) the first term is 6 and the common difference is 7. (b) the first term is 8 and the common difference is - 3. (c) the first term is -4 and the common difference is It,
4. If a, b, c forms an arithmetic sequence, show that b = a ; c.
5. The first three terms of an arithmetic sequence are 7, x, 35. Find x.
6. If a, b, c, d, e are in arithmetic sequence, show that a + e = b + d = 2c.
The nth term of an arithmetic sequence may be found by letting a represent the first term and d the common difference.
The sequence is then: a, a+ d,a + 2d,a +3d, a+ 4~ .... The nth term of this sequence may be obtained by examining the following table:
First term Second term Third term Fourth term Fifth term ... nth term Tt Tz T3 T4 Ts ... T,,
a a+d a+ 2d a+ 3d a+ 4d ... a+ (n- l)d
Can you see the pattern? The coefficient of din each term is one less than the number of the term.
Example (i):
The nth term of the arithmetic sequence a, a + d, a + 2d, a + 3d, ...
is given by T,, = a + (n - l)d.
Find the twentieth term of the sequence 5, 8, 11, 14, ....
Solution: Here a = 5 and d = 3 Using T,, = a + (n - l)d
T20 = 5 + 19 x 3 = 62
Therefore the twentieth term is 62.
Example (ii): Which term of the arithmetic sequence 15, 11, 7, ... is - 33?
Solution: T,, = a + (n - l)d
-33 = 15 + (n - 1) x ( -4) -33 = 15- 4n + 4
4n = 52 n = 13
Therefore -33 is the thirteenth term.
26
Example (iii): Find the first term and the common difference of an arithmetic sequence whose fifth term is 31 and whose twelfth term is 73.
Solution: T5 =a+ 4d = 31
T12 = a + lld = 73 .'. 7d = 42
.·. d = 6 and a= 7 Therefore the first term is 7 and the common difference is 6.
Example (iv): Insert four numbers between 8 and 23 such that the six numbers form an arithmetic sequence.
Solution: Since we are inserting four numbers between 8 and 23 then 8 is the first term and 23 is the sixth term of the arithmetic sequence. Using T,, = a + (n - l)d
23 = 8 + 5d .'. d = 3
Hence the four numbers are 11, 14, 17, 20 and the sequence is 8, 11, 14, 17, 20, 23.
1. Find the term indicated in each of the following arithmetic progressions. (a) lOth term of 7, 11, 15, 19, . . . (e) 13th term of 4, 61, 9, .. . (b) 16th term of 18, 15, 12, . . . (f) 55th term of 5, 7, 9, .. . (c) 9th term of -9, -4, 1, . . . (g) 40th term of -4, -8, -12, ... (d) 20th term of 11, 16, 21, . . . (h) 8th term of 5 + 2b, 5 + 5b, 5 + 8b, ...
2. What is the nth term of the sequence (a) 1, 3, 5, 7, 9, ... ? (b) 2, 4, 6, 8, 10, ... ?
3. What is the general term, T,, of the sequence 7, 14, 21, 28, ... ?
4. Find the nth term of the arithmetic sequence 20, 17, 14, ... .
5. The nth term of an arithmetic sequence is given by T,, = 4n + 8. Write the first four terms and the common difference.
6. The general term of an arithmetic progression is 2n - 1. Write the first three terms and the 60th term.
7. Which term of the sequence 7, 12, 17, ... is 372?
8. Is 279 a term in the arithmetic sequence 13, 17, 21, ... ?
9. Find how many terms there are in the sequence 9, 12, 15, ... , 195.
10. If the nth term of an arithmetic sequence is given by T,, = 13 - 2n, find the first three terms, the common difference and the 1 OOth term.
11. If the first term of an arithmetic sequence is 9 and the fourth term is 27, find the common difference.
12. In an arithmetic progression T7 = 20 and T13 = 38, find the first term and the common difference.
13. In an arithmetic sequence T10 = 5 and T17 = 54, find the first three terms.
14. Find T50 of an arithmetic sequence in which T5 = 13 and T9 = 25.
15. Find the number of terms in an arithmetic sequence with a= 5, d = 2 and the last term 43.
27
16. The fourth term of an arithmetic sequence is 116 and the common difference is 9. Find (a) the first term (b) the 30th term.
17. The first three terms of a sequence are 3, 7, 11, .... What is the first term to exceed 200?
18. Write down the numbers 24 and 44, then insert three numbers between them so as to give five numbers in arithmetic progression.
19. Insert six numbers between 43 and -6 such that the eight numbers form an arithmetic sequence.
20. The first three terms of an arithmetic sequence are 48, 41, 34. (a) Write down a formula for the nth term. (b) If the last term of the sequence is -29, how many terms are there in the sequence?
21. After starting full-time work a girl saves $15 in the first week, $19 in the second week, $23 in the third week and continues to increase her savings each week by the same amount until the twelfth week. How much is she then saving each week?
22. The temperature of the water in a boiler is rising at a constant rate. Readings taken every 15 minutes are as follows 25°C, 28°C, 31 oc and so on. The last reading taken was 91 °C. How many readings were taken in all?
23. The sum of the first three terms of an arithmetic sequence is 24 and the sum of the next three terms is 51. Find the first term and the common difference of this sequence.
24 f 2ab h h 1 1 1 . . h . . . I x = -- s ow t at -, -, -b are m ant metlc progressiOn. a+ b a x
Let us consider the sum of the arithmetic sequence a + (a + d) + (a + 2d) + · · · + (l - 2d) + (! - d) + l
where a is the first term, dis the common difference and lis the nth term. Note that the terms next to the last will be (l - d), (! - 2d), etc. Thus if S11 stands for the sum of n terms of this sequence
sll = a + (a + d) + (a + 2d) + . . . + (l - 2d) + (! - d) + l also S11 = l + (!- d) + (l- 2d) + · · · + (a + 2d) + (a + d) +a The sum does not change
when the order of the terms is reversed.
Adding the two series, term by term, we have 2S11 = (a + !) + (a + !) + (a + 1) + + (a + !) + (a + !) + (a + 1)
that is 2S,, = n(a + !) . n .. sll = 2(a + !)
The sum of n terms of an arithmetic sequence, given the first term and the last term is : n
sll = 2(a + !)
Note also that since 1 is the nth term, l = a + (n - l)d. n
Then S11 = l [a + a + (n - 1 )d]
.·. sn = ~ [2a + (n - l)d]
The sum of n terms of an arithmetic sequence, given the first term and the common difference is: n sll = 2[2a + (n- l)d]
28
Example (i): Find the sum of 20 terms of the series 7 + 12 + 17 + 22 + Solution: Here a= 7 and d = 5.
Using S" = ~[2a + (n - l)d]
20 s2o = 2 [14 + 19 x 5]
= 10 X 109 = 1090.
Example (ii): Find the sum of the series -15- 8 - 1 + 6 + 13 + Solution: Here a = -15, d = 7 and 1 = 188 but we do not known. :. Using T,, =a + (n- l)d
188 = -15 + (n - 1)7 203 = 7n - 7
Hence n = 30 30 n
Now S30 = 2 (-15 + 188) using S, = l(a + l) = 15 X 173 = 2595.
Example (iii):
+ 188.
How many terms in the series 48 + 44 + 40 + · · · need to be taken to give a sum of 308?
Solution: n
Using S, = 2[2a + (n- l)d] where a= 48, d = -4 and S, = 308
n we have 308 = 2[96 + (n - 1) x ( -4)]
616 = n[96 - 4n + 4] 616 = lOOn - 4n2
:. n2 - 25n + 154 = 0
:. (n - ll)(n - 14) = 0 :.n= llor14.
Note that the sum of 11 terms = the sum of 14 terms = 308. What does this tell you about T12 + T13 + T14?
In this method the quadratic equation sometimes yields solutions which are fractional or negative. Since n represents the number of terms, only positive integer solutions have meaning.
1. For the series : (a) 5 + 12 + 19 + 26 + (b) 6 + 8 + 10 + 12 + .. . (c) 20 + 17 + 14 + 11 + .. . (d) -17 - 8 + 1 + 10 + .. . (e) 1 t + 3 + 4± + 6 + · · ·
find the sum of 10 terms. find the sum of 18 terms. find the sum of 14 terms. find the sum of 30 terms. find the sum of 24 terms.
29
2. Show that the sum of n terms of the series 1 + 2 + 3 + 4 + · · · is given by S, = ~(1 + n)
and hence find the sum of all the integers from 1 to 100.
3. What is the sum of the first 30 positive even numbers?
4. Find the sum of 20 terms of an arithmetic sequence if the first term is 9 and the last term is 99.
5. Find the sum of all the positive multiples of 7 which are less than 100.
6. Find the sum of all the positive integers less than 100, which are not divisible by 6.
7. The first term of an arithmetic sequence is 3 and the twentieth term is 136. Find the common difference of the sequence and the sum of 20 terms.
8. Show that the sum of n terms of the series 1 + 3 + 5 + 7 + · · · is given by S, = n2 and hence find the sum of the first 50 odd numbers 1 + 3 + 5 + 7 + · · · + 99.
9. Find the sum of the series 2 + 3 + 4 + · · · + 399.
10. The nth term of an arithmetic sequence is T,, = 3n + 1. Find the sum of 40 terms in this sequence.
11. The nth term of an arithmetic series is 28 - 3n, find T1 , T40 and S40 .
12. How many terms must be taken in the series 5 + 11 + 17 + 23 + · · · to make the sum 208?
13. How many terms in the series -3 + 0 + 3 + 6 + 9 + · · · are needed to give a sum of 105?
14. Find the sum of all the integers between 100 and 200 which are multiples of 9.
15. A boy earns $10 the first week and 50 cents more each week than in the preceding week. What is the total sum of money he earns in 20 weeks?
16. A ball, rolling down a slope, rolls 16 em in the first second, 48 em in the second second, 80 em in the third second and so on. At this rate how far will it roll in 8 seconds?
17. A drill test costs $200 for the first 10 metres, $250 for the next 10 metres, $300 for the next 10 metres and so on. What will the drilling cost for a depth of 180 metres?
18. A man saves $2000 in the first year of a savings programme and increases his yearly savings by $200 each year thereafter. How much has he saved at the end of 15 years?
19. The cost of building a multi-storey building is $300 000 for the first floor, $375 000 for the second floor, $450 000 for the third floor and so on. Find the total cost if the building is 8 storeys high.
20. The sum of 6 terms of an arithmetic series is 45, the sum of 12 terms is -18. Find the first term and the common difference.
21. Find the sum to 20 terms of the series whose nth term is 311 - 1.
22. How many terms of the series 23 + 19 + 15 + 11 + · · · must be added to give a sum of 50?
23. The sum of n terms of a sequence is given by S, = 11 2 + 311. Find the first three terms of the sequence and show that it is an arithmetic progression. What is the common difference?
24. The sum of 11 terms of a certain series is given by S,, = 211 + 311 2• Show the series is in arithmetic
progression and find: (a) the common difference (b) the 20th term (c) the sum of20 terms.
25. The numbers x 1 , x 2 , x 3 , x4 , •.. , x, are such that for 11 a positive integer x, = 411 + 1. Show the numbers form an arithmetic sequence and find the sum to 30 terms.
26. The sum of the first 10 terms of an arithmetic series is 100 and the sum of the next 10 terms is 300. Find the series.
30
27. There are 10 apples in a row 5 metres apart. The first apple is 5 metres from a basket. How far does a boy run who starts at the basket and returns the apples to the basket, one by one?
28. A man moves a load of soil for top dressing an oval by emptying barrow-loads in a line 20 metres apart, with the first heap 20 metres from the load of soil. How far does he walk if he empties 24 barrowfulls and returns to the load each time? ·
29. From a length of rod 600 em in length, 25 pieces are cut off, each 1 em longer than the preceding piece. If the rod is exactly used up find the length of the first piece cut off.
Sigma notation is a shorthand method for indicating the sum of a series. The symbol L means sum. 6
L ( 4n - 2) means sum the series for which T,, = 4n - 2 for values of n from 1 to 6. Thus in expanded n=l
form: 6
I ( 4n - 2) = 2 + 6 + 10 + 14 + 18 + 22 n=l
Example: 40
Evaluate L: (2n + 5).
Solution: 40
n=1
L: (2n + 5) = 7 + 9 + 11 + 13 + · · · + 85 n=1
Since this is an arithmetic sequence we can useS" = ~(a + /). 40 40
Thus 11~1 (2n + 5) = S40 = 2 (7 + 85)
= 20 X 92 = 1840.
Evaluate: 12 50 12
1. L: (5n - 1) 4. L: (3n - 4) 7. L (6n + 1) n=1 n=1 n=S
20 18 30
2. L: (3n + 5) 5. L (2n + 9) 8. L: (15 - n) n=1 n=1 n=10
16 30 20
3. L: (30- 2n) 6. L (12- n) 9. L: (2n - 6) n=1 n=l n=4
15
10. L (3n) n=1
14
11. L: (8 - n) n=4
60
12. L (5n - 1) n=1
The sequence 2, 6, 18, 54, 162, ... is a geometric sequence or geometric progression. In a geometric sequence, each term after the first is obtained from the preceding one by multiplying it by a constant number. This constant number is called the common ratio of the sequence.
The value of the common ratio may be found by dividing any term by the one immediately preceding it. In the geometric sequence above, the common ratio is 3; (! = 1l = i~ = \6
/ etc.). The common ratio can obviously be used to find the next term in a geometric sequence. In the
sequence above, the next term would be 162 x 3 = 486.
31
Example (i): Find the common ratio and the fifth term of the geometric sequence 32, I6, 8, 4, ....
Solution: Let the common ratio be r then, r = ~~ = t. Fifth term T5 = 4 x ! = 2.
Example (ii): Is the sequence 8I, -27, 9, -3, ... geometric?
Solution: T2 = -27 T3 9 r· -3
____± =-T1 8I Tz -27 T3 9
I I I 3 3 3
Since the sequence has a common ratio (-!)it is geometric.
Example (iii): If x - 3, x and x + I2 form a geometric sequence, find the value of x and the common ratio.
Solution: Each term divided by the preceding are gives the common ratio.
Hence X X+ I2
X- 3 X
.'. x 2 = x 2 + 9x - 36
.'. 9x = 36 ,',X= 4.
Now the common ratio is r = _x_ X- 3 4 -I
= 4.
I. Find the common ratio and the next term for each of the following geometric sequences. (a) I, 4, I6, . . . (d) I6, I2, 9, . . . (g) I6, 20, 25, ... (b) I2, 6, 3, :. . (e) 4, 6, 9, . . . (h) 256, -64, I6, ... (c) 2, -6, I8, . . . (f) I,!, i, . . . (i) a, ax, ax2
, •••
2. (a) The first term of a geometric sequence is 4 and the common ratio is 3. Write down the first four terms of the sequence.
(b) A geometric sequence has a first term of 8 and a common ratio of!. Find the first five terms.
3. Which of the following sequences are in geometric progression? For those sequences that are geometric, find the common ratio. (a) 5, 10, 20, . . . (c) 54, 36, 27, .. . (b) I2, I6, 24, . . . (d) 8I, 72, 64, .. .
(e) x, 2x, 4x, ... (f) a, a2
, a 3, ..•
4. State the condition required for a, b and c to be in geometric progression. Using this statement show that b2 = ac.
5. If the three given terms form a geometric sequence, find x in each case. (a) 5, 35, x (c) x, 8, I2 (e) I, x, 9 (b) 4, 5, X (d) 3, X, 75 (f) X, -I2, 9
6. If x, x + I, x + 4 are in geometric progression, find x.
7. If y, y + 4, y + I2 form a geometric sequence, find y.
32
Any sequence in geometric progression can be written a, ar, ar2
, ar3, ar4
, •••
where a is the first term and r is the common ratio. The general term can be found by examining the following table:
First term Second term Third term Fourth term Fifth term Tl Tz T3 T4 Ts
a ar ar2 ar3 ar4
Sixth term T6
ar 5
Notice that the index of r in any term is one less than the number of the term.
• 0 ••••
...
...
For the geometric sequence a, ar, ar2, ar3
, •.• the nth term is given by T,, = ar"-1
Example (i): Find the seventh term of the geometric progression 54, - 18, 6, ....
Solution:
a = 54 r = - 6- = _! ' -18 3
Now T7 = ar6
=54 X (-D6
1 = 2 X 33 X-36
2 27
The seventh term is l 7 •
Example (ii): The third term of a geometric sequence is 8, the sixth term is -1. Find (a) the common ratio (b) the first term (c) the eighth term.
Solution: (a) Since T3 = 8, ar2 = 8
and T6 = - 1, ar5 = - I Equation (2) --;- equation (1) gives r 3 = -!
.'. r = -! The common ratio is -!.
(b) Now substituting r = -!in equation (1) ax i = 8
.'.a= 32 The first term is 32.
(c) The eighth term
The eighth term is -i. 33
nth term T,,
ar"-1
(1) (2)
Example (iii): Which term of the geometric sequence 2, 6, 18, ... is equal to 486?
Solution: Let ar"-1 = 486 then 2 x 3"-1 = 486
3"-1 = 243 3"-1 = 35
,', 11 - 1 = 5 n=6
Thus the sixth term is equal to 486.
Note: It is often easier in exercises of this type to write out the sequence as far as the term required. That is in the example above using the calculator or otherwise we continue the sequence 2, 6, 18, 54, 162, 486, . . . to the required term.
Example (iv): Insert three numbers between 3 and 48 such that the five numbers form a geometric sequence.
Solution: Since we are inserting three numbers between 3 and 48, then 3 is the first term and 48 the fifth term of the geometric sequence.
Thus a = 3 and ar4 = 48 ,', 3 X r 4 = 48
r 4 = 16 l' = ±2.
Thus two geometric sequences satisfy the conditions stated in the question 3, 6, 12, 24, 48, ... and 3, -6, 12, -24,48, ....
1. In a geometric sequence if:
2.
(a) a= 5 andr = 2, find T6 •
(b) a= t and r = -2, find T9 •
(c) a = 81 and r =%,find T7 .
(d) a = 0·8 and r = 1·5, find T4 •
In the following geometric sequences find the term indicated.
(a) 12, 6, 3, ... Ts 1
(f) J2' 1, J2, ... Ts
(b) 4, 6, 9, " . T6 (g) 16,-12, 9, ... T7 (c) 1, It, 1{6 , •.• Ts (h) 1, -J3, 3, ... T6 (d) J, g, 4t, ... T6 (i) m-4 , m- 2 , 1, ... T1o
(e) a, ab2, ab4
, .•. T1o (') m 2 3 J -,m ,m n, ... 11
T6
3. If the second term of a geometric sequence is t and the common ratio is-!-, find the eighth term.
4. (a) Write down the first three terms of a sequence whose nth term in 2". Is this a geometric sequence? What is the common ratio?
(b) If the nth term of a sequence is 3 x 2"-1, write down the first three terms. Is this a geometric sequence?
5. Find the common ratio in each of the following geometric sequences, given that: (a) T1 = 6 and T4 = 48 (c) T3 = 3 and T10 = 6561 (b) T2 = 63 and T7 = 2
77 (d) T4 = - 1 and T9 = ll3
6. Find two geometric progressions having: (a) 7 as the first term and 112 as fifth term. (b) 5 as the third term and 80 as seventh term.
34
(c) 2 as second term and 1458 as eighth term. (d) 54 as third term and r~ as ninth term. (e) 2 as first term and 1o§- as fifth term.
7. Which term of the sequence 3, 6, 12, ... is equal to 384?
8. If the first term of a geometric progression is 486 and the common ratio is i, find which term is equal to 18.
9. If 324 is the nth term of a geometric sequence with first term l 4 and common ratio 6, find n.
10. In any geometric progression, show that the product of the first term and the last term is equal to the product of the second term and the last term but one.
11. Insert two numbers between 54 and 2 so that the four numbers are in geometric sequence.
12. If five numbers are inserted between 2 and 1458 to form a geometric sequence, find the common ratio.
13. If x, x + 2, x + 6 are in geometric progression, find x and hence find the common ratio.
14. Given that x, 6, 8, yare in geometric sequence, find x andy.
15. The sequence p, 12, 9, q is geometric. Find p and q.
16. Find the sixth term of a geometric sequence, if a = 5 and T4 = 0·04.
17. Insert three numbers between 5 and 80 so as to give five numbers in geometric sequence.
18. If in a geometric sequence the sum of the second and third terms is 20 and the sum of the fourth and fifth terms is 320, find the common ratio and the first term. Assume that the common ratio is positive.
19. Find the smallest term of the sequence!,!, i, ... which is greater than 300.
20. Find the smallest term of the sequence 400, 200, 100, ... which is greater than 3.
21. Find x and the common ratio in each case if: (a) 1, x, a form a geometric sequence.
b2 (b) a, x,- form a geometric sequence.
a
The sum S11 of n terms of any geometric series can be written S11 = a + ar + ar2 + · · · + ar"- 2 + ar"-1 (1)
:. rS11 = ar + ar2 + ar3 + · · · + ar"- 1 + ar" (2) (multiplying both sides by r) Subtracting equation (2) from equation (1) all terms on the right hand side except two will disappear
i.e. S11 - rS11 = a - ar" S11 (1 - r) = a(l - r")
a(l - r") thus S = --'----'-
" 1 - r
Note that if r > 1 the formula can be conveniently written in the equivalent form S11 = a~·"_-11 ).
The sum S11 of the first n terms of the geometric series a + ar + ar2 + · · · + arn-l is given by a(1 - r") a(r" - 1)
Sn = 1 - r or S" = r - 1
35
Example (i): Find the sum of the first 9 terms of the series 32 + 16 + 8 +
Solution: The series is geometric with a = 32, r = t, and since r < 1
a(l - r") useS,=
1 -r . - 32[1 - (t)9
] .. Sg- 1 - _t
2
= 64[1 - (t)9]
= 63·875 (from the calculator)
Example (ii): Find the sum of the first 8 terms of a geometric series if a = 2 and r = )2. Solution:
a(r" - 1) Use S, =
1 1'-
. - 2[()2)8 - 1]
.. S8- )2 _ 1
2(16 - 1)
)2- 1 30 )2 + 1
------,=,------ X ~,------)2-1 )2+1
= 30)2 + 30
Example (iii): How many terms of the geometric series 1 + 3 + 9 + 27 + · · · are needed to give a sum of 1093?
Solution: a(r" - 1) s, = ---'------'-
1'- 1 where a = 1, r = 3, S, = 1093
1093 = 1(3"- 1)
2186 = 3"- 1 2187 = 3"
37 = 3" .'. n = 7
.'. Seven terms are needed to give a sum of 1093.
Example (iv): 8
Find I 3 x 2"-1 .
n=1
Solution: The series is 3 + 6 + 12 + · · · + 384. Since this is a geometric series we can use
a(r" - 1) S, =
1 where a = 3, r = 2, n = 8
/'-
Thus ± 3 x 2"-1 = S8 = 3(28
- 1) n=1 2- 1
= 765
36
I. Sum the series: (a) I + 3 + 9 + · · · to 8 terms (b) i + 1 + I + · · · to 7 terms (c) -5 - 10- 20- · · · to 9 terms
(d) I + J3 + 3 + · · · to 8 terms (e) 9 - 6 + 4 - · · · to 6 terms (f) x + xy2 + xy4 + · · · to 7 terms
2. The second term of a geometric series is 8 and the fifth term is -1. Find the sum of the first 6 terms.
3. The third term of a geometric series is 1 and the eighth term is 16. Find the sum of the first 8 terms.
4. The series i - i + l 2 - • • • is geometric. (a) What is the common ratio? (b) Find T4
(c) Find the sum of the first 8 terms.
5. The sum of the first 4 terms of a geometric series is 390 and the common ratio is i. Find these 4 terms and add them to check your answer.
6. The sum of the first 6 terms of a geometric series is 315 and the common ratio is t. Find the six terms.
7. During vacation a student agrees to work for 14 days if he is paid one cent for the first day and if his salary is doubled on each succeeding day. How much would he earn on the 14th day? How much would he earn over the whole 14 days?
8. How many terms of the series 2 + 6 + 18 + 54 + · · · are needed to give a sum of 728?
9. How many terms of the series 9 + 18 + 36 + · · · are needed to give a sum of 1143?
10. What is the least number of terms of the series 7 + 14 + 28 + · · · that need to be taken to give a sum greater than 1000?
11. Evaluate: 6 7 10 8
(a) L 4 x 3"-1 (b) I ( -sy•-1 (c) L 8 x (1·5)''-1 (d) I 16 X (0·75)"-1
f1=1 r~=1 r~=1 r~=1
An important practical application of geometric sequences is in connection with compound interest, superannuation and time payments.
We can use our knowledge of geometric sequences to develop a general formula for finding the amount $A that $P becomes when invested for n years at r% p.a. compound interest.
$100 amounts to $(100 + r) in one year
.·. $1 amounts to$ (1 + 1 ~0) in one year
Applying this to our investment of $P we have
$P amounts to $P ( 1 + 1 ~0) in the first year
$P (1 + 1 ~0) amounts to $P (1 + 1 ~0) 2
in the second year
( ~"\2 ( r_\3 $P I + IOO) amounts to $P I + 100) in the third year
37
The successive amounts, in dollars, form a geometric sequence.
p ( 1 + 1 ~0 > p ( 1 + 1 ~0 r. p ( 1 + 1 ~0 r. .. . We can see this sequence is geometric with a first term of P (1 +
1 ~O) and a common ratio of
(1 + 1~0} The nth term of this geometric sequence is given by T,, = ar"-1
i.e. r.. = p ( 1 + 1~0) ( 1 + 1~o)"-1 = p( 1 + 1~0)"
Thus when $Pis invested for n years at r% p.a. compound interest, it amounts to $A where
A= p( 1 + 1~0)" Example (i): Find what $800 amounts to in 6 years if it is invested at 8% p.a., compounded yearly.
Solution:
A= P(1 + 1 ~0)" = 8oo(1 + 1 ~0r = 800 X 1·086
= 1269·4994 (from the calculator) Thus after six years $800 amounts to $1269·50.
In the example above the interest was compounded annually. That is, at the end of each year the interest is calculated and the principal for the next year is increased by this amount. Interest is often compounded quarterly, monthly or even daily. In the example above, if the interest was compounded quarterly instead of yearly, there would be 24 periods of time and the interest for each period would be a quarter of the annual rate.
Example (ii):
In general, the compound interest formula is
( r ~"
A"= p 1 + 100)
where r is the rate% per period, P is the principal
and A 11 is the amount after n periods
A man borrows $4000 at 18% p.a. for 3 years, the interest to be compounded monthly. Find to the nearest dollar the amount owing after 3 years.
Solution: Three years is 36 monthly periods so n = 36. 18% p.a. is 1· 5% for each monthly period so r = 1· 5.
Now A" = P ( 1 + 1 ~O )", thus the amount owing after 36 months is
A 36 = 4000 (1 + :~~) 36
= 4000(1·015)36
= 6836·558 (from the calculator) Thus the total amount owing is $6837 (to the nearest dollar).
38
1. Find what $2400 amounts to when invested for 4 years at an interest rate of 9% p.a. if the interest is compounded annually.
2. To what sum will $1500 amount if invested for 3 years at 12% p.a. compounded yearly?
3. Find the total amount required to pay off a loan of $16 000 plus interest at the end of 8 years if the interest is compounded half-yearly and the rate is 14% p.a.
4. Calculate the amount, when: (a) $4500 is invested at 10% p.a. for 6 years if the interest is compounded quarterly. (b) $6000 is invested at 1· 5% per month for 2 years, the interest being compounded monthly. (c) $8600 is invested for 3 years at 10% p.a., if the interest is compounded
(i) yearly (ii) quarterly (iii) monthly.
Example: A man invests $750 at the beginning of each year in a superannuation scheme. If the interest is paid at the rate of 8% p.a. on the investment (compounded annually), how much will his investment be worth after 20 years?
Solution: The first $750 will be invested for 20 years and will amount to $750 x 1·0820
•
The second $750 will be invested for 19 years and will amount to $750 x 1·0819.
The third $750 will be invested for 18 years and will amount to $750 x 1·0818•
And so on until the last $750 which will be invested for 1 year and will amount to $750 x 1·08. At the end of 20 years his investment will be worth
$750(1·0820 + 1·0819 + 1·0818 + ... + 1·08) The series in parentheses can be written
1·08 + 1·082 + 1·083 + ... + 1·0820
This is a geometric series with a first term of 1·08 and a common ratio of 1·08. Thus its sum is 1·08(1·0820
- 1) 1·08 - 1
Therefore the value of the total investment= $750 x l·08(1·0820
- 1) 0·08
= $37 067 (to the nearest dollar)
1. A woman invests $600 at the beginning of each year into a superannuation fund for a period of 10 years. Interest is paid at the rate of 7% p.a. and is compounded annually. Find how much her investment is worth at the end of the 10 year period.
2. A man deposits $1000 annually to accumulate at 9% p.a. compound interest. How much will he have to his credit at the end of 15 years?
3. An employee pays $500 at the beginning of each year into a pension fund. It is agreed that at the end of 25 years he will receive what he has paid together with 8% p.a. compound interest. What amount should the employee receive?
4. A superannuation scheme available to a self-employed person pays compound interest at the rate of 8· 5% p.a. for any money invested. If a dentist pays $2000 into the scheme at the beginning of each year, how much is he entitled to receive after 30 years?
39
5. An insurance company offers 10% p.a. compound interest on investments in its superannuation fund. If a contributor pays in $Pat the beginning of each year, show that after n years his superannuation entitlement is $11P(1·1" - 1).
Example: A man borrows $5000 at an interest rate of 18% p.a. (1· 5% per month) where the interest is compounded monthly on the balance owing. The loan is to be repaid in equal monthly instalments. What should the amount of each instalment be, to pay off the loan in 4 years?
Solution: Let the monthly instalment be $P and the amount owing after n months be $A 11 •
After 1 month and after paying the first instalment of $P the amount owing A1 = 5000 X 1·015- p
After 2 months, A2 = A1 x 1·015 - P = (5000 x 1·015- P)1·015- P = 5000 X 1·0152 - 1·015P - P = 5000 X 1·0152
- P(1 + 1·015) After 3 months, A3 = A 2 x 1·015 - P
= 5000 x 1·0153 - P(1·015 + 1·0152
) - P = 5000 x 1·0153
- P(l + 1·015 + 1·0152)
Continuing this pattern: Afternmonths,A11 = 5000 x 1·015"- P(l + 1·015 + 1·0152 + ··· + 1·015"-1)
Therefore after 48 months, A48 = 5000 x 1·01548 - P(1 + 1·015 + 1·0152 + . · · + 1·01547) But the loan is repaid in full after 48 months, thus A48 = 0. Hence 0 = 5000 x 1·01548 - P(1 + 1·015 + 1·0152 + ... + 1·01547)
, p _ 5000 X 1·01548
. . - 1 + 1·015 + 1·0152 + ... + 1·01547
Now the denominator is a geometric series in which a= 1, r = 1·015 and n = 48 thus its sum is 1 (1·01548 - 1) 1·01548 - 1
1·015 - 1 0·015
.'. P = 5000 X 1·01548 X --,---.,--,--0--.'0~1-5-1·01548 - 1
= 146·87 Each monthly instalment must be $146·87.
Note: The simple interest rate (r%) which is equivalent to the true interest rate used in the above example is found as follows.
100/ Using r = -pt
where I is the total interest paid p is the principal t is the time in years
100(146·87 X 48 - 5000) r = -----'------------"-5000 X 4
= 10·25 .'. The equivalent simple interest rate is 10·25% p.a.
1. A man borrows $2000 at 1· 5% per month interest where the interest is compounded monthly on the balance owing. If he pays off the loan in equal monthly instalments over 3 years, what is the amount of each monthly instalment?
40.
2. Mrs. Thompson borrows $3000 and agrees to repay the loan and interest in 12 equal quarterly instalments. The interest is 12% p.a. (3% per quarter) and is calculated quarterly on the balance owing. Find the amount of each quarterly instalment.
3. Mr. Lucas borrowed $12 000 from the bank to finance an extension to his home. He agrees to pay back the loan in 8 yearly payments at 14% p.a. compound interest. (a) What is the amount of each yearly payment? (b) Calculate the simple interest rate equivalent to this compound interest.
4. Mr. Read borrowed $7000 to buy a new car. He has to pay off the loan over 4 years in equal monthly instalments, allowing for an interest rate of 15% p.a. which is compounded monthly. Find the amount of each instalment.
5. A loan of $4000 is to be repaid by equal annual instalments. Compound interest at the rate of 10% p.a. is calculated yearly. If the annual instalment is $P, show that: (a) the amount owing after the first instalment is made at the end of the first year is
$ ( 4000 X 1 . 1 - P) (b) the amount owing at the end of the second year is
$4000 x 1·1 2- P(1 + 1·1)
(c) the amount owing at the end of11 years is $4000 x 1·1"- P(l + 1·1 + 1·1 2 + · · · + 1·1"-1
)
(d) if the loan, including interest charges, is exactly repaid at the end of 11 years; then 400 X 1·1"
P= 1·1"-1
Consider the series t + -!:- + -§- + l6 + l2 + l4 + · · ·
The sum of 11 terms S,, = ! + (!)2 + (!)3 + (!)4 + . . . + (!)" Now the sum of two terms S2 = i = 1 - (!)2
the sum of three terms S3 = i- = 1 - (!) 3
the sum of four terms S4 = t~ = 1 - (!)4
the sum of five terms S5 = ~i = 1 - (!) 5
the sum of six terms S6 = ~~ = 1 - (!)6
It seems that the more terms we take the closer the sum tends to 1. Now using the formula for the sum of n terms of a geometric series we have a = !, r = !. Thus S = t[l - (t)"]
II 1-! = 1- (!)"
Now as 11 becomes larger and larger, (!)" becomes smaller and smaller. If 11 is very large (t)" ---+ 0 [ (!)" tends to zero]
and S,---+ 1 When the sum of a series tends to a particular value in this way, we say that this value is the limiting
sum of the series. In the example above the limiting sum is 1.
I 1 r . . S a(l - r") n genera 10r a geometnc senes , = . 1 - r
Now if lrl < 1 (i.e. if -1 < r < 1) then r" becomes smaller and smaller as n becomes larger and larger.
We say the limit of r" as n approaches infinity equals zero. In symbols lim r" = 0 for lrl < 1.
n-+oo
Hence the limiting sum is 1 ~ ,. .
41
If S is the limiting sum of a geometric series with I r I < 1, then:
S=-a-1- r
Note: A geometric series with lrl > 1 can have no limiting sum since the terms grow larger and larger as n increases.
Example (i): Find the limiting sum of the series 50 + 30 + 18 + Solution: a = 50, r = !. Since lrl < 1 and the series is geometric we have a limiting sum.
S=-a-1 -,.
50 1 - ~
=50 X~ = 125
Example (ii): Find the limiting sum of the geometric series in which a = 15 and r = -0·2.
Solution: Since lrl < 1 the geometric series has a limiting sum.
S=-a-1 - ,.
15 1- (-0·2) 15 1·2
= 12·5
Example (iii): Express 0· i2 as a fraction in its lowest terms.
Solution: This recurring decimal can be written
12 12 12 12 0·121212 ... = 102 + 104 + 106 + 108 +
The right hand side is a geometric series where a = lio and r = 160 and we require its limiting sum
S=-a-1 - ,.
12 TOO
1 - 16o = lo~ x 19o9o _g - 99 _ __±__ - 33 • • 4
Thus 0·12 = 33.
1. Use your calculator to see what happens to r" when (a) lrl < 1 and n increases
(b) lrl > 1 and n increases
(c) lrl = 1 and n increases.
42
2. (a) Which of the following series have limiting sums? (i) 6 + 3 + 1-!- + 0 0 0 (iii) l6 + t + ! +
(ii) 12 + 18 + 27 + . . . (iv) 1 + t + ~ + ... (b) Find the limiting sums where they exist.
3. For what range of values of x will the series 1 + x + x 2 + x 3 + · · · have a limiting sum? What is this sum when x = -!-?
4. Find the limiting sum of the series: (a) 1000 + 100 + 10 + .. . (d) 60 - 30 + 15 -(b) 81 + 27 + 9 + · · · (e) 20 + 15 + 12 + (c) 40 + 20 + 10 + . . . (f) 125 + 25 + 5 +
5. Find the limiting sum in the series where: (a) a = 5, r =! (b) a = 1, r = 1
4
(g) 24 - 8 + 2~ - 0 0 0
(h) 2 + J2 + 1 + 0 0 0
(i) No + 1o1~oo + 1 oo1o5 ooo +
(c) a = 4, r = 0·1
6. If the limiting sum of a geometric series, with common ratio ;i:, is 16; find the first term.
7. A geometric series has a common ratio oft and a limiting sum of 25, find the first term.
8. A geometric series has a first term of 8 and a limiting sum of 12, find the common ratio.
9. Find the common ratio of a geometric series with a first term of 125 and a limiting sum of 100.
10. If 1 + x + x2 + · · · has a limiting sum of 5, find x.
11. If 2 + 2y + 2y2 + · · · has a limiting sum ofl t, find y.
12. Express the following recurring decimals as fractions in their lowest terms: (a) 0·4 (b) 0·25 (c) 0·67 (d) 0-43
13. A geometric series has a = 6 and r = t. Find the difference between its limiting sum and the sum to five terms.
14. Express as a decimal the difference between the limiting sum of the series lo + lo + 2 ~ 0 + · · · and its sum to 3 terms.
15. A ball is dropped 12 metres onto a flat surface, rebounds 9 metres and continues to bounce to i of its previous height. If it is allowed to bounce until it stops, through what total distance does it travel?
16. The height of a tree was 10 metres and it increased by 2 metres during the next year. If in each succeeding year the growth is~ of that in the previous year, find the limiting height.
17 F . d h 1' 0 0 f h 0 5 5 5 h 1 4 . m t e 1m1tmg sum o t e senes - + 2 + 3 + · · · w en x = · . X X X
18. Can there be a geometric series with (a) a limiting sum of~ and a first term of 4? (b) a limiting sum of 4 and a first term of~?
00 00
19. Find (a) L 5 x (!)"-1 (b) I elY' n=1 n=1
20. The infinite geometric series x - tx + ~x - · · · has a limiting sum of ;i:. Calculate the value ofx.
21. If the sequence 4, -2·5, ... is geometric, find: (a) an expression for the nth term. (b) an expression for the sum of n terms and find its value when n = 5. (c) the limit of S" as n---+ co.
43
1. Find the missing terms in the following arithmetic sequences: (a) 2, *, 8, *, * (c) 17, 22, 27, *, *, * (b) 6, *, *, 12, * (d) *, *, 14, *, *, -2
2. Find whether 2863 is a term of the sequence 5, 8, 11, ....
3. What is the value of the first term to exceed 350 in the sequence 7, 13, 19, 25, ... ?
4. The third term of an arithmetic sequence is 7 and the ninth term is 19. Find the first term and the common difference.
5. If the sum of n terms in an arithmetic series is 100, the first term 2 and the common difference 3, find the number of terms.
6. Find the limiting sum of the series 16 + 8 + 4 + · · · 7. The limiting sum of the series 1 + x + x2 + x 3 + ···is 10, find x.
8. For a geometric sequence T3 = 18 and T6 = 144, find the first term and the common ratio.
9. The sum ton terms of an arithmetic series is 5n2 - lln for positive values of n. What is the com
mon difference of the sequence?
10. The numbers 2, a, bare in arithmetic progression and a, b, 9 are in geometric progression. Find a and b.
11. For what range of values of y will the series y + y 2 + y 3 + ... have a limiting sum? Find this sumify = !.
12. Express 0· 57 as a geometric series and then find its limiting sum.
5 00
13. Find (a) I (5n - 4) (c) I 27 X cw-l n=l n=l n=l
14. The third term of a sequence is 2-i and the sixth term is 9. Find the first six terms when the sequence is (a) arithmetic (b) geometric.
15. Show that the three numbers a - d, a, a + dare in arithmetic progression. Hence, find three numbers in arithmetic sequence whose sum is -12 and whose product is 36.
16. Insert two numbers between 64 and 27 so that the sequence of four numbers is (a) in arithmetic sequence (b) in geometric sequence.
17. A man invests $1500 at the beginning of each year in a superannuation fund. If interest is paid at the rate of 10% p.a. on the investment (compounded annually), how much will his investment be worth after 20 years?
18. Mr. Ross borrows $6000 and agrees to repay the loan and interest in 16 equal quarterly instalments. The interest is 8% p.a. and is calculated quarterly on the balance owing. Find (a) the amount of each quarterly instalment. (b) the simple interest rate which is equivalent to the interest rate used in the calculation above.
44
1. Solve: (a) lx- 41 > 2 (b) -1 < 3 - 2x <-5 _ (c) x(4 - x) = 0
2. From a pack of cards, the four kings are turned face down and thoroughly mixed. One card is selected at random. What is the probability of selecting: (a) the king ofhearts? (b) a black king? (c) the king of hearts or a black king?
3. The four points (5, 5), (10, 13), (25, 17) and (35, 5) reprvsent the corners of a field. If each square unit represents 100m2
, find the area of the field in hectares.
4. (a) Differentiate x3 + 2 and find the gradient of the tangent to the curve y = x 3 + 2 at the point (1, 3).
(b) Find the equation of the tangent to the curve y = x3 + 2 at the point (1, 3) and show that it passes through the origin.
5. The n'th term of a sequence is given by T,, = 3n - 2. Show that the sequence is arithmetic and find the twentieth term and the sum of 20 terms of this sequence.
6 F . d A . d d . 'f . A 84·6 sin 48°40' . m , m egrees an mmutes, 1 sm = 79
.4
1. Differentiate: (a) x4 - 2x2 (b) 2x(x + 5)
2x4 - 3x2
(c)---x
2. Find rational numbers a and b such that (j5 - 2) 2 - J5 1 = a + bJ5.
5-2
3. Two dice are rolled. What is the probability of throwing (a) a total of 5? (b) a total of 8?
4. In the given diagram, the dimensions are as shown. Find (a) the length BD
15 em A
(b) the area of 6.DBC (c) the area of quadrilateral ABCD. 8 em
B 21 em
5 s· 1'f ( ) X2 + 5x x
2 + 4x - 5 . 1mp 1 y: a x2 + 1 Ox + 25 x x2
3 2 (b)-----
x+5 x-5
D
c
6. Find the perpendicular distance of the point (4, -!) from the line 5x - 12y + 26 = 0. Is the point (4, -±)on the same side of the line as the origin or on the opposite side?
1. Given that E = Jm - n, find p when r = 950, m = 0·732, n = 0·569. Give answer correct to r m + n , -
one decimal place.
45
2. ABCD is a parallelogram in which the lengths of AB and AD are 5 em and 12 em respectively and angle ABC is 50°. Calculate: (a) the length of the diagonal AC (b) the size of the angle BA C (c) the area of the parallelogram ABCD.
3. Given the series (a) 4 + 11 + 18 + · · ·. Find the sum of 40 terms. (b) 18 + 6 + 2 + · · · . Find the limiting sum.
4. (a) If sin 0 =!and tan 0 = -~'find 0 for oo ~ e ~ 360°.
(b) Solve tan3 e = 1·268, finding all possible values of e between oo and 360°.
5. Find the equation of the normal to the curve y = x 2 - 3x + 2 at the point (2, 0). Where does
this normal cut they-axis?
6. Use a tree diagram to find all the two-digit numbers which can be formed using the digits 4, 5, 6 and 7 without repetition. What is the probability that any number chosen at random from these should be less than 50?
1. Solve the equations: (a) 2x2 + 7x - 4 = 0 (b) 15 - 2xl = 9
2. Simplify: (a) sec2 A - tan2 A
30
3. Evaluate: (a) I ( 4n - 3) n=1
(b) 2 sec2 A - 2
4 tan A
00
(b) I 3 X rn n=1
4. In the diagram, AB = 5, AD = 2, BD = 4, BC = 4. Calculate: (a) LADB (b) length DC (c) area !:::.BDC
A
(c) y = x 2- 2x
y=2x-3
2 D
B
4
c
5. Find the stationary points on the curve y = x 3 - x 2
• In each case state whether the point is a maximum or a minimum.
6. (a) Show that the points (3, 8), (6, 14) and ( -3, -4) are collinear. (b) Show that the line 3x + 4y + 15 = 0 is a tangent to the circle with centre at the origin and
radius 3 units.
1. (a) Indicate by shading, the region 9 < x 2 + y 2 < 25. (b) Draw sketch graphs (showing the main features) of:
(i) y = lxl (ii) y = x2 - 4 (iii) y = rx
2. The sum of the first three terms of an arithmetic sequence is 27 and the sum of the next three terms is 54. Find the first term and the common difference of this sequence.
46
3. Two artillery guns are situated 5 kilometres apart at positions A and B. They are both aiming at a target T. The angles TAB and TBA are respectively 74° and 66°. Find the distance of the target from the closer gun.
4. A bag contains 4 black and 2 white marbles. If two marbles are drawn at random in succession, the first not being replaced before the second is drawn, find the probability of drawing a white followed by a black.
5. Find the value of S if S = a(;.~~_- 11 ) and a = 18, r = 1· 5, n = 30. Give answer to nearest whole
number.
6. Find the maximum value of 2x(1 - x) by use of the calculus.
1. Solve: (a) 2x - 3x2 = 0 (b) -2 :;:; 8 - 2x :;:; 4
2. Using the data in the diagram, find: (a) area of LABC (b) length of BC.
B
(c) 3x + y = 5 X- 2y = 4
A
18
3. Find the stationary points on the curve y = x 3 - 7x2
- 5x and establish their nature.
40 00
4. Find the value of (a) L (2n + 1) (b) L 4 X 3-11
11=1 11=1
c
5. AD and BE are altitudes of LABC; BC = 5 em, CA = 6 em and AD= 4·5 em. Find the length of BE.
6. The functionf(x) is defined as follows: j(x) =X+ 1 if -2 :;:; X < 3 f(x) = 4 if x ~ 3
Draw a sketch graph of the function for the given domain and find f(- 2) + /(2) - /(5).
1. For the geometric series% + ~ + 287 + · · · find
(a) the common ratio (b) the limiting sum.
2. The angle of elevation of the top of a building from a point P on level ground is 52°46'. If Pis 60 metres from the foot of the building, find the height of the building in metres, correct to one decimal place.
3. A and Bare two acute angles and it is given that sin A = ! and cos B = l 3 . Without using tables or calculator find the exact values of: (a) cos A (b) sin B (c) tan A . cot B
47
4. The two perpendicular lines 3x + 2y = 16 and 2x + ay = b
intersect at the point (4, 2). Find the values of a and b.
5. The numbers 1, 2, 3, and 4 are marked on four cards (one number on each) and placed face downwards on the table. Two cards are selected at random, one after the other, to form a two-digit number. What is the probability that the number formed is (a) even? (b) greater than 32?
6. The curve y = ax2 + 4x - 5 has a gradient of 10 when x = 2. Find the value of a.
1. Solve: (a) 13 - 5xl = 8 4 (b)-- X= 3 X
2. The perimeter of a rectangle is 60 em. If the length is x em, find: (a) an expression for the width in terms of x. (b) an expression for the area of the rectangle in terms of x. (c) the maximum area of this rectangle, by the use of the calculus.
3. (a) Given S = -1
a , find r when S = 32, a = 24. -r
(b) Express the recurring decimal 0· 34 in rational form.
(c) -7 < 5x- 2 ~ 7
4. Find the equation of the straight line through the point of intersection of 2x - 3y + 5 = 0 and 5x + y - 7 = 0, which also passes through the origin.
5. Show, by shading, the region represented by: X + 3y - 4 ~ 0, X - y - 4 ~ 0 and X ;;::: 0.
What is the area of the shaded region?
6. A rhombus has sides 8 em in length and its shorter diagonal is 10 em in length. Find the area of the rhombus.
48
The graphs of two quadratic polynomials.
tic • I
CHAPTER 15
The algebraic expression ax2 + bx + c, where a -=!= 0, is a quadratic polynomial, a polynomial of the second degree or a quadratic expression. In all quadratic polynomials to be studied the coefficients a, b and c will be rational (usually integers) and the domain of the variable x will be the set of real numbers.
The graph of y = ax2 + bx + cis a parabola and it can be seen from the graph that for each real number x there is one and only one value of y. This means that y = ax2 + bx + c is a quadratic function.
y
Looking at the graph of y = ax2 + bx + c we can see that the values of x which make ax2 + bx + c = 0, occur where the graph cuts the x-axis. These values of x are the roots of the quadratic equation ax2 + bx + c = 0 and are sometimes called the zeros of the polynomial.
If a > 0 the graph of y = ax2 + bx + c is concave upwards and has a minimum value. Three distinct cases arise.
(i) The graph cuts the x-axis in two points. In this case the v quadratic equation ax2 + bx + c = 0 has two distinct roots.
(ii) The graph touches the x-axis in one point. In this case the quadratic equation ax2 + bx + c = 0 has one solution (equal roots).
50
y
X
X
(iii) The graph neither cuts nor touches the x-axis. In this case the quadratic equation ax2 + bx + c = 0 has no real roots.
y
X
If a < 0 the graph of y = ax2 + bx + c is concave downwards, has a maximum value and three cases arise corresponding to those above.
y y
y
X
X
X
TWO ROOTS EQUAL ROOTS NO REAL ROOTS
Example: Sketch the graph of y = x2 + x - 6.
Solution: (i) To find where the graph cuts the x-axis first solve x2 + x - 6 = 0. This is equivalent to the
equation (x + 3)(x - 2) = 0 Hence x = _:_ 3 or 2.
Since the roots are - 3 and 2 the graph of the function cuts the x-axis at - 3 and 2.
(ii) The axis of symmetry of the parabola passes through the point on the x-axis mid-way between these two values. Thus its equation is
-3 + 2 X=
2 i.e. x = --!-.
(iii) The coefficient of x 2 is positive hence the parabola is concave upwards and has a minimum value at x = --!-. This minimum value is given by y = ( --!-) 2 + ( --!-) - 6 = - 6!. Hence the coordinates of the vertex of the parabola are ( --!-, -6!).
(iv) It is always easy to find the point where the parabola cuts they-axis and this is a useful check point for the sketch graph.
When x = 0, y = -6, thus the graph cuts they-axis at the point (0, - 6).
51
Draw sketch graphs of the following quadratic functions by finding, the points where the curve cuts the x-axis, the coordinates of the vertex, and the point where the graph cuts they-axis.
1. y = (x- 2)(x + 1) 6. y = x 2- 6x + 8 11. y = 2x2 + x- 10
2. y = (x + 2)(x + 1) 7. y = x 2 + 2x + 1
3. y = x(x + 3) 8. y = x 2 - x - 30
4. y = (x + 3) (x + 4) 9. y = 20 + 8x - x 2
5. y = (5 - x)(3 + x) 10. y = x 2 - 6x
The graph of y = ax2 + bx + c cuts the x-axis at the points whose x values are the roots of the quadratic equation ax2 + bx + c = 0.
This graph can be used to solve the quadratic inequalities ax2 + bx + c > 0
and ax2 + bx + c < 0.
Example (i):
Solve the inequality x 2 + x - 6 < 0.
Solution: y = x 2 + x - 6 = (x + 3) (x - 2) Sketchthegraphofy = (x + 3)(x- 2).
The parabola is concave upwards and cuts the x-axis at x = -3andx = 2.
From the graph it can be seen that x 2 + x - 6 is negative for values of x between - 3 and 2.
Thus the solution of x 2 + x - 6 < 0 is {x: -3 < x < 2}.
Example (ii): Solve the inequality 4 - 3x - x 2 < 0.
Solution: y = 4- 3x- x 2 = (4 + x)(1 - x) Sketch the graph of y = (4 + x)(l - x).
The parabola is concave downwards and cuts the x-axis at x = - 4 and x = 1.
From the graph it can be seen that 4 - 3x - x 2 is negative for values of x less than -4 and for values of x greater than 1.
Thus the solution of 4 - 3x - x 2 < 0 is {x: x < -4} u {x: x > 1}.
52
12. y = 3x2 + 5x - 2
13. y = 8 + 2x - x 2
14. y = 9x2 - 12x + 4
15. y = 9 - 3x - 2x2
tv I
Solve the following quadratic inequalities.
1. (x + 2)(x - 2) > 0 6. (2x - 1) (x + 4) < 0 11. 21- 4x- x 2 > 0
2. (x - l)(x - 3) < 0 7. (3 - x)(4 + x) > 0 12. 10 + 7x + x 2 < 0
3. (x + 4)(x - 2) < 0 8. (x - 5) (2x + 5) > 0 13. 2x2 - 5x - 12 > 0
4. (x + 2) (x + 5) > 0 9. x 2- 5x + 4 < 0 14. 3x2 + x- 2 < 0
5. (7 - x)(3 + x) < 0 10. x 2- 5x- 14 > 0 15. 14 + 3x - 2x2 > 0
By completing the squares in the general quadratic equation ax2 + bx + c = 0 the following formula for solving a quadratic equation was found earlier.
The roots of the quadratic equation ax2 + bx + c = 0 are given by the formula: -b ± )b2
- 4ac X= 2a .
Example: Solve (a) 2x2
- 3x - 7 = 0 and (b) 3x2 - lOx + 8 = 0 indicating whether the roots are rational or irrational in each case.
Solution: (a) If 2x2
- 3x - 7 = 0,:--,;---b + )b2
- 4ac then x = ----="--':----2a
where a = 2, b = - 3, c = -7
thus x = 3 ± .J9 + 56
4 _3±.[65 - 4
(b) If 3x2 - lOx + 8 =~0.,.----------:--_
-b + )b2 - 4ac
then x = -2a
where a= 3, b = -10, c = 8
thus x = 10 ± .JlOO - 96 6
10 ± J4 6
. 3+.[65 3-.[65 .. x =
4 or
4 10 ± 2
The roots are irrational. 6
.·. x = 2 or It The roots are rational.
1. Solve by formula the following quadratic equations indicating whether the roots are rational or irrational in each case. (a) x 2 + 2x - 9 = 0 (b) 2x2 + 5x + 1 = 0 (c) 2x2 + 7x - 4 = 0
(d) x 2 - x - 3 = 0
(e) 3x2 + 2x - 1 = 0 (f) 2x2 + 7x + 3 = 0
(g) x 2 + 3x + 1 = 0 (h) 5x2 + x - 2 = 0 (i) 3x2 + 5x + 2 = 0
2. The quadratic equation x 2 + 2x + 7 = 0 has no real roots. Try to solve this equation using the formula. What do you notice?
3. The equation 9x2 - l2x + 4 = 0 has one solution only. Use the quadratic formula to solve
this equation. What do you notice about the value of (b 2 - 4ac)?
53
4. The equation 6x2 - 11x + 3 = 0 has rational roots. Use the quadratic formula to solve this
equation. How does the value of(b2- 4ac) tell you that the roots must be rational?
5. The equation 5x2 - 3x - 1 = 0 has irrational roots. How does the value of (b 2
- 4ac) tell you that the roots must be irrational?
When using the quadratic formula we have seen that if the number under the square root sign (b2
- 4ac) is negative there are no real roots, if it is zero the two roots are equal, and if it is positive there are two unequal roots which may be rational or irrational.
Since the number under the square root sign discriminates between the nature of the roots, it is called the discriminant and is usually denoted by A (delta).
For the general quadratic equation ax2 + bx + c = 0 the discriminant is given by A= b2
- 4ac. (i) If A < 0 the quadratic equation has no real roots. (ii) If A = 0 the quadratic equation has equal roots which are rational.
(iii) If A > 0 the quadratiC equation has 2 unequal roots. (a) If A is also a perfect square these two roots are rational. (b) If A is not a perfect square these two roots are irrational.
Example: Use the discriminant to find the nature of the roots of the quadratic equations (a) 4x2
- 12x + 9 = 0 (b) 4x2 + 11x - 3 = 0 (c) 4x2 + 3x + 1 = 0
Solution: (a) A = b2
- 4ac = ( -12)2
- 4(4)(9) = 144- 144 =0
Since A = 0 the roots are equal and rational.
(b) A = b2 - 4ac
= (11) 2 - 4(4)( -3)
= 121 + 48 = 169 = 132
Since A > 0 and a perfect square the roots are unequal and rational.
(c) A = b2 - 4ac
= (3)2 - 4(4)(1)
= 9- 16 = -7
Since A < 0 the equation has no real roots.
Using the discriminant find whether the roots of the following quadratic equations are real or unreal, rational or irrational, equal or unequal.
1. 4x2 - 4x + 1 = 0
2. x 2 + 2x + 3 = 0
3. 6x2 + x - 2 = 0
4. 3x2 - 5x + 5 = 0
5. x 2 - 3x - 7 = 0
6. 4x2 - 16x + 16 = 0
54
7. 2x2 - llx - 21 = 0
8. x 2- 3 = 0
9. 2x2 - 5x - 4 = 0
Example: Find the values of k for which the equation 2x2
- 5x - k = 0 has: (a) equal roots (b) real roots (c) unreal roots (d) real and different roots.
Solution:· ~ = b2
- 4ac = (-5)2 - 4(2)(-k) = 25 + 8k
(a) For equal roots~ = 0 i.e. 25 + 8k = 0
(c) For unreal roots~ < 0 i.e. 25 + 8k < 0
k = -3t (b) For real roots ~ ;?: 0
i.e. 25 + 8k ;?: 0 8k;?: -25
k;?: -3t
8k < -25 k < -3t
(d) For real and different roots ~>0
i.e. 25 + 8k > 0
I. Find k if 3x2 - 8x + k = 0 has equal roots.
2. For what values of a has ax2 + 3x - 4 = 0 unreal roots?
3. Find values of k for which 5 + 4x - x 2 = k has equal roots.
4. If (2k + 3)x2 - 4kx + 4 = 0 has equal roots find k.
5. Find the values of k for which x 2 + 3x - k = 0 has:
8k > -25 k > -3t
(a) equal roots (b) real roots (c) unreal roots (d) real and different roots.
6. If the equation x 2 + kx + 36 = 0 has real roots, what values can k take?
7. Find the values fork for which x 2 + (k - l)x = 2k + I has equal roots.
8. Show that 4x2 + 3kx + P = 0 has no real roots whatever the value of k.
9. Prove that kx2 - (k + I)x + 1 = 0 has rational roots if k is rational.
10. Arrange the equation x + _! = k into the normal quadratic form and find values for k if the equation has real roots. x
The connection between the roots of the quadratic equation ax2 + bx + c = 0 and the coefficients a, b and c can be found as follows.
Let rx and f3 be the two roots of ax2 + bx + c = 0.
That is the equation x 2 + [J_ x + !!_ = 0 has roots rx and f3. a a
Now also the equation with roots rx and f3 is (x - rx) (x - [3) = 0
i.e. x 2 - rxx - f3x + rxf3 = 0
i.e. x 2 - (rx + f3)x + rxf3 = 0
Therefore x 2 - (rx + f3)x + rxf3 = 0 and x 2 + [J_x + !!_ = 0 are equivalent equations.
a a
This means that rx + f3 = _[J_ and rxf3 = !!_, a a
55
If a and fJ are the roots of ax2 + bx + c = 0, b c
then a + fJ = -- and afJ = -. a a
Example (i): If a and fJ are the roots of the quadratic equation 2x 2
- 5x + 1 = 0, find the values of: 1 1
(a) a+ fJ (b) afJ (c) - +- (d) a2 + {1 2 (e) (a- 3)(fJ 3) a fJ
Solution:
(a) a + fJ = _'!_ = - (- 5) = 21 a 2
c 1 (b) afJ =- =-
a 2 1 1 fJ+a 5.1 (c) - + - = -- = - -,.- - = 5 a fJ a{J 2 2
(d) a2 + {1 2 =(a+ {J) 2 _ 2afJ =G)
2
- 2G)= 2:-1 = 5*
(e) (a - 3)(fJ - 3) = afJ - 3a - 3{1 + 9 = afJ - 3(a + fJ) + 9 =!- 3{i) + 9 =2
Example (ii) : Form the quadratic equation whose roots are -1! and 4.
Solution: The quadratic equation whose roots are a and fJ is
x 2 - (a + fJ)x + afJ = 0
Now a + fJ = - H + 4 = 2! and afJ = ( -1!)(4) = -6 . ·. The quadratic equation with roots - 1! and 4 is
x 2 - (i)x + (- 6) = 0
i.e. x 2 - h - 6 = 0
i.e. 2x2 - 5x - 12 = 0
Example (iii) : Find the value of k in the equation kx2
- 5x + 4k = 0 when four times the sum of its roots is equal to their product.
Solution: Let the roots be a and {J,
then b 5 a+ fJ = -a= k
a{J = .£ = 4k = 4 a k
Now 4(a + fJ) = afJ
i.e. 4(~) = 4 ~ = 1
k=5
56
1. Write down the sum and product of the roots of the following quadratic equations. (a) x 2 - 2x + 5 = 0 (c) kx2 - 2x + k = 0 (e) 4x2 = 5x - 1 (b) 3x2 + x - 4 = 0 (d) 4x2 - kx + k - 1 = 0 (f) 7x = 3 - 2x2
2. If a and f3 are the roots of the quadratic equation x 2 - 5x + 2 = 0 find the value of: 1 1
(a) a + f3 (d) a2 + {3 2 (g) a2 + {32
(b) af3 (e) (a - 2)({3 - 2) (h) '!:_ +I}_ f3 a
1 1 (c) - +
a f3 (i) (a - {3) 2
3. If a and f3 are the roots of 2x2 + 3x - 4 = 0, find the value of: (a) a + f3 (c) a2 + af3 + {32 (e) af3 - 2a - 2{3 + 1 (b) af3 (d) (a- 1)({3- 1) (f) a2{3 3 + a3 [32
4. Form the quadratic equation whose roots are: (a) 3 and 5 (c) 1 and -7 (e) -4 and -} (b) 4 and -2 (d) } and 1} (f) J3 and -J3
5. Given the quadratic equation x2 - (2 + k)x + 3k = 0, (a) write down the sum and product of the roots in terms of k. (b) find the value of kif:
(i) the sum of the roots is 5. (ii) the product of the roots is 12.
(g) (2 + y"j) and (2 - J'}_) (h) (5 - y'2) and (5 + y'2)
(iii) the product of the roots is four times the sum of the roots.
6. For what value of n will one root of the equation (n - 2)x2 + (n + 2)x + 2n + 1 = 0 be the reciprocal of the other?
7. For what value of k will the equation x2 - (k + 2)x + (k - 4) = 0 have: (a) one root equal to zero? (b) one root equal to 4? (c) one root which is the reciprocal of the other?
8. If one root of 3x2 - 8x + k = 0 is three times the other root, find k.
9. Given that one root of x 2 - 1·3x - 8·88 = 0 is 3·7, find the other.
10. If a and f3 are the roots of-1- = nx, find in terms of m and n:
x+m (a) a+ {3 (b) af3 (c) (a- {3)2
11. Find the value of k if the sum of the roots of the equation, x2 - (4 - k)x + (k - 2) = 0, is equal to the reciprocal of the product of the roots.
12. Find the value of k if the roots of the equation, 4x 2 - 20x + k = 0, differ by 2.
13. a and f3 are the roots of the equation x2 - (m2 + n2)x + mn = 0. If 3(a + {3) = 10a{3, show n
that m = 3 or m = 3n.
14. If a and {3 are the roots of x2 + hx + k = 0, find the relationship between hand kif: 1 1
(a) a= 2{3 (b) ~ + {j = 3
57
Consider the graph ofj(x) = ax2 + bx + c. The roots of the quadratic equation ax2 + bx + c = .0 are given by the x values of the points of intersection with the x-axis.
Algebraically these roots are given by ·
x = - b ± --/ b2 - 4ac
2a
Th t . - b + J b2
- 4ac d - b .J b2 - 4ac a IS X = - an X = - - --'-~--
2a 2a 2a 2a The axis of symmetry of the parabola passes through the point on the x-axis which is mid-way between
these two roots. Hence the equation of the axis of symmetry is x = - b and since the vertex of the 2a
parabola occurs on the axis of symmetry, the maximum or minimum value of f(x) is given by the value ofj(x) when
x = ~: i.e. !(~:)
The axis of symmetry of the graph of the quadratic function f(x) = ax2 + bx + c
is given by the equation -b
x=-. 2a
The maximum or minimum value of the quadratic function occurs when
x = - b and hence is given by 1(-b) . 2a 2a
Note: The maximum or minimum value of a quadratic function can also be found by using the calculus as in Chapter II.
Example (i): Find the maximum or minimum value of f(x) = x 2
- 6x + 5 and sketch the graph of the function.
Solution: Since the coefficient of x 2 is positive the graph is concave upwards andf(x) has a minimum value.
The minimum value occurs when -b
x = 2a where a = 1, b = - 6
6 =2=3
.". Minimum value of f(x) is given by /(3) /(3) = 9 - 18 + 5 = -4
Therefore the minimum value of f(x) is -4 occurring when x = 3.
58
Example (ii): Find the maximum or minimum value ofj(x) = -x2 - 2x + 5 and sketch the graph of the function.
Solution: Since the coefficient of x 2 is negative the graph is concave downwards andf(x) has a maximum value.
The maximum value occurs when -b
x = - where a = - 1 b = -2 2a ' 2
= -2 = -1
.'. Maximum value ofj(x) is given by f(-1) f(- 1) = - 1 + 2 + 5 = 6
Therefore the maximum value ofj(x) is 6 occurring when X= -1.
1. Show that the minimum value of x 2 - 2x + 3 occurs when x = 1 and hence determine this
minimum value.
2. (a) How do you know that the quadratic expression 3 + 8x - 2x 2 has a maximum value? (b) Show that the axis of symmetry of the parabolaf(x) = 3 + 8x- 2x2 is given by x = 2. (c) Find the maximum value of 3 + 8x - 2x2
•
3. Find the maximum or minimum values of the following quadratic functions and sketch the graph of each function. (a) f(x) = x 2
- 2x + 6 (b) f(x) = x 2 + 2x - 5 (c) f(x) = x 2 + 3x - 7 (d)f(x) = -x2
- 5x- 3 (e) f(x) = 2x2 + 8x - 1
(f) f(x) = -x2 + 6x- 5 (g) f(x) = 2x2 - lOx - 3 (h) f(x) = 4x2
- 4x + 1 (i) f(x) = 7 - x - 3x2
(j) f(x) = 3x2 + 4x + 1
4. Find the coordinates of the vertex of the parabola y = 2x2 - 6x + 7. Sketch the curve and give the minimum value of the expression.
5. If the minimum value of the expression x 2 - 4x + k is 5, find the value of k.
6. The parabola y = 3x2 - kx + 2 is symmetrical about the line x = !. Find the value of k and the
minimum value of the expression.
7. The height in metres of a projectile for a horizontal displacement of s metres is given by h = 108 + 40s - s2
• Find the maximum value for the height h.
8. Find the two numbers whose sum is 12 and which have a maximum product. Let the two numbers be x and 12 - x.
59
CASE 1 POSITIVE DEFINITE
We say that the quadratic expression ax2 + bx + c is positive definite if it is positive for all real values of x.
For this to happen as can be seen in the diagram: (i) the parabola y = ax2 + bx + c must be concave upwards and
hence the coefficient of x 2 must be positive. i.e. a > 0. (ii) the parabola must not cut or touch the x-axis. Thus the equation
ax2 + bx + c = 0 has no real roots. i.e. ~ = b2 - 4ac < 0.
The expression ax2 + bx + cis positive definite if a > 0 and b2
- 4ac < 0.
Alternatively: If a quadratic expression has a minimum value which is greater than zero, then the expression must be positive definite since it will be positive for all values of x.
CASE 2 NEGATIVE DEFINITE
We say that the quadratic expression ax2 + bx + c is negative definite if it is negative for all real values of x.
For this to happen as can be seen in the diagram: (i) the parabola y = ax2 + bx + c must be concave downwards
and hence the coefficient of x 2 must be negative. i.e. a< 0.
(ii) the parabola must not cut or touch the x-axis. Thus the equation ax2 + bx + c = 0 has no real roots. i.e. ~ = b2
- 4ac < 0.
60
The expression ax2 + bx + c is negative definite if a < 0 and b2
- 4ac < 0.
Alternatively: If a quadratic expression has a maximum value which is less than zero, then the expression must be negative definite since it will be negative for all values of x.
CASE 3 As can be seen in the diagrams a quadratic expression can take both positive and negative values. For this to happen the parabola y = ax2 + bx + c must cut the x-axis in two distinct points and thus the equation ax2 + bx + c = 0 in each case has two distinct roots. i.e. 11 = b2
- 4ac > 0.
The expression ax2 + bx + c is indefinite if b2 - 4ac > 0.
CASE 4 If 11 = b2
- 4ac = 0 then the quadratic expression is zero for one
-b value of x; at x = 2;; and is:
(i) positive for all other values of x if a > 0.
(ii) negative for all other values of x if a < 0.
61
Example: Investigate the sign of these quadratic expressions for varying values of x. (a) -2x2 + x- 5 (c) x 2 + x- 2 (b) x 2
- 6x + 9 (d) 3x2 + 5x + 4
Solution: (a) - 2x2 + x - 5
~ = b2- 4ac
= 1-40 = -39
Thus ~ < 0 and a < 0. Therefore the expression is negative definite.
(b) x 2- 6x + 9
~ = b2- 4ac
= 36- 36 =0
Thus~= 0 and a> 0.
Therefore the expression is zero at x = - b = 3 and positive for all 2a
other values of x.
(c) x 2 + x- 2 ~ = b2
- 4ac =1+8 =9
Since the ~ > 0 the expression takes both positive and negative values. To find where y = x 2 + x - 2 cuts the x-axis we solve the equation
x2 +x-2=0 (x + 2)(x - 1) = 0
x = -2 or 1 Hence: x 2 + x - 2 > 0 for x < - 2 and x > 1
x 2 + x - 2 < 0 for - 2 < x < 1 x 2 + x - 2 = 0 for x = -2 and x = 1
62
(d) 3x2 + 5x + 4 Ll = b2
- 4ac = 25- 48 = -23
Thus Ll < 0 and a > 0. Therefore the expression is positive definite.
1. Show that 5x2 - 3x + 2 is positive for all values of x.
2. Show that 5x - 3 - 6x 2 is negative for all values of x.
3. By considering the value of the discriminant and the sign of the coefficient of x2 classify the following quadratic functions as positive definite, negative definite or indefinite. (a) 3x2 + 2x - 1 (c) x 2
- 2x + 5 . (e) 2x2 - lOx + 11 (g) x 2 + 6
(b) 3x - 2x2 - 4 (d) 2x - x2
- 5 (f) x 2 - 3x (h) 20 + 4x - x 2
4. Investigate the sign of these qudratic functions for varying values of x and draw a sketch graph in each case. (a) x 2
- 6x + 5 (b) 2x2 + x + 3
(c) 5x - 3x2 - 4
(d) x 2 + 2x - 3 (e) x2 + x- 12 (f) 5x2
- 3x (g) 2x- x 2
(h) 8 - 6x - 5x2
5. For what values of x are the following expressions positive? A sketch graph may help in each case. (a) (x- 7)(x + 3) (c) (x- l)(x- 6) (e) (2x- 1)(3 - x) (g) 2x2
- 9x + 10 (b) x(x - 4) (d) (x - 3)2 (f) (x - 4)(4 - x) (h) 3 + 2x - x 2
6. For what values of xis: (a) (x- 1)(x + 4) < 0 (b) x(x - 5) > 0
(c) x 2 + 4x - 12 < 0 (d) 16 + 6x - x 2 > 0
7. For what values of xis the expression x 2 - 2x greater than 15? Hint: Find what values of x makes
x 2 - 2x - 15 positive.
8. Find values of k for which: (a) 3x2
----: 2x + k is positive definite. (b) - 2x2 + 3x + k is negative definite. (c) kx2
- 6x + 2 is positive definite. (d) x 2 + kx + 16 is positive definite. (e) x 2
- (k + 1)x + (2k - 1) is positive definite.
If the two quadratic expressions a1x 2 + b1x + c1 and a2x 2 + b2x + c2 are equal for all values of x we say the expressions are identically equal and we write
a1x 2 + b1x + c == a2 x 2 + b2x + c2
63
Following are two important results:
(i) If two quadratic expressions are identically equal then the coefficients of like powers are equal. Thus if a1x 2 + b1x + c1 = a2x 2 + b2x + c2
then a1 = a2 , b1 = b2 , c1 = c2
(ii) If a quadratic equation in x is true for more than two values of x then it is an identity and thus is true for all values of x.
Example (i): If 3x2 + 4x + 2 = Ax(x- 1) + B(x + 1) + C, find A, Band C.
Solution: Method 1 3x2 + 4x + 2 = Ax2
- Ax+ Bx + B + C = Ax2 + x(B - A) + (B + C)
Equating coefficients A = 3 B - A = 4 B + C = 2 Solving A= 3,B = 7, C = -5.
Method 2 The identity 3x2 + 4x + 2 = Ax(x - 1) + B(x + 1) + C is true for all values of x. Therefore we can substitute any value for x which will make the calculation easy. Substituting x = 0 in the identity we have 2 = B + C Substituting x = 1 in the identity we have 9 = 2B + C Solving B = 7 and C = -5 Substituting x = -1 in the identity we have 1 = 2A + C
Hence A= 3 Therefore A = 3, B = 7, C = -5.
Example (ii): Express 5x2 + 2x - 3 in the form A(x + 1)2 + B(x + 1) + C.
Solution: Let 5x2 + 2x - 3 = A (x + 1 )2 + B(x + 1) + C Without multiplying out we can see that the coefficient of x 2 on the R.H.S. must be A. Hence equating coefficients of x2
, A = 5 Now substituting x = -1 in the identity we have Substituting x = 0 in the identity we have
i.e.
Hence 5x2 + 2x - 3 = 5(x + 1)2 - 8(x + 1).
0 = c -3 =A+ B + C -3 = 5 + B + 0
.". B = -8
1. If 5x2 - 2x + 9 = ax2 + bx + c, find a, band c.
2. If 6x2 - 2x + 9 = Ax(x - 1) + B(x - 1) + C, find A, Band C.
3. If2x2 + x + 1 and a(x + l)(x - 2) + b(x - 1) + care equal for all values of x, find a, band c.
4. Express 3x2 + 4x + 5 in the form A(x - 1 )2 + B(x - 1) + C.
5. Express 3x2 - 5x + 2 in the form A(x + 1)2
- B(x + 1) + C.
6. Write 5x2 - 2x + 6 in the form Ax(x - 1) + B(x - 2) + C.
7. If x 2 = ax(x - 1) + bx for more than two values of x, find a and b.
64
8. Find a, band c if: (a) x 2 = a(x + 1)2 + bx + c (b) x 2
- 2x = ax(x + 1) + b(x + 1) + c (c) 4x2
- 1 = a(x + 1)2 + b(x + 1) + c (d) x 2 = a(x + 2)2 + b(x + 2) + c (e) a + b(x - 1) + cx(x - 1) = 5x2 + x - 2
9. Find values for A and B if: (a) 4x2
- 12x + 9 = (Ax + B)2
(b) 3(x - 2)2 = A(x2 + 4) + Bx
Certain equations that look difficult to solve may be reduced to a quadratic form and readily solved.
Example (i): Solve x4
- 5x2 + 4 = 0.
Solution: Let m = x 2
then m 2 - 5m + 4 = 0
(m - 4)(m - 1) = 0 Thus m = 4 or m = 1 i.e. x 2 = 4 or x 2 = 1
x = ±2 or x = ±1
Example (ii): Solve (2x - 1)4 + (2x - 1)2
- 6 = 0.
Solution: Let m = (2x - 1)2
then m2 + m- 6 = 0 (m - 2) (m + 3) = 0
Thus m = 2 or i.e. (2x - 1)2 = 2 or
m = -3 (2x - 1)2 = -3
2x- 1 = ±ji 2x=1±~
1 ± J2 X=------'--
2
Example (iii): Solve 9x - 10(3x) + 9 = 0.
Solution:
This equation has no real roots
We note that 9 = 32 and that (32Y = (Y) 2. Hence the equation can be rewritten.
(3x) 2 - 10(3x) + 9 = 0
Let m = 3x then m2
- 1Om + 9 = 0 (m - 9)(m - 1) = 0
Thus m = 9 or m = 1 i.e. 3x = 9
x=2 or or
3x = 1 x=O
65
1. By first substituting m for x2 solve the equation x 4 - 10x2 + 9 = 0.
2. Show that by making the substitution m = x 3, the equation x 6
- 9x3 + 8 = 0 becomes m2
- 9m + 8 = 0 and hence solve the original equation.
3. Solve the following equations giving all real roots in exact form. (a) x4
- 17x2 + 16 = 0 (c) x4 - 13x2 + 36 = 0 (e) x 6
- 26x3 - 27 = 0
(b) x4- 3x2
- 4 = 0 (d) x4 + 2x 2- 15 = 0 (f) x 8
- 17x4 + 16 = 0
4. Solve 4x - 5(2x) + 4 = 0. Make the substitution m = 2x.
5. By substituting m for (x + 1)2 in the equation (x + 1)4 - 6(x + 1)2 + 8 = 0, find its roots.
6. Solve the following equations giving all real roots in exact form. (a) 4x- 6(2x) + 8 = 0 (d) (x- 1)4 - ll(x- 1)2 + 18 = 0 (b) 9x- 12(3x) + 27 = 0 (e) (2x + 1)4
- (2x + 1)2 - 12 = 0
(c) (5x) 2 - 26(5x) + 25 = 0 (f) (x - 3)4
- 18(x - 3)2 + 32 = 0
66
The slope of the tangents determines the nature of turning points.
CHAPTER 16
Given a function y = f(x), the first derivative is a new function of x. If we differentiate this new function we have the second derivative of the original function.
The second derivative is therefore :X (t) and this is normally written as ~J. The following illustrations show some commonly used notations for first and second derivatives.
(i) y = x 3 - 4x2 (ii) f(x) = x 3
- 4x2 (iii) f(x) = x 3 - 4x2
dy = 3x2 - 8x f'(x) = 3x2 - 8x !!_(x3 - 4x2) = 3x2 - 8x ~ ~
d2y d2 - = 6x - 8 f"(x) = 6x - 8 -(x3
- 4x2) = 6x - 8 ~2 ~2
Example (i): Find the second derivative of y = 4x3
- 3x2 + 2x.
Solution:
First derivative
Second derivative
Example (ii):
dy = 12x2 - 6x + 2 dx
d2 _____l = 24x - 6 dx2
If f(x) = x 3 + x 2 - 3x + 4, find the value of x for which f"(x) = 0.
Solution: f'(x) = 3x2 + 2x - 3 f"(x) = 6x + 2
If f"(x) = 0, then 6x + 2 = 0 and x = -1
1. Find the second derivative of the following functions:
(a) 4x5 (d) _?: X
(b) x4 - 2x2 + 6x - 3 (e) x 3
- 2x2 + x - 1
(c) 8 - 7x- 2x3 (f) (x + 2)(x2 + 1)
2. If f(x) = x 3 - 4x + 2, find /"(2).
3. Given that f(x) = (6 + x)(2 - x 2), find the value of f"(1).
4. If f(x) = ax2 + bx + c, show that f"(x) is independent of x.
(g) (2x + 3)4
(h) _x_ X+ 1
(i) -J4x + 2
5. Given that f(x) = x 3 - x 2 + x - 1 evaluate /(2) + /'(2) - f"(2).
. dy d2y 2y- X 6. GIVen that y = x 2
- x, prove that- - - 2 = --. dx dx x
68
7. If f(x) = 3x3 - 4x2 + 3x - 6, find the value of x for which f"(x) = 0.
8. Find the equation of the tangent to the curve y = x 3 + 3x2 + 4x + 2 at the point where~~ = 0.
9. Find the value of x for which ::2 (4x3 - 6x2 + 9) = 0.
10. If y = JX, find the value of x for which y + ~~ = 0.
If y = f(x), the value of k at a point on the curve gives the gradient of the tangent at that point.
The value of fx ( k), that is ~~, at a point on the curve is the rate of change of the gradient of the
tangent at that point with respect to x.
J ..• /
0
If ~~ is positive in a given domain then the
gradient of the tangent is increasing. As x increases in the curve above the gradient is increasing. It is changing from negative through zero to positive.
In the domain shown the curve is concave upwards.
If ~~ is positive in a given domain, then
the curve is concave upwards
Example (i):
0
If ~~ is negative in a given domain then the
gradient of the tangent is decreasing. As x increases in the curve above the gradient is decreasing. It is changing from positive through zero to negative.
In the domain shown the curve is concave downwards.
If d2y. . . . d . h dx2 1s negative m a glVen omam, t en
the curve is concave downwards.
By considering the sign of f"(x) show that the curve f(x) = 3x2 - 2x + 5 is concave upwards at all points.
Solution: If f(x) = 3x2
- 2x + 5 f'(x) = 6x- 2 f"(x) = 6
The second derivative is equal to 6 for all values of x, so it is always positive. Thus the graph of f(x) is concave upwards at all points.
69
Example (ii): Consider the variation in sign of the second derivative of y = 2x3
- 3x2 - 36x + 4 and use this
information to describe the shape of its graph.
Solution: y = 2x3
- 3x2 - 36x + 4
dy = 6x2 - 6x - 36 dx
d2y - = I2x- 6 dx2
d2y Therefore - 2 < 0 for x < !
dx d2y 1
and dx2 > 0 for x > 2 .
Hence the graph of the function is concave downwards for x < ! and concave upwards for x > ! as illustrated.
I. Show that the curve y = 3 - 2x - 2x2 is concave downwards for all values of x.
2. Show that the curve y = 4x2 - 6x + 2 is concave upwards for all values of x.
3. Find the domain for which the curve y = x 3 - 8x2 + 6x - 3 is concave upwards.
4. Find the values of x for which the curve y = 4x2 - x 3 is concave downwards.
X
5. By considering the sign of f'(x) and f"(x) sketch the shape of the curve y = f(x) in the given domains. (a) y = x 3 from x = I to x = 3. (b) y = JX from x = I to x = 4. (c) y = x 3
- 3x2 + 6x- 2 from x = 2 to x = 3. (d) y = 2x2
- x4 from x = I to x = 2.
6. Sketch the curve y = f(x) if: (i) f(O) = 3, f(2) = 7 and if in the domain x = 0 to x = 2, f'(x) > 0 and f"(x) < 0.
(ii) f(l) = 5, f(3) = -I and if in the domain x = 1 to x = 3, f'(x) < 0 and f"(x) > 0.
Whenever the concavity of a curve changes at a point then that point is called a point of inflexion.
In the diagram the curvature changes from concave upwards to concave downwards at point P. Thus P is a point of inflexion and the tangent at P cuts the curve at P.
d2 At A the curvature is concave upwards and { > 0.
dx d2
At B the curvature is concave downwards and----{ < 0. dx
d2 At P the curvature changes from concave upwards to concave downwards and { = 0.
dx
70
TEST FOR A POINT OF INFLEXION For a point of inflexion on the curve y = f(x), two conditions must be satisfied.
(i) ~J = 0 at the point
(") d2y h . h . 11 dx2 c anges s1gn at t e pomt
i.e. ~Jz > 0 on one side of the point and ~J < 0 on the other side of the point.
Example: Find the point of inflexion on the curve y = 2x3 + 3x2 + 6x - 7.
Solution:
~ = 6x2 + 6x + 6 and
d2ydxz- 0 when
i.e. 12x + 6 = 0
X=
When xis a little less than -t, ~J < 0.
When xis a little more than -t, ~J > 0.
1 2
Hence there is a point of inflexion when x = -t. Ifx = -t, y = 2(-t)3 + 3(-t)2 + 6(-t)- 7
= -9t Thus the point ( -t, -9t) is a point of inflexion.
1. Find the point of inflexion for the curve y = x3 - 3x2
- x + 5.
2. For the curve y = 2x3 - 12x2
- 5x - 3 find the point of inflexion.
3. Find the point on the curve y = x3 + x 2 where the curvature changes sign.
4. Show that the function y = x 2 + 5x - 2 has no point of inflexion.
5. Show that the curve y = x4 - 24x2 + 2x - 1 has two points of inflexion. Find these points.
6. For the curve y = x3 - 2x + 1, find:
(a) dy (b) d2y . (c) the equation of the inflexional tangent.
dx dx 2
7. Find the equation of the tangent at the point of inflexion for the curve y = 5 - 4x + 3x2 - x3.
8. Show that the curve y = (x + 1)3 has a point of inflexion at ( -1, 0). Show that ( -1, 0) is also a stationary point on the curve.
A stationary point which is not a turning point is a point of inflexion at which the tangent is horizontal. Such a point is called a horizontal point of inflexion.
9. Show that the curve y = x 5 has a horizontal point of inflexion at the origin.
10. Find the stationary point for the curve y = (2x - 1)3 and show that it is a horizontal point of inflexion.
71
In Chapter 11, by considering the sign of the gradient (~) in the neighbourhood of a stationary
point, we were able to determine whether it was a maximum or a minimum turning point. The sign of the second derivative gives another method for determining the shape of a curve in the
region of a stationary point.
If P is a stationary point on a curve and ;;;_ is positive at P, the curve must be concave upwards.
Thus there is a minimum turning point at P.
Similarly if Pis a stationary point on a curve and;;;_ is negative at P, the curve must be concave
downwards. Thus there is a maximum turning point at P.
If f'(c) = 0 and f"(c) > 0, y = f(x) has a minimum turning point at x = c. If f'(c) = 0 and f"(c) < 0, y =:= f(x) has a maximum turning point at x = c.
Example: Find the maximum and minimum turning points of the functiony = x 3
- 12x + 12 using the second derivative test.
Solution: y = x 3
- 12x + 12
dy = 3x2 - 12 and dx
For stationary points~= 0 and this occurs when
3x2- 12 = 0
i.e. when x = ±2 d2y
When x = 2,-2 > 0. Therefore there is a minimum turning point at (2, -4). dx
d2y When x = -2, - 2 < 0. Therefore there is a maximum turning point at (- 2, 28).
dx
Note: (i) The second derivative test is not always as easy to use as in the above example. Sometimes the second derivative is difficult to find.
(ii) For a function such as y = x4 both~ = 0 and;;;_ = 0 at x = 0. In such a case the
second derivative test does not apply. (iii) In general, when working examples use either the gradient method of Chapter 11 or the
second derivative method, whichever appears the easier to apply.
Use the second derivative test to find the maximum and minimum turning points of the following functions:
1. y = 2x2 - 4x + 3 5. y = x4
- 32x + 48
2. y = 6 - 2x - x 2 6. y = x3 + 3x 2 - 9x - 11
3. y = x 3 - 3x2 + 2 7. y = x2 + ~
4. y = 2x3 + 2x2 - 2x - 1 8. y = x4
- 2x2 + 2
72
Example (i): Sketch the curve y = 4 + 3x - x3
, showing any turning points or points of inflexion.
Solution: y = 4 + 3x- x3
dy d2y (i) First and second derivatives: - = 3 - 3x2 and - = -6x.
dx dx2
(ii) Stationary points: ?x = 0 when 3(1 - x2) = 0
i.e. when x = ± 1 There are stationary points at (1, 6) and ( -1, 2).
d2y (iii) Testing stationary points: At (1, 6), dx2 < 0, hence (1, 6) is a maximum turning point.
d2y At ( -1, 2), dx 2 > 0, hence ( -1, 2) is a minimum turning point.
(iv) Points of inflexion: ~J = 0 when - 6x = 0 i.e. when x = 0
When x is a little less than zero, ~J > 0.
When x is a little more than zero, ~;;. < 0.
Thus there is a point of inflexion at (0, 4).
(v) Plotting other points to assist, draw the sketch of the curve:
Example (ii): Draw a neat sketch of the function f(x) = x 4 - 2x2 showing turning points and points of inflexion.
Solution: f(x) = x 4 - 2x2
(i) First and second derivatives: f'(x) = 4x3 - 4x and f"(x) = 12x2
- 4 = 4x(x2 - 1) = 4(3x2
- 1)
(ii) Stationary points: f'(x) = 0 when 4x(x2 - 1) = 0 i.e. when x = 0 or x = ± 1
There are stationary points at (0, 0), (1, -1) and ( -1, -1).
73
(iii) Testing stationary points: At (0, 0), f"(x) < 0, At (I, -I), f"(x) > 0,
hence (0, 0) is a maximum turning point. hence (I, -I) is a minimum turning point.
At (-I, -I), f"(x) > 0, hence (-I, -I) is a minimum turning point.
(iv) Points of inflexion: f"(x) = 0 when 4(3x2 - I)= 0
i.e. when X= +_I_ -J3
f"(x) changes sign as x passes through~ and - ~·
Hence there are points of inflexion at ( ~' -~)and (-~' -~} (v) Even function: x 4 - 2x2 is an even function, hence its graph is symmetrical about they-axis.
(vi) For /x/large: When /x/ is large (positive or negative x), f(x) is very large and positive.
(vii) Plotting other points to assist, sketch the curve:
Example (iii): 4
Sketch the curve y = x + -. X
Solution: 4
y=x+x
Av
I Maxj(O, 0)
Min (-1, -1) Min(1, -1)
dy = I - _4 and d2y- 8 (i) First and second derivatives: - -dx x 2 dx2 - x3
(") s . . dy 0 n tatzonary poznts: dx = when I - _±_ = 0 x2
i.e. when x = ±2 There are stationary points at (2, 4) and (- 2, - 4).
X
d2y (iii) Testing stationmy points: At (2, 4), dx2 > 0, hence (2, 4) is a minimum turning point.
d2y At ( -2, -4), dx2 < 0, hence ( -2, -4) is a maximum turning
point.
74
(iv) Points of inflexion: dd2
{ = 0 when 8
3 = 0 and this equation has no solution. Hence there are X X
no points of inflexion.
(v) Points of discontinuity and asymptotes: There is a point of discontinuity at x = 0 and they-axis (x = 0) is an asymptote.
For large values of x (positive or negative), the i term becomes small andy ~ x, soy = x . X 1s an asymptote.
(vi) Odd function: x + _i is an odd function (since f( -x) = - f(x)) and hence the curve has point X
symmetry about the origin.
(vii) Plotting other points to assist, sketch the curve:
/
Max (/2, -4) /
/ /
-4
2
/ /
Note: This curve could also be sketched by using the addition of ordinates method performed on 4
the graphs of y = x andy = -. X
1. Draw neat sketches of the following functions showing turning points and points of inflexion. (a) y = x 3
- 12x + 12 (c) y = x 2 - 4x + 3 (e) y = x3
- 6x2 + 9x - 5 (b) y = 8 + 6x2
- x3 (d) y = (x - 2)3 (f) y = x4 - 8x2 + 16
1 2. Sketch the curve y = x + -.
X
3. Locate the maximum, minimum and inflexion points for the curve y = 2x3 + 2x2 - 2x - 1
and sketch the curve.
4. Sketch the curve y = 18x2 - x4
_:_ 81, showing significant points.
75
5. Findthepointsonthecurvey = 2x3- 3x2
- I2x + 20atwhichyhasamaximumoraminimum value.
Use your results to make a neat sketch of the curve.
6. (a) Show that if y = x3 - 3x2
- 9x + 20, then dy = 3(x - 3)(x + 1). dx
(b) Find the stationary points and distinguish between them. (c) Find the point of inflexion. (d) Sketch the curve.
7. Show that the curve y = x4 - 32x + 2 has only one stationary point and determine its nature.
Show that the curve has no point of inflexion and draw a sketch of the curve.
8. Show that the curve y = x + ~ has a minimum value turning point at (2, 3) and no points of X
inflexion. Show that the curve has asymptotes y = x and the y-axis. Hence or otherwise draw a neat
4 sketch of the curve y = x + 2 . X
Example (i): Find two positive numbers whose sum is 20 and whose product is a maximum.
Solution: Let one of the numbers be x, then the other number is 20 - x. If y is their product, then
y = x(20- x) = 20x - x 2
dy d 2y - = 20 - 2x and - = -2 dx dx2
dy = 0 dx
when 20 - 2x = 0 i.e. when x = 10
Hence there is a maximum product when x = 10 because~~ is always negative.
The maximum product is 10(20 - 10) = 100.
Example (ii): Find the dimensions of the cylindrical can of greatest volume that can be made from 600n cm2 of sheet metal.
Solution: Since the surface area of a cylinder is given by A = 2m·2 + 2nrh.
2nr2 + 2nrh = 600n h = 600n - 2m·2
2nr 300
=--r r
Using the formula for the volume of a cylinder, V = nr2h, and substituting for h
v = nr2 e~o -r)
= 300nr- nr3
dV d2 V dr = 300n - 3m·2 and dr2 = - 6nr
76
dV = O dr
when 300n - 3nr2 = 0
i.e. when ,.2 = 300n = 100 3n
giving r = 10 (as r must be positive) d2 V 300
If r = 10, dr 2 < 0 and h = lo - 10 = 20.
Hence Vis a maximum when r = 10 and h = 20. The maximum volume of the can is 2000n cm3.
1. In the triangle ABC, the angle Cis a right-angle and the sum of the sides CA and CB is 12 em. Find the maximum area of the triangle.
2. A rectangle has a perimeter of 32 em. What dimensions would the rectangle have for maximum area?
3. Find the dimensions of the cylindrical can of greatest volume that can be made from 150n cm2
of sheet aluminium.
4. An open tank is to be constructed with a square base of side x metres and with four rectangular sides. The tank is to have a capacity of 32 cubic metres.
(a) Show that the height of the tank is 3; metres.
X
(b) Find the least area of sheet metal from which the tank can be constructed.
5. A magazine advertisement is to contain 50 cm2 of lettering with clear margins of 4 em each at top and bottom and 2 em at each side. Find the overall dimensions if the total area of the advertisement is to be a minimum.
6. The cost per hour of running a pleasure cruiser is$(:; + 10), where Vis the speed in knots.
(a) For a trip of D nautical miles show that the cost is$[~(:; + 10) J . (b) What is the most economical speed for running the cruiser on this trip?
7. (a) The sum of the radii of two circles is 100 em. If one of the circles has a radius of x em, show that the sum of the areas of the two circles is given by
A = 2n(x2 - lOOx + 5000).
(b) Find the value of x for which A is least.
8. A piece of wire is 20 em long. It is cut into two portions, the length of one portion being x em. Each portion is then bent to form a rectangle in which the length is twice the breadth. (a) Show that the sum of the areas of the two rectangles is given by
A = /8 (2x2 - 40x + 400).
(b) For what value of xis this area a minimum?
Consider the following functions.
y = x2
+ 2x } y = x 2 + 2x + 3
_ 2 + 2 1 In all cases Y-X x- d y = x 2 + 2x + 7 j' = 2x + 2 y = x 2 + 2x + C x
77
Now suppose you have the reverse problem. You are given:= 2x + 2 and you wish to find y.
By the process of anti-differentiation, y = x 2 + 2x + C because whatever the value of the constant Cit will vanish on differentiation. The function x 2 + 2x + Cis called the primitive of 2x + 2.
Geometrically given : = 2x + 2, then
the primitive function y = x 2 + 2x + C represents a family of curves as shown in the diagram; where each curve corresponds to a different value for C.
:1
In simple cases the primitive can often be found by inspection. If: = x2 we know the primitive
must contain x3• Now since the derivative of x3 is 3x2 the derivative of tx3 must be x 2
• Hence the primitive function is y = h 3 + C.
x"+1 In general the primitive of x" is -- + C, n i=
n + 1
This means to find the primitive of x" we raise the index by I, divide by the new index and add a constant.
Example (i): Find the primitive function if:
(a) dy = xs dx
x6 y=-+ c
6
78
(b) dy = 2x3
dx
y = 2(~4
) + c 4
=~+ c 2
Example (ii): Find the primitive of 5x3 + 3x2
- 4x + 1.
Solution:
If dy = 5x3 + 3x2 - 4x + 1 dx
then y = 5 (~4
) + 3 (~3
) - 4 ( ~) + x + c i.e. y = -;i-x4 + x3
- 2x 2 + x + C
1. Find the primitive function for each of the following. Check by differentiation.
(a) dy = x6 (c) dy = 2\2 (e) dy = 3x + 2 (g) dy = h4 ~ ~ ~ ~
(b) dy = 3x5 (d) dy = 5 (f) dy = x2 - x + 5 (h) dy = 2 - 3x3
~ ~ ~ ~
2. Find the primitive of: (a) x 2
(b) 7x3 (c) 8 (d) 2x + 6
3. Find f(x) given that f'(x) is: (a) 3x - 5 (b) 2x2
- x + 2
(e) x4 - x
(f) h3 + 2
(c) 10 - 4x - 2x3
. ~ 4f 4. GIVen -d = 4t2
- 4t, show that s = - - 2t2 + C. t 3
dv 5. If dt = 5t - 3, find v.
6. If~= 4nr2, find V.
d -2 7. Ifdy = x- 3 ,showthaty =.£_+C.
X -2
8. If Jx = 2x-t, show that y = 4x1 + C and check by differentiation.
9. Find the primitive function if:
(a) dy = x-2 (d) dy = x-t dx dx
(b) dy = x- 5 (e) dy = 3x-4
dx dx
(c) dy = xt (f) dy = -4[3 dx dx
Example (i):
dy 3 (g) - = xz
dx
(h) dy = 6x- 2
dx
(i) dy = 5x-6
dx
(g) ~X - 1 (h) 7x - tx3
(d) 6x4 - 2x 2 + 5x
(j) dy = x-t dx
(k) dy = 2xt dx
(1) dy = -4x-t dx
The gradient function of a curve is 3x2 - x + 1 and the curve passes through the point (2, 13). Find
the equation of the curve.
79
Solution:
dy = 3x2 - x + 1 dx
2
... y = x3 - x2 + x + C
Since the curve passes through the point (2, 13) 22
then 13 = 23 -- + 2 + C
2 13=8-2+2+C
thus C = 5 2
The equation of the curve is y = x3 - ~ + x + 5.
Example (ii):
Find the equation of the curve y = f(x) given that~;~ = 6 + 2x
and when x = 2, : = 20
also when x = 0, y = 2.
Solution: d2y -=6+2x dx2
hence dy = 6x + x 2 + C dx
but dy = 20 when x = 2 dx '
giving dy = 6x + x 2 + 4 dx
thus 20 = 12 + 4 + C hence C = 4
hence x3
y = 3x2 + 3 + 4x + C
but y = 2 when x = 0, thus 2 = C 3
Thus the equation of the curve is y = ~ + 3x2 + 4x + 2.
1. Find the equation of the curve which passes through the point (1, 8), given that its gradient function is 2x + 5.
2. The gradient function of a curve is 3 - 2x and the curve passes through the point (2, 1). Find its equation.
3. Given: = 6x - 4 andy = 5 when x = -1, find an express~on for y.
4. Given~ = 2t 2 - 1 and that s = 10 when t = 3, express sin terms oft.
5. The gradient function at any point x on a curve is 3x2 - x + 5. What is the equation of the curve if it passes through the point (2, 20)?
6. For a certain curve the general expression for the gradient is given by: = 3x2 - 4x + 1.
What is the equation of the curve if it passes through (0, 4)?
80
7. The gradient of a curve is given by : = 3 + 2x - x 2• What is the equation of the curve if:
(a) it passes through the origin? (b) it passes through the point (3, 0)? (c) it passes through the point (2, 5)?
8. Given 1x = xt + 2 andy = 14 when x = 4, express yin terms of x.
9. The gradient function of a curve is given by: = [ 2• If the curve passes through the point
(!, 4), find its equation.
10. If~~ = 3 - 4t and v = 15 when t = 3, express v in terms oft. Now if v = ~;and s = 4 when
t = 0, find s in terms oft.
11. Find the equation of the curve given: d 2 d
(a) d3 = 3x and when x = 1, Jx = 4 andy= 6.
d 2 d (b) dx~ = 2x + 3 and when x = 2, Jx = 12 also when x = 1, y = 5.
. d2 x dx 12. GIVen that dt 2 = 12t - 2, that dt = 0 when t = 0 and that x = 4 when t = 1, find x when
t = 5.
81
I. Given V 2 = k 2 [~ - lJ find V when k = 36·9, r = 0·00746 and a = 0·0853. Give answer to r a
three significant figures.
2. The path of a projectile is given by the equation lOOy = x(300 - x). Find the maximum height of the projectile and its range. You may use either graphical means or the calculus.
3. If the roots of the quadratic equation 2x2 - 5x + 4 = 0 are a and {3, find the value of:
(a) a + f3 (b) af3 (c) l + _!_ (d) a2 + f3 2
a f3 4. A parallelogram has adjacent sides of 9 em and 13 em which meet at an angle of 72°. What is
the area of the parallelogram?
5. Find the derivative of:
(a) (x2 + 2)(5 - 3x) (b) x2
- 2x 2x + 3
(c) (5x - 2)4
6. A bag contains 3 black, 4 red and 2 white marbles. One marble is drawn at random and not replaced, then another marble is drawn. By drawing a tree diagram or otherwise, find the probability that: (a) two red marbles are drawn. (b) a black and a white marble are drawn in any order.
1. The volume of one litre of a certain liquid decreases by evaporation to Vlitres after n days according to the formula V = (1 - r)". Find the value of r, correct to one decimal place, if V = 0·51 and n = 3.
2. If 3x2 - 7x + 5 = Ax(x - 1) + Bx + C, find A, Band C.
3. Draw a sketch graph of y = (x + 2)(x - 3) and use it to find (a) the roots of (x + 2)(x - 3) = 0. (b) the solution of the inequality (x + 2)(x - 3) < 0. (c) the minimum value of (x + 2)(x - 3).
4. Starting with 6, how many consecutive multiples of 6 must be added to make their sum exceed 1000.
5. Find the equation of the tangent to the curve y = x2 + 22 at the point where x = I.
X
6. In the diagram, ABCD is a quadrilateral AD = 5 m, AB = 6 m, BC = 9 m, CD = 7 m and L BCD = 60°. Calculate: (a) BD (b) LDAB (c) area of ABCD
82
A
7
B
c
l. Solve the equations: (a) 3x - 12x2 = 0 (b) 32x- 10·3x + 9 = 0
2. The efficiency, E percent, of a particular spark plug when the gap is set to x mm, is given by E = 800x - 1600x2
• Find the gap setting which gives maximum efficiency. 40
3. Evaluate: (a) L 6n - 1 n=l
4. Differentiate: (a)~ X + 1
00
(b) L 5 X 21- 11
n=l
5 3 (c) 2JX-- +JX xz
5. (a) Find the perpendicular distance of the point (1, -1), from the line 4x - 3y + 8 = 0. (b) Find the equation of the perpendicular from (1, -1) to the line 4x- 3y + 8 = 0.
6. A survey note book contains the information in the diagram. (a) Calculate the distances BC and BD. (b) Calculate the size of L CBD. (c) Calculate the distance CD.
A
AB=100 m
c
D
1. (a) Without solving, determine the nature of the roots of the quadratic equation 2x2 + 3x- 2 = 0. (b) Solve the inequality 2x2 + 3x - 2 < 0.
2. Find the stationary points of the function y = x 3 - 12x and determine their nature.
3. Given that the gradient function of a curve is x2 - 4x + 1 and that the curve passes through
the point ( -1, 2), find the equation of the curve.
4. (a) Show that the lines 9x - 2y + 20 = 0 and 3x + y - 10 = 0 intersect at a point P on the y-axis.
(b) Find the equation of the line m joining (- 2, 1) and ( 4, - 2). (c) Find, as a surd, the perpendicular distance from P tom.
5. Two ordinary dice are thrown. What is the probability that (a) they both show 6? (b) at least one of them shows a 6? (c) they show a total of 6?
6. Find the values of m for which the equation x2 + (m - 2)x + 4 = 0 has equal roots.
1. The fourth term in an arithmetic sequence is 128 and the common difference is 7. Find: (a) the first term (b) the 30th term (c) the sum of the first 30 terms.
2. A number is selected at random from the integers 2 to 20. Find the probabilities that it is: (a) a perfect square (b) a prime (c) either a prime or a perfect square.
83
3. During a survey, two points A and Bare ends of a base line. AB = 85 metres. Compass bearings are then taken as shown on the survey sketch. Find (a) the size of LBAC.
(b) the size of LABC. Calculate (c) the distances AC and BC.
A
080°
85 m B
4. Sketch the graph of y = 2x2 + x - 1 showing where it cuts the x-axis and the coordinates of its turning point. (a) read off the solution to 2x2 + x - 1 = 0. (b) write the solution of 2x2 + x - 1 < 0. (c) what is the minimum value of 2x2 + x - 1?
5. (a) If a and f3 are the roots of the equation 2x2 - 5x + 3 = 0, find the value of (a + 1)([3 + 1).
(b) Find values for A, Band C such that x 2 + x + 1 = A(x - 2)2 + B(x - 2) + C.
6. Find the equation of the curve y = f(x) give that~~ = 2x, the gradient is 2 when x = 1 and the
curve passes through the point (3, 6).
1. Draw separate sketches of the following relations. (a) xy = 1 (b) y = lxl (c) y = x 2
- 1
2. The first three terms of an arithmetic sequence are 40, 33, 26. (a) Write down a formula for the nth term.
(d) 4 < x 2 + y 2 < 25
(b) If the last term of the sequence is - 3 7, how many terms are there in the sequence? (c) Find the sum of the terms in this sequence.
3. Find the equations of the tangents toy = (x + 2)(x - 1) at the points where the curve cuts the x-axis.
4. In 6ABC, AB = 10 em, AC = 8 em and LBAC = 60°. (a) Find the length of BC. (b) What is the area of 6ABC?
d2y dy 5. Given dx2 = 6x- 2, also when x = 1, dx = 6 andy= 2, find y when x = 3.
6. (a) Find the stationary points on the curve y = x 3 - 3x + 2 and state whether they are maxima
or minima. (b) Find the point of inflexion on the curve. (c) Sketch the curve.
84
A circle is the locus of all points equidistant from a given point called the centre.
CHAPTER 17
In chapter 7 we had a brief look at the idea of a locus but we did not use the word locus. Locus is an attempt to describe either in algebraic or geometric terms the path of a point which moves so that it always obeys a given condition.
Example (i):
/ A point P moves so that its distance from the y-axis is always equal to its distance from the x-axis. What is the locus of P?
Solution: The condition is PA = PB (see diagram).
----/:P(•.vl
This means y=x /i A
Algebraically the locus of Pis the equation y = x. Because y = x represents a line though the origin and at 45 degrees to the axes then the locus could be described geometrically in these terms. Note: Directly by geometry we could prove that L PO A = L PO B and hence that all points P do lie on a line which bisects the angle between the axes. But usually it is easier to work algebraically.
Example (ii): What is the equation of a circle with centre at the origin and radius 4 units?
Solution: The circle is (by definition) the set of points where each is 4 units from the origin. Hence in the diagram PO = 4 units for any point P. So PO= 4
, )xz + yz = 4 x 2 + y 2 = 16
0 X
P(x, y)
I v I I I
1. Find the locus of a point which moves so that its distance from the x-axis is always twice its distance from they-axis.
2. Find the locus of a set of points which are such that their distance from the x-axis is always 2 units more than their distance from the y-axis.
3. What is the equation of the circle, centre the origin, radius 5 units?
4. Write down the equation for each of these loci: (a) the distance from the x-axis is the square of the distance from they-axis. (b) the distance from the y-axis is three times the distance from the x-axis. (c) the distance from the x-axis equals the distance from the line x = -2. (d) the distance from the line y = 2 equals the distance from the line x = -1.
86
5. A set of points is such that each is 5 units above the x-axis. What is the equation of all such points?
6. What is the equation of a circle whose centre is at the origin and whose radius is 2 units?
7. Can you describe geometrically all the points which obey the algebraic condition x 2 + y 2 = 9?
8. Draw a diagram and shade in the region where: (a) all points obey the inequality x 2 + y 2 < 4. (b) all points obey x2 + y2 < 16 andy > x.
Remember: A locus is the path traced out by a point as it moves in obedience to some given law or condition.
Study these examples which show the idea.
Example (i): A point P(x, y) moves so it is always equidistant from the points A( -1, 6) and B(3, 2). Find the equation of its path (locus).
___. ~P(x,y)
// \ / \
0
Solution: We are looking for the condition satisfied by all the points where P A = P B.
Now P A = J (x + 1)2 + (y - 6)2 and P B = J (x - 3)2 + (y - 2)2
.. So the condition P A = P B becomes J(x + 1)2 + (y - 6)2 = ),-(x---3-)2;;--+-(_y_--2)-oo2
which gives (x + 1)2 + (y - 6)2 = (x - 3)2 + (y - 2)2
i.e. x 2 + 2x + 1 + y 2 - 12y + 36 = x 2
- 6x + 9 + y 2 - 4y + 4
which simplifies to 8x - 8y + 24 = 0 or x- y + 3 = 0
The path is the line x - y + 3 = 0 (or y = x + 3)
\ \ \
\ B(3, 2)
Note: We have found the algebraic condition which is x - y + 3 = 0. We can translate this into more precise geometric terms. This is the equation of a line and it has a gradient of 1. It can be shown that the line joining A and B has a gradient of -1 so we could go further and prove that the locus is a line perpendicular to AB. Actually we could show it is the perpendicular bisector of AB (see Exercise 17.2 below).
Example (ii): Find the locus of a point which is always 5 units distant from the point A(3, 6). Note: As the path of such a point is obviously a circle this question is also asking for the equation of a circle whose centre is (3, 6) and whose radius is 5 units.
87
Solution: Let the point be P(x, y) as shown.
Now PA = -J(x- 3)2 + (y- 6)2
The condition P A = 5 becomes -J (x - 3)2 + (y - 6)2 = 5 or (x - 3)2 + (y - 6)2 = 25 The answer may be left in this form, or alternatively simplified as follows
x 2 - 6x + 9 + y 2
- 12y + 36 = 25 i.e. x 2
- 6x + y2 - 12y + 20 = 0
/ /
A(3, 6)
<-, / /
/ /
/
,. P(x, y) /
1. Find the locus equation of a point P which moves so as to be equidistant from the two fixed points A(1, 2) and B(7, 4).
2. Find the equation of the locus of a point that moves so that it is always equidistant from the points A( -1, 6) and B(5, -4). Show that the path of this point is a line and show that this line is perpendicular to the line AB. Also show that the midpoint of AB is on the locus and hence that the locus is the perpendicular bisector of AB.
3. Find the equation of a circle whose centre is at the point ( -1, 6) and whose radius is 3 units. Note: this question asks for the locus of a point whose distance from (1, 6) is always 3 units.
4. Find the equation of a circle, centre (0, 2) and radius 2 units. Should this circle pass through the origin? Check by substitution in the locus that this is so.
5. A point moves so that it is always equidistant from the points (- 6, 0) and (6, 0). Show that the locus is they-axis.
6. The point P(x, y) moves so that its distance PAfrom the point A(1, 5) is always twice its distance P B from the point B( 4, -1 ). Show that the equation of the locus is x2 + y 2
- lOx+ 6y + 14 = 0.
7. Find the algebraic condition obeyed by all points P(x, y) which are always three times further from the point ( -2, -3) as from the point (4, -1).
8. Find the expression for the distance PA if A is the point ( -3, 4) and Pis any point (x, y).
Show that the locus of the point P(x, y) which moves so that its distance PM from the x-axis is equal to its distance P A from the point (- 3, 4) is 8y = x 2 + 6x + 25.
A __... ,./
(-3, 4)
0
__... -1 P(x, y)
__... I
I I I I
9. Find the equation of the locus of a point which moves so that its distance from they-axis is equal to its distance from the point (2, 3).
88
10. What is the equation satisfied by all points whose distance from the line y = -2 is the same as their distance from the point (0, 2)?
11. (a) What is the expression for the perpendicular distance of the point P(x, y) from the line y = x?
(b) What is the equation of the locus of a point which moves so that its perpendicular distance from the line y = x is equal, to its perpendicular distance from the x-axis? P(x, y)
12. A( -6, 4), B(4, 4) and C(O, 0) are the vertices of a triangle. Find the equations of the perpendicular bisectors of each of the sides of this triangle. Find the point where the perpendicular bisectors of AB and AC meet and show that the perpendicular bisector of BC also passes through this point.
If the point where the perpendicular bisectors meet is called R, find the length of RA and find the equation of the circle with centre Rand radius RA.
Show by substitution that points B and C lie on this circle. Note: This circle is called the circumcircle of the triangle ABC.
13. A and Bare the two fixed points (0, 4) and (5, 0). Pis a variable point (x, y). (a) Find expressions for the gradients PA and PB. (b) Express algebraically the fact that PA and PB are
perpendicular. (c) Show in this way that the locus of a point P which
moves so that the angle LAP B is a right angle is x2 + y 2
- Sx - 4y = 0.
14. A point P(x, y) moves so that it is always further from the point (- 2, 4) than from the point (6, -1). Find the inequality which expresses this condition. Draw a rough sketch and shade the region in which the point P must lie.
We use the ideas oflocus to find the equation of the general circle. Suppose Pis a point (x, y) and Cis a fixed point (a, b)
and the distance PC is r units. We proceed to find the locus of P as it moves so that PC is always r units. That is, we will find the equation of the circle whose centre is at (a, b) and whose radius is r units.
Proof: PC= .j(x - a)2 + (y - b) 2
The condition PC = r becomes .j(x - a)2 + (y - b) 2 = r or (x - a) 2 + (y - b) 2 = r 2
89
P(x, y)
C(a, b)
0
This is called the central form of the equation of the circle, centre (a, b), radius r. Expanding: x 2
- 2ax + a2 + y 2 - 2bx + b2 = r2
or x 2 - 2ax + y 2
- 2by + C = 0 where C = a2 + b2 - r2
This is called the general form of the equation of a circle. Notice that there is a connection between the coefficients of x andy and the coordinates of the centre.
(x - a)2 + (y - b)2 = r2 is the central form for the equation of a circle with centre (a, b) and radius r.
x 2 + y 2 - 2fx - 2gy + C = 0 is the general form for the equation of a circle where the
centre is at the point (J, g).
Example (i): What is the equation of a circle whose centre is at (3, - 2) and whose radius is 4 units?
Solution: A general circle is (x - a)2 + (y - b)2 = r 2
In this case this becomes (x - 3)2 + (y + 2)2 = 16 The answer may be left in this form or expanded as
x 2 - 6x + 9 + y 2 + 4y + 4 = 16
i.e. x 2 - 6x + y 2 + 4y - 3 = 0
Example (ii): Show that the equation x 2 + y 2 + lOx - 8y + 5 = 0 represents a circle and find its centre and radius.
Solution: We proceed to complete the squares as follows
x 2 + y 2 + lOx - 8y + 5 = 0 becomes (x2 + lOx+ D) + (y2
- 8y + D) = -5 Now adding 25 and 16 to both sides to complete the perfect squares, we have,
(x2 + lOx+ 25) + (y2 - 8y + 16) = -5 + 25 + 16
or (x + 5)2 + (y - 4)2 = 36 This is the equation of a circle with centre (- 5, 4) and radius 6 units. Note: Remember that to complete the square of an expression like (x2 + lOx + D) it is necessary to add the square of half the coefficient of x.
Example (iii): Shade the region where the following inequalities both hold
x 2 + y 2 - 6x + 4y < 3 and y ;?! 0.
Solution: We consider x 2 + y 2
- 6x + 4y = 3 i.e. (x2
- 6x + D) + (y 2 + 4y + D) = 3 i.e. (x2
- 6x + 9) + (y2 + 4y + 4) = 3 + 9 + 4 i.e. (x - 3)2 + (y + 2)2 = 16 This is the equation of a circle whose centre is (3, - 2) and whose radius is 4. Now consider the inequality x 2 + y 2
- 6x + 4y < 3. Substitute the point (3, 0) which is obviously inside the circle
x 2 + y2 - 6x + 4y = 9 + 0 - 18 + 0
= -9 which is less than 3.
90
I \ \
' ' ' \ \ I \
• (3, -2) 1
/ ...... - ---,/
I I
/
I
Hence the point (3, 0) satisfies the inequality. So the inequality represents points inside the circle. Because y ?: 0 also the required region is those points inside the circle and on and above the x-axis
as shown.
1. What is the equation of a circle with centre (3, - 5) and radius 3 units?
2. What is the centre and radius for each of these circles? (a) (x - 2) 2 + (y - 5)2 = 9 (c) (x + 4)2 + (y - 1)2 = 4 (b) (x + 1)2 + (y + 3)2 = 16 (d) x 2 + (y - 6)2 = 36
3. What is the equation of each circle? (a) centre at origin, radius 3 units (b) centre at (1, 4), radius 5 units
(c) centre at ( -3, -2), radius 1 unit (d) centre at (4, 0), radius 4 units
4. Find the equation of a circle with centre at the point (- 1, 6) and radius 5 units. Show that this circle passes through the point (2, 2).
5. Show that the points where the circle, whose centre is (2, 4) and radius is 6 units, cuts they-axis are (0, 4 + 4.j2) and (0, 4 - 4J2).
6. It is obvious geometrically that the circle with centre at (2, 4) and radius 3 units will not cut or touch the x-axis. Show algebraically that the circle and the line y = 0 have no common points.
7. By completing the squares, find centre and radius for the circles whose equations are: (a) x 2 + y 2
- 2x + 6y + 1 = 0 and (b) x 2 + y 2 + 4x + 8y + 11 = 0
8. Sketch the graph of the equation x 2 + y 2 - 2x + 4y - 4 = 0. On the same axes sketch the
line y = 2x - 4. Show algebraically that the line passes through the centre of the circle.
9. Referring to the graph drawn in question 8, shade the region where x 2 + y 2 - 2x + 4y < 4. Also shade the common region defined by x 2 + y 2
- 2x + 4y < 4 and y > 2x - 4.
So far we have met the parabola as the graph which results from equations of the form y = ax2 + bx + c. In this section we will attempt to derive the equation of a parabola from its basic definition as a particular locus.
Definition: A parabola is the path traced out by a point which moves so that its distance from a given point (later to be called the focus) is equal to its perpendicular distance from a given line (later to be called the directrix).
IfF is a given point and if AB is a given line then the locus of all points P which are such that FP = PM will be a parabola.
91
F
A B
Let the directrix AB be the line y = -a (where a is any constant).
Let the focus F be the fixed point (0, a) Consider any point P(x, y).
The condition P F = PM means )(x - 0)2 + (y - a) 2 = y + a squaring x 2 + (y - a)2 = (y + a) 2
i.e. x 2 + y 2 - 2ay + a2 = y 2 + 2ay + a2
or x 2 = 4ay.
i I
I dM A l~a
I
I
The equation of a parabola with focus at (0, a) and directrix y = -a is x 2 = 4ay. Since there is no constant term, this parabola passes through the origin.
B
Note that if a = t this is the familiar curve y = x 2. If a has other values the parabolas will be of
differing curvature but all with vertex at the origin.
The parabola is an important curve in mathematics and in real life. 1. If a cone is cut by a plane parallel to one sloping edge
the section will be the shape of a parabola. The parabola is one of the so called conic sections.
2. It can be proved that ifF is the focus of a parabola and PT is a tangent at any point P and if PN is a normal (i.e. perpendicular to the tangent PT) and if PR is parallel to they-axis then the angles LFPN and LNPR (marked a and b) are equal. This means that if light emanates from a source at the focus it will, on reflection, make a parallel beam to the y-axis. The parabolic shape is used in car headlights and in searchlights.
92
T R
3. A chain hanging freely forms a shape called a catenary. But if a rope or chain supports a load (as in a suspension bridge) where the load is uniformly distributed in a horizontal direction then the shape will be a parabola.
Can you find further applications of the parabolic shape?
Example (i): What is the focus and directrix of (a) y = 2x2
Solution:
(b) y = -ixz
The standard parabola is x 2 = 4ay and has focus (0, a) and directrix y = -a. (a) y = 2x2 (b) y = -ix2
Hence x 2 = ~ 2
So 4a = i a=i
Focus = (0, i) Directrix is y = - i
1
Hence x 2 = -2y 4a = -2 a= -i
Focus = (0, -i) Directrix is y = t
I v=! --------r--------I
"'-~-"""-"~~~,~~'
t F ~~~~~~~ -~-~ I o ------r -- ----v-==~
x 2 = 4ay is concave up if a is positive.
The parabola is concave down if (a) a is negative or (b) if x 2 = -4ay (a positive)
Example (ii): A parabola has its vertex at the origin and its focus at the point (0, 2). Find its equation and the equation of its directrix.
Solution: A parabola with this orientation has the standard equation x 2 = 4ay.
The focus of such a parabola is at the point (0, a). So in this case a= 2.
Thus the equation is x 2 = 8y and the directrix is the line y = -2.
93
F(0,2)
X
l. A parabola has the equation x 2 = 4y. Where is its focus and directrix?
2. What is the focus and directrix for each of these parabolas? (a) x 2 = 8y (c) x 2 = 2y (b)x2 =-16y (d)y=x2
(e) y = 4x2
(f) y = -2x2
3. What is the focal length ofthe parabolay = !x2 ? Where is the focus?
4. A parabola has its vertex at the origin and its focus at F(O, -4). What is its equation?
5. In the formula x 2 = 4ay, what distance does the constant a represent? Find the equation of a parabola whose vertex is at the origin, whose axis of symmetry is the y-axis, and whose focal length is!. (The parabola is concave up.)
6. Find the equation of the following parabolas whose vertices are at the origin. (a) Directrixisy = -2 (c) Focusat(O, -2) (b) Focus at (0, 1) (d) Directrix is y = 4
Example (iii): A parabola is drawn with its axis parallel to they-axis and with its vertex at the point (2, 4) and focus at the point (2, 7). Find its equation.
Solution: 1-..
The distance from focus to vertex is equal to the distance from the vertex to the directrix.
: "-....,.., F(2, 7) I 'e I I
Thus the directrix is the line y = 1. M: 1V(2. 4) Y=1
Let P(x, y) be any point on the curve and draw PM perpendicular to the directrix.
...J._---+-----1 I I
By definition of a parabola PM = PFr------,:-------;c (y - 1) = )(x - 2) 2 + (y - 7) 2
By squaring each side (y - 1)2 = (x - 2) 2 + (y - 7) 2
J
y 2 - 2y + 1 = x 2
- 4x + 4 + y 2 - 14y + 49
12y = x 2 - 4x + 52
or y = /2 (x2 - 4x + 52).
Example (iv): What is the minimum value and the line of symmetry of the parabola y = x 2
- 4x + 8.
Solution: Method A
y = x 2- 4x + 8
dy -=2x-4 dx
dy = 0 when x = 2 dx when x = 2, y = 4 - 8 + 8
=4 Thus minimum value is 4. The axis of symmetry is x = 2.
Note: We have used our knowledge of the symmetry properties of the curve and that the turning point must be a minimum because the parabola is concave up. The turning point could be tested by normal means.
Method B y = x 2
- 4x + 8 we complete the square as follows
y = (x2 - 4x) + 8
= (x2 - 4x + 4) + 8 - 4
= (x- 2)2 + 4 The minimum value is 4. The axis of symmetry is x = 2. Because the perfect square (x - 2)2 cannot be negative. And its least value occurs when x = 2.
94
1. A parabola has its vertex at the point (3, 1) and focus at the point (3, 3). (a) What is the focal length? (b) What is the equation of the directrix? (c) What is the equation of the parabola?
2. A parabola has the line y = - 3 as its directrix and the point (0, 1) as its focus. (a) Where is the vertex? (b) What is the equation of the parabola?
3. Find the equation of these parabolas: (a) Focus at ( -1, 6) and vertex at (-1, 4). (b) Focus at (0, 3) and directrix y = -1.
' (c) Focus at (1, 4) and directrix y = 8.
4. Find the least value of the parabola y = x 2 - 6x + 12. Where does it occur?
5. Find the axis of symmetry and maximum value of y = 12 - 4x - x2.
6. Suppose a parabola is drawn sideways; that is, with the x-axis as its axis of symmetry. Also suppose the focal length is a units, that is, the focus is at (a, 0) and the directrix is the line x = -a as shown.
From the definition that the distance of P(x, y) from the focus is equal to the distance from the directrix, show that the equation is y 2 = 4ax.
I I I L,,
~a I I I I I I I I
7. Given a parabola with axis of symmetry along the x-axis and with equation y 2 = 8x find: (a) the focal length and (b) the position of the focus.
8. A parabola is drawn with the x-axis as its axis of symmetry and with its focus at (1, 0) and its vertex at the origin. What is its equation and the equation of the directrix?
9. Sketch the graph x = y2 marking its focus on the graph.
10. Sketch y2 = 2x and indicate the focus.
11. On the graph for question 10 shade the region for which y2 < 2x.
In the questions of the previous sections numerical or specific data was given. In this section we look at quite general proofs.
Example (i): Show by co-ordinate methods that the locus of all points equidistant from two fixed points is the perpendicular bisector of the line joining the two given points.
Solution: This question is intended to be general with the proof applying to any given points. However the proof is still general if the points are orientated in some convenient way with respect to the axes. So let us choose as the two given points, the points (a, 0) and (-a, 0) where of course a can be any given number.
95
Proof: Take two fixed points A( -a, 0) and B(a, 0) and let P be any point (x, y). An equation is required for the condition
PA =PB i.e. PA 2 = PB 2
i.e. (x + a) 2 + y2 = (x - a)2 + y2
i.e. x2 + 2ax + a2 + y2 = x2 - 2ax + a2 + y2
Thus 4ax = 0 or x = 0 The locus is they-axis which is perpendicular to AB and passes through the mid-point of AB.
Example (ii): Find the locus of the point P(x, y) which moves so that the lines P A and P B joining P to two fixed points A and B are perpendicular at P.
Solution: Let A and B be the points (-a, 0) and (a, 0).
Gradient of PA = ( y ) x+a
Gradient of PB (say m2) = ( y ) x-a
As these lines are perpendicular m1 x m 2 = -1
i.e. y x y = -1 (x +a) (x- a)
y2 2 2 = -1
x -a y2 = a2 _ x2
x2 + y2 = a2
/ /
/
(-a, 0)
/ /
/
l
/
0
P(x, y)
"" / \ \
\ \ \
\ B
(a, 0)
This is the equation of a circle with centre at the mid-point of AB (the origin) and radius a. That is, it is a circle on AB as diameter.
1. Take A and Bas the points (0, b) and (0, -b) on they-axis and repeat Example (i) proving that in this case the locus of points equidistant from A and B is the x-axis and hence the perpendicular bisector of AB.
2. Take the origin and the point A(2a, 0) as two fixed points. Assume P is a general point which moves so that always PA l_ PO. Show that the locus of Pis a circle on OA as diameter.
96
3. A and Bare two fixed points and a point P(x, y) moves so that the distance PAis always double the distance P B. Choose suitable points for A and B and find the equation of the locus proving in consequence that the locus is a circle whose centre is on the line AB.
Hint: The final equation and all the algebra may be simpler if the origin (0, 0) is on the locus. Remember that P A = 2P B so think where it might be best to place A and B so that the origin is one point of the locus.
4. Show that the locus of a point, which moves so as to always be three times further from one fixed point than from another fixed point, is a circle.
97
The area under the curve y = f(x) from x =a to x = b is given by fbt(x) dx.
a
CHAPTER 18
In the early history of calculus there were two fundamental problems: the gradient of the tangent to a curve y = f(x) at a point, which has already been studied as differentiation, and the calculation of the area of a region under a curve y = f(x). This latter problem has been called the operation of Integration. Historically, this problem was solved much earlier than that of differentiation.
Consider now a function y = f(x) which is continuous v and above the x-axis from x = a to x = b. Thus f(x) is defined for all values of x in the interval a ~ x ~ b. The problem is to find the area under the curve, above the x-axis and between the ordinates x = a and x = b.
Now f(x), in the interval a ~ x ~ b, has a minimum value m and a maximum value M. These may occur at the end points of the interval as in figure 18-1 where m = f(a) and M = f(b) or they may occur within the interval as in figure 18-2.
y
v=f(x)
F 1-I I I
D
m
J A 0 a b X
Figure 18-1
In both cases the area of the rectangle ABCD = AD . AB = m(b- a)
and the area of the rectangle ABEF = BE . AB = M(b- a)
and it is clear that :
0
y
M
D
A
0 a
I f(a) 1
a
Figure 18-2
the area of the rectangle ABCD < the area under the curve < area of rectangle ABEF . . Thus m(b- a)< A< M(b- a).
A
: f(b)
I I
b X
X
Now divide the interval from x = a to x = b into three equal sub-intervals, each of width b ~ a.
100
y v=f(x) y
0 X 0
In both diagrams, for each of these sub-intervals,J(x) has a minimum value, say m1 , m2 , m3 and the rectangles formed with these values as heights lie below the curve y = J(x) and have areas m 1(b- a) m2(b- a) m3(b- a)
3 3 ' 3 Again, for each sub-interval,f(x) has a maximum value M1 , M 2 , M3 and rectangles formed with
these values as heights have areas M 1(b- a) M2(b- a) M3(b- a) 3 ' 3 ' 3 °
It can be seen that : the sum of the areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles.
Thus (m 1 + m2 + m3)(b ~a)< A< (M1 + M2 + M3)b ~a.
Discussion Example: Now let us consider the area bounded by the curve y = x 2
, the x-axis and the ordinates x = 0 and x = 1. We will divide the interval from x = 0 to x = 1 into (a) two equal sub-intervals (b) five equal sub-intervals (c) ten equal sub-intervals and show that the sum of the areas of the lower rectangles and the sum of
the areas of the upper rectangles approach each other more closely as n increases.
(a) Two sub-intervals 1 - 0
Width of each rectangle = -2- = 0·5.
The minimum values are m1 = f(O) = 02
and m2 = /(0·5) = 0·5 2•
Thus the sum of the areas of the lower rectangles is 0·5[02 + (0·5)2
].
The maximum values are M1 = /(0·5) = 0·52 and M2 = f(l) = 12
•
Thus the sum of the areas of the upper rectangles is 0·5[(0·5)2 + 12
].
The sum of the areas of the lower rectangles < area under the curve < the sum of the areas of the upper rectangles. Thus 0·5[02 + (0·5) 2
] < A < 0·5[(0·5)2 + 12]
i.e. 0·125 < A < 0·625.
101
y
0.5 X 0
X
(b) Five sub-intervals 1 - 0
Width of each rectangle = -5- = 0·2.
The heights of the lower rectangles are f(O), /(·2),/(·4),/(·6),/(·8). The heights of the upper rectangles are /(·2), f(·4),f(·6),f(·8),f(l) and since: the sum of areas of lower rectangles < A < sum of areas of upper rectangles ·2{02 + (·2)2 + ('4) 2 + (·6) 2 + ('8) 2
} < A < ·2 { ('2) 2 + (·4) 2 + (·6) 2 + ('8)2 + 12
} and by calculator 0·24 < A < 0·44.
(c) Ten sub-intervals 1 - 0
Width of each rectangle = -w = 0·1.
y v=x 2
0 .2 .4 .6 .8 X
Thus ·1 {02 + (·1)2 + (·2)2 + (·3)2 + (-4)2 + (·5)2 + (·6)2 + (·7)2 + (·8)2 + ('9) 2} < A <
·1{(·1)2 + (·2)2 + (·3)2 + (-4)2 + ('5)2 + (·6)2 + (·7)2 + ('8)2 + (·9)2 + 12} and by calculator
0·285 < A < 0·385. Hence as n increases the bounding values approximate more closely to each other and each
approximates more closely to A, the area under the curve. If 20 sub-intervals are taken we find the bounds are 0·30875 < A < 0·35875. Note an approximation for A could be found by taking the average of these two bounds.
That is A = 0·30875 + 0·35875 . 2
=i= 0·33375. This value compares favourably with the exact value for the area which will later be shown to
be 0·3.
(i) For the function y = x 3 divide the interval from x = 0 to x = 1 into: (a) two equal sub-intervals (b) five equal sub-intervals (c) ten equal sub-intervals and form lower and
upper rectangles on these sub-intervals as in the discussion example above.
(ii) Show that the sum of the areas of the lower rectangles and the sum of the areas of the upper rectangles approach each other more closely as n increases.
(iii) Find a close approximation for the area under the curve from x = 0 to x = 1 by averaging the lower and upper bounds in part (c).
102
y
X
Suppose now that the interval a ~ x ~ b is divided into n equal sub-intervals, where n is a large number and the upper and lower rectangles drawn.
y
y=f(x)
0 a >< >< >< >< >< >< >< b X
<l <l <l <l <l <l <l b=a+n!.x + N M <t M' N' + + ...
"' + I -I I
"' "' "' .s .s 1: ....-+ + + "' "' "'
The width of each rectangle is b - a, Since n is large b - a is small and we let b - a be ~x. Now n n n
the heights of the lower rectangles are: m1 , m2 , m3 , ••. , m,.
and the heights of the upper rectangles are: Mu M 2 , M 3 , .•. , Mil'
Thus (m 1 + m 2 + m3 + · ·. + m,.)~x <A < (M1 + M 2 + M 3 + ... + M,,)~x As n increases :
(i) the width of each rectangle ~x becomes very small, (ii) the sum of the areas of the rectangles becomes very close to A, the area under the curve
and (iii) m 1 =i= M 1 =i= f(a) m 2 =i= Mz =i= f(a + ~x) m3 =i= M3 =i= f(a + 2~x)
m,. =i= M,, =i= f[a + (n - l)~x] = f(b - ~x). This process may be continued so that as n -+ oo, ~x -+ 0 and the sum of the rectangles -+A. This may be written as
A = lim {f(a) + f(a + ~x) + f(a + 2~x) + · · · + f(b - ~x)}~x t.x->0
b-t.x
= lim I f(x)~x L\x-tO x;:::;a
A special symbol has been introduced to represent this operation. Using this symbol the area under the curve is written as :
A =I: f(x)dx, which reads as:
the definite integral off(x) as x goes from the lower limit x = a to the upper limit x = b. This integral represents the area enclosed by the x-axis, the ordinates at x = a, x = b and the
curve. The integral symbol is an old form of S, the first letter of the word sum, and the word integrate means to find the total sum.
Integration is an operation which involves the formation of a sum and the taking of its limit.
103
The area enclosed by the x-axis the ordinates x = a, x = b and the curve is given by:
A= ff(x)dx.
The process of approximating the area under a curve by taking the sum of the areas of a large number of rectangles is very tedious, as we have found earlier when calculating the area under the curve y = x 2 between x = 0 and x = 1.
Fortunately there is a relationship between the definite integral and the primitive function which leads to simple method of finding the area under a curve and this method yields an exact result.
Consider now a function y = f(x) where f(x) is: (a) continuous (b) above the x-axis (i.e. a positive curve) for all values of x > a where a is a constant. Let L be on the x-axis at x = a. Choose M on the x-axis with abscissa x and let the area LMQP =
A(x). N is on the x-axis at (x + Llx) so that MN =
Llx and the area MNRQ is LlA(x) which lies between the area of the rectangle MNTQ and the area of the rectangle MNRS. That is MQ . MN < LlA(x) < NR. MN
f(x) . Llx < LlA(x) < f(x + Llx) . Llx LlA(x)
Hence f(x) < AX < f(x + Llx)
As Llx ~ O,J(x + Llx) ~ f(x).
Hence LlA(x) ~ f(x) as Llx ~ 0. Llx
Thus lim LlA(x) = dA(x) = f(x). Ax-->O Llx dx
y
A(x)
L M 0 a X
)(
<l + )(
N
X
Hence if F(x) is a known primitive ofj(x) [i.e. F'(x) = f(x)] then A(x) = F(x) + C for some value of the constant C. To find the value of C we note that since A(x) is measured from x = a.
A(a) = 0 also since A(x) = F(x) + C then A(a) = F(a) + C i.e. 0 = F(a) + C thus C = -F(a) Hence A(x) = F(x) - F(a) As this is true for all values of x where x > a
A(b) = F(b) - F(a) since b > a.
y
v=f(x)
A(b)
0 a b X
From the previous section in this chapter, the area under the curve y = f(x), above the x-axis and between the ordinates from x = a to x = b was defined as the definite integral off(x)from x = a to x = b and this was written as
104
A= Ib f(x)dx
and so A = Ib f(x)dx
= F(b) - F(a) where F(x) is a primitive function ofj(x).
The area enclosed by the curve y = f(x), the x-axis and the ordinates x = a and x = b is given by
A = J: f(x)dx = F(b) - F(a)
where F(x) is a primitive function ofj(x).
Thus, if the primitive function of a given function f(x) can be found the definite integral can be evaluated and the area under the curve found. Basically this is what is known as the Fundamental Theorem of the Calculus. It links together the summation process as used by Archimedes some 2000 years ago with the operation of differentiation.
Example (i): y
Find the area under the curve y = x 2 above the x-axis from X= 0 to X= 1.
Solution:
A= Ib f(x)dx
3
A primitive ofj(x) = x 2 is F(x) = ~
A= fol xzdx
= F(1)- F(O) = ~- 0 = 0·3.
For brevity this may be set out as follows
A = fol xzdx
0
{
n+l
Since in general the primitive of x" is nx + 1
+ C
X
= [~]~ =~-0 _1_
l We now find the value of the function in the brackets at the upper limit x = 1 and subtract the value at the lower limit x = 0. This is equivalent to F(1) - F(O).
- 3·
The area is ~ of a square unit.
105
Example (ii):
Evaluate the definite integral I3
2x3dx.
Solution: I3
2x3dx = [~4
]: 0ince ~4
is a primitive of 2x3)
= (~4)- (~4) = 40·5 - 0·5 = 40.
l. (a) Find the area under the curve y = x 3, above the
x-axis from x = 0 to x = l. (b) Compare your answer with the approximation
found in Exercise 18.1.
2. Find the shaded area below the curve y = !x2
between the ordinates x = 1 and x = 3.
y
v=x3
y
0 3
3. Find the areas under the following curves, above the x-axis and between the given ordinates. (a) y = x 2
; x = 0 to x = 2 (f) y = 3x; x = 1 to x = 5 (b) y = 4x3 ;x = Otox = 1 (g) y = JX;x = Otox = 1 (c) y = 2x; x = 1 to x = 3 (h) y = 2x2
; x = 1 to x = 3 (d) y = 5x4
; x = 0 to x = 1 (i) y = -jx2; x = 1 to x = 3
(e) y = 2x3; x = 2 to x = 3 (j) y = -jJX; x = 0 to x = 4
4. Evaluate the following definite integrals.
(a) I3
x 2 dx (d) J: 3dx
(b) I2
3x2dx (e) J~1 x4
dx
(c) I3
4xdx (f) {1
x- 2dx 2
106
J2 1
(g) 3 dx 1 X
J4 3
(h) 1
x 2dx
J10 J10
(i) 1
dx = 1
1dx
X
X
5. Calculate the exact area bounded by the arc of y = x2, the x-axis and the ordinate x = 3.
6. Find the area enclosed by the curve y = xt, the x-axis and the line x = 9.
7. The shaded area is equal to I:1
-x3dx. Find v= -x
3
v
its value.
-1 X
8. Find the areas given by the following definite integrals.
(a) A = J- 2
-3xdx (b) A = Io -~3
dx -5 -2
9. Show that 11
x2dx = J: t
2dt = ~·
10. (a) Iff 2xdx = 15, find the value of k, given that k > 0.
f"' dx (b) If
1
JX = 6, find the value of m.
We have found that a close connection exists between the definite integral and the primitive function. That is: lb f(x) = [F(x)]~ where F(x) is a primitive ofj(x).
If F(x) is a primitive ofj(x) then the general primitive ofj(x) is F(x) + C. It is convenient to have a notation for this process of finding the primitive function.
We shall denote the general primitive ofj(x) with respect to x by the symbols I f(x)dx, using the
integral sign without the upper and lower limits.
We write I f(x)dx = F(x) + C.
This is called the indefinite integral ofj(x) with respect to x.
IMPORTANT RESULTS
1. x"dx = -- + C I x"+1
n + 1
2. I kx"dx = k I x"dx
3. I dx = I 1dx = x + C
4. I {f(x) ± g(x) }dx = I f(x)dx ± I g(x)dx
107
Example (i):
Find J (x 3 - 2x2 + 1)dx.
Solution:
J (x3
- 2x2 + 1)dx = J x 3
dx- 2 J x 2dx + J dx
x4 2x3
4 3 +X+ C.
Note: Only one constant is needed as it represents the sum of the individual constants.
Example (ii): Find the indefinite integral of x t + x -t.
Solution:
J (x! + x-!)dx = J x!dx + J x-!dx
x!+l x-!+1 = -1--1 + . 1 1 + c
2 + -2 + 2 3 1
= 3x 2 + 2x 2 +C.
Example (iii):
Find J .jX(7x2 + 5x - 3)dx.
Solution:
J .jX(7x2 + 5x - 3)dx = J x1(7x2 + 5x - 3)dx
= J (7xJ + sxt- 3x!)dx
2 ]_ 2 ;_ 2 ~ = 7. 1x 2 + 5. 5x 2
- 3. 3x 2 + C
= 2x1 + 2xt- 2x1 + C = 2x1(x2 + x- 1) + C.
Example (iv):
Find f (3x - 2)(x + 3)dx.
Solution:
J (3x - 2)(x + 3)dx = J (3x2 + 7x - 6)dx
7 = x3 + 2 x2
- 6x + C.
Example (v):
Find-the indefinite integral J x 3
x-; 2
dx.
Solution:
J x3
x-; 2
dx = J ( x - ~2) dx
= x2 +~+C. 2 X
108
L
>
1. Find the indefinite integral of the following functions with respect to x. (a) x7 (d) x 2
- 2 (g) 2[3 (j) x 0·2
(b) 4 (e) 2x3 - 3x2 (h) JX (k) JX3
(c) 9x8 (f) xt (i) ax10 (1) x + x-t
2. Find:
(a) J (4x 2 - 2x)dx
(b) J (2x4 + 3x3)dx
(c) J (x2 - x- 2)dx
(d) I (2x + 1)2dx
(e) J (3x2 - 2x + 1)dx
(f) J (x 5 - 4x 3 + x 2
- x + 1)dx
(g) J (x- 1)(x + 1)dx
(h) I x3 + 2:2 + x
3. Find the following indefinite integrals.
(i) I x 2JXdx
(j) I (3x2 + ~3) dx
(k) I (xt + x-f)dx
(1) I c~ + [1.3) dx
(a) J (12x2
- 4)dx (e) J 2u2(4u
3 + 3u2 -. 1)du . (i) J2du
(b) I 2X
2
; 3
X dx (f) J (21 + ~ y dt , · (j) J v: _=-; dv ~ ~~· (c) J (3t 2
- 2t + S)dt (g) J (x2 - .jX)2 dx (k) I 3x:; hJx ' 1'~'
(d) JU2 + t)dt (h) f 0 + ~)G- ~)dx (1) f (1- jU)(3 + 2;)~u ,- -} __ \ I
4. If: = 3(x2 - 1), find a general solution fo/y.Af y = 1 when x = 1, what is the numerical
value of the constant .of integration?
5. The gradient of a curve is given by: = 2x - 3. What is the equation of the curve if:
(a) it passes through the origin? (b) it passes through (1, - 2)? (c) it passes through (1, 2)?
6. If: = x 2 - 5x + 4 and when x = 1, y = 0, what is the value of y when x = 4?
7. Verify that (a) J xJXdx =/= J xdx. J JXdx
(b) I x(x + 1)dx =/= x . J (x + 1)dx.
A FURTHER IMPORTANT RESULT
Since d~[(ax + b)"+1 + CJ = a(n + 1)(ax +b)".
then (ax+ b)"dx = + C. f (ax+ b)"+1
a(n + 1) /
109
Example (i):
J(2x + 3tdx
Solution:
J(2x + 3)4dx = (2x + 3)
5 + C
2 X 5
= (2x + 3)5 + C
10
Example (ii):
Find J (x !x 2)3.
Solution:
J dx = Jcx + 2)- 3dx
(x + 2)3
= (x + 2)-2 + C 1 X ( -2)
1 2(x + 2)2 + C
Perform the following integrations.
1. J (5x - 1)3dx 4. f (3 :x6x)2
2. J (2x + 4) 5dx 5. J )4x + 1dx
3 J dx · (2x + 5)4
6 J dx · )7x- 1
Example: Evaluate the following definite integrals.
(a) J1
3
(3x2 + 2x + 1)dx (b) l 4
(x + : 2)dx
Solution:
(a) t3
(3x 2 + 2x + 1)dx = [x3 + x 2 + xJi
= (33 + 32 + 3) - (1 3 + 12 + 1) = 36
110
7. J (3 - 2x)5dx
8. J )3- 3xdx
9 J dx · (1 - 2x) 3
1. Evaluate the following definite integrals.
(a) t4
2xdx (d) L (x2 + x + 1)dx
(b) 12
dx (e) f1
(x2 + 2x + 1)dx
(c) 12 (xz + :3 )dx (f) t2 x3 - ~;2 + 1 dx
2. Evaluate:
(a) f :1
3xdx
(b) J4 dx x2
3
(c) t3
4t3dt
(d) f :1
(x3 - 2)dx
(e) t4
(x2 + .jX)dx
(f) J:3 (6- x- x 2)dx
(g) t3
(2x2 - x + 1)dx
(h) l 2
2t(t2 - 1)dt
3. Show that {3
(x2 + x)dx = - {1
(x2 + x)dx.
(i) f :1
(2u + 1)2
du
(j) f ~: (x - 1)(x - 2)dx
f -1( 1)2 (k) _2
r--;: dr
(1) f 1
2
(3x - 2x2 + ~;) dx
4. (a) Evaluate 12
(x2 + x + 1)dx. (b) What is the value of 12
(r2 + r + 1)dr?
5. By evaluation of each side show that f 1
(x2 - 2x)dx = Io (t 2
- 2t)dt + r 1
(u 2 - 2u)du.
-1 -1 J 0
Note: The definite integral is a number and is independent of the variable used.
6. Show by evaluation that:
f2
(3x2 + 2x + 1)dx + f3 (3x2 + 2x + 1)dx = f3
(3x2 + 2x + 1)dx.
1 J1 2 l 7. (a) Evaluate
0
(a + bx + cx2 + dx3)dx.
(b) Evaluatei-{f(O) + 4 .f(!) + f(1)} wheref(x) =a+ bx + cx2 + dx3, and hence show that
i{f(O) + 4. f(!) + f(1)} = I1
f(x)dx.
111
The area enclosed by the curve y = f(x), the x-axis and the ordinates x = a and x = b is given by
A= rf(x)dx or A= r ydx.
Before evaluating a definite integral which gives such an area it is important to sketch the graph of the function because it is essential to know where the curve cuts the x-axis.
Example (i):
y
0
y=f(x}
a b X
Find the area between the curve y = 3 + 2x - x 2 and the x-axis and show that this area is given by I :1
(3 + 2x - x2 )dx. Evaluate this definite integral and show that it is equal to
Jo (3 + 2x- x 2 )dx + f3
(3 + 2x- x 2 )dx. -1 0
Solution: y = 3 + 2x - x 2
= (3 - x)(l + x) The parabola cuts the x-axis at -1 and 3. The required area, A, is given by:
A = r1
(3 + 2x - x 2 )dx
= [3x + x2 - x3] 3 3 -1
= (9 + 9 - 9) - (- 3 + 1 + i) =lOt.
The area is lOt square units.
y
0
Jo f3 [ x3Jo [ x3]3 _
1
(3 + 2x - x 2)dx + 0
(3 + 2x - x 2)dx = 3x + x 2 - 3 _
1 + 3x + x 2
- 3 0
= [0- -(It] + [9- OJ = 10~.
Thus J3
(3 + 2x - x 2)dx = Jo (3 + 2x - x 2)dx + 13
(3 + 2x - x 2)dx. -1 -1 0
This result illustrates the general property:
ff(x)dx = r f(x)dx + J: f(x)dx where a < b < c
112
X
Example (ii): Find the area between the curve y = (x - l)(x - 2) and the x-axis.
Solution: The graph cuts the x-axis at v x = 1 and x = 2. Hence the required area A is given by:
A = r (x- 1)(x- 2)dx = r (x2 - 3x + 2)dx
= [x3 - 3x2 + 2x]z 3 2 1
= (2i- 6 + 4) - (~ - I!+ 2) 1
-6·
The area is i of a square unit. The negative sign indicates that the area is in the negative region, below the x-axis. The measure of an area under the x-axis is given by the absolute value of the definite integral. In this case A = 1-il = f,.
When an area is below the x-axis, then its magnitude is given by
A= I rf(x)dxl.
Example (iii): Find the area between the curve y = x(x + 1)(x - 2), the x-axis and the ordinates x = 1 and x = 3.
Solution: The curve y = x(x + 1)(x - 2) cuts the x-axis at x = -1, 0 and 2. It can be seen that the required shaded area contains one part below and one part above the x-axis. These must be found separately.
A= A1 + A2 = If ydxl + f ydx
= 1 r (x3 - X
2 - 2x)dx 1 + f (x3
_ x 2 _ 2x)dx
= I [~4 _ ~3 _ x2I I + [ ~4 _ ~3 _ x 2 J:
y
= IC4 - 2i - 4) - ct- ~- 1)1 + c2ot - 9 - 9) - (4 - 2i- 4)
= 1- 1 / 2 1 + 4 g = 6!.
X
Note: It is clear that r ydx would not give the total area A 1 + A2 • The value of r ydx is in fact
3~ which would give the difference between the two areas A1 and A2 •
113
1. Sketch the graph of y = x 2 - 1 and find the area:
(a) bounded by the graph, the x-axis and the ordinates x = 2 and x = 3. (b) enclosed by the graph below the x-axis.
2. Find the areas enclosed by the following parabolas and the x-axis. Draw a sketch in each case. (a) y = x 2
- 16 (b) y = 4 - x 2 (c) y = x 2 - 5x + 6 (d) y = (2 - x)(l + x)
3. Find the area bounded by the graph of each of the following functions, the x-axis and the given ordinates. Sketch the graph in each case. (a) f(x) = x2
- 4; x = 2 to x = 3 (b) f(x) = x3
; x = -2 to x = -1 (c) f(x) = x2
- 3x + 2; x = 2 to x = 4
4. (a) Evaluate J :1
x 3 dx.
(d) f(x) = x 2- x- 2; x = -3 to x = -1
(e) f(x) = 2 - x - x 2; x = 0 to x = 2
(f) f(x) = x(x + l )(x + 2); x = -1 to x = 0
(b) Find the area bounded by the graph of y = x 3, the x-axis and the ordinates x = - 1 and
X= 2.
5. Find the area bounded by the graph of each of the following functions, the x-axis and the given ordinates. (a) y = x 2
- 4; x = 0 to x = 3 (b) y = x 2
- 2x - 3 ; x = - 1 to x = 5
6. It can be seen by symmetry that areas A 1 and A2 are equal, so if we require the area A1 + A2
its magnitude can be found by evaluating either
2 Jo x 2 dx or 2 l\2 dx. -2 Jo
Find the shaded area A1 + A2 .
(c) y = x 3; x = -1 to x = 1
(d) y = x(x2 - 1); x = -l to x = 2
y
A,
-2 0 2
Note: All even functions are symmetrical about the y-axis and hence this method can be used for even functions if the required area is also symmetrical about they-axis.
7. Show that the area under the curve y = x 2 + 1, above the x-axis and between the ordinates
x = -4 and x = 4is given ~y A = 2 I4
(x2 + 1)dx and find this area.
8. Find the area enclosed by the curve y = 25 - x 2 and the x-axis.
9. y = x 3 is an odd function and thus has point symmetry about the origin. Hence the areas A1 and A2 are equal and the magnitude of the area A1 + A2 can be found by evaluating:
2 I2
x3dx
Find the shaded area A1 + A2 •
114
-2
A,
y
2 X
10. Draw a sketch of the curve y = x(x - l)(x + 1) and show that the area enclosed by the curve
and the x-axis is given by A = 2 J 0
(x3 - x)dx and find this area.
-1
11. Find the area bounded by the curve y = 2x3, the x-axis and the ordinates x = - 1 and x = 1.
12. Show that the x-axis and the curve y = 4x(x - l)(x - 2) encloses two regions. Find their areas.
13. Find the area bounded by the graph y = 4x2 - 1, the x-axis and the ordinates x = -! and
x=l.
14. Find the area enclosed by the arc of the curve y = x 2 - x 3 and the x-axis.
15 .. Find the area enclosed by the curve y = x4, the x-axis and the ordinates x = -3 and x = 3. . '
The area between the curve y = g(x), they-axis and between y = a, y = b may also be found.
y
y=g(i<)
b 1----------/
A
0
Example:
X
The first step is to find x as a function of y, say x = f(y), then it follows using the pattern established for the x-axis:
The area between the curve, they-axis and the abscissae y = a and y = b is given by:
A= rf(y)dy or A= r xdy
Find the area bounded by y = x3, they-axis andy = 0 andy = 8.
Solution: y
8 f------1
X
The area is 12 square units.
As y = x3
X= y1
and A= f f(y)dy
= f ytdy
= [~y~: = ;f(16 - 0) = 12.
115
1. Calculate the area bounded by the graph of y = ix2, the y-axis and the lines y = 0, y = 1,
lying in the first quadrant.
2. Find the area bounded by y2 = 4x, they-axis and the line y = 4.
3. Calculate the area enclosed by the curve y = 4x2, they-axis and the lines y = 1, y = 4, lying
in the first quadrant.
4. Find the area between y = --;, the y-axis and the lines y = 1, y = 2. Leave the answer as a X
surd.
5. What is the area between the curve x = I - y2 and they-axis.
6. The diagram shows the curve y = x 2• Find the shaded y
areas At and A2 and show that At + A2 is 7 square units.
X
7. Calculate the area enclosed between the curve y = JX, they-axis and the lines y = 1 andy = 4.
8. Find the area between the curve y = ~'they-axis and the abscissae y = 1 andy = 3.
y
/
0 a b X
Two curves y = f(x) andy = g(x) intersect on the ordinates x = a and x = b. The problem is to find the area A between these two curves. A= {area under y = g(x)}- {area under y =f(x)}
A = r g(x)dx - f f(x)dx.
which can be written
A = f {g(x)- f(x)}dx.
If it is not obvious which curve lies above the other over the interval, the use of absolute value again ensures a positive area.
Thus A =I r {g(x) - f(x)}dx,.
This result is true for all positions of the curves (above or below the x-axis), provided they do not cross each other between x = a and x = b.
116
Example (i): Find the area of the region bounded by f(x) = x 2 and g(x) = 3x - 2.
Solution: y
X
The area is i of a square unit.
Example (ii):
These curves intersect whenf(x) = g(x). That is when x 2 = 3x - 2
x 2- 3x + 2 = 0
(x - 1)(x - 2) = 0 x = 1 or x = 2
The points of intersection are (1, 1) and (2, 4).
A = I r { (3x - 2) - x2
} dx I
= I ~~2 - 2x - ~] ~I = 1(6 - 4 - !) - (~ - 2 - t)l = IC -~) - C -i)l _1 - 6•
Find the area bounded by the curves y = x andy = x 3.
Solution: y
-1
X
The graphs intersect when x 3 = x x 3
- x = 0 x(x2
- 1) = 0 x(x - 1)(x + 1) = 0 and X = 0, X = 1, X = -1.
Thus the graphs intersect at the points ( -1, -1), (0, 0), (1, 1).
As the two functions are each odd, the two symmetrical areas are equal but of opposite sign.
Hence the required area is given by:
A = 21 f (x - x3)dx I
= 2\[~2- ~4II = 21ct- !)I -1 - 2·
The area bounded by the two curves is ! square unit.
1. Find the area enclosed by the curve y = !x2 and the line y = x.
2. Calculate the area between the curves y = x 3 andy = 4x lying in the first quadrant.
3. Find the area of the region bounded by the graphs of y = x 2 andy = x 3•
4. Find the area bounded by f(x) = 2x - x 2 and g(x) = 2 - x.
5. Find the points of intersection of y = 2x2 and y = 2x + 4. Hence find the area between the two curves.
117
6. (a) Calculate the area bounded by the parabolas, y = 8 - x 2 andy = x 2•
(b) Find also the area between the two curves between x = 0 and x = 2.
7. Calculate the area bounded by the graphs of y = 6 - x 2 andy = 3 - 2x.
8. Find the area bounded by the curves y = JX andy = x 2•
9. Find the area bounded by y = x 2, y = x, x = 1 and x = 2.
10. Find the area between y = x 2 andy = - x 2 from x = 0 to x = 2.
11. Find the area bounded by the graphs of x = y 2 and x = y + 2.
12. Calculate the area bounded by x = yiand x = y2 in the first quadrant.
13. Find the area of the region between the graphs of y = 2x - x 2 and 3y = x 2 - 4x + 6.
14. Find the total area of the region bounded by f(x) = x 3 - x and g(x) = 1 - x 2
•
15. Find the area between the two curves y = 5 andy = x(6 - x).
In the previous section it was possible to find the exact area under a curve by means of the definite integral when the primitive function was known. However, there are some functions whose definite integrals cannot be easily found. In such cases, especially with the advent of the electronic calculator and the computer, numerical methods exist which are capable of giving a very close approximation to the value of the required area. Two such methods are: (a) Trapezoidal Rule. (b) Simpson's Rule.
TRAPEZOIDAL RULE: The method makes use of the formula for calculating the area of a trapezium.
Area of Trapezium = half (sum of lengths of parallel sides) x (distance between them). Consider now a curve y = f(x) defined on the interval from x = a to x = b. At A, x = a and AD = f(a). Y
At B, x =band BC = f(b). If CD is joined then ABCD is a trapezium with area
A = !AB(AD + BC) = !(b- a){f(a) + f(b)}, and
this is an approximation to the area under the curve y = f(x).
fb· 1
Thus, a f(x)dx ~ l(b - a){f(a) + f(b)}
This is the formula for one application of the Trapezoidal Rule.
Example: Find (a) the exact area under y = x 2 from x = 1 to x = 2.
(b) an approximate area using the trapezoidal rule.
118
0
: f(a) A
a
y=f(x)
c .//··1
I
: f(b) I I B
b X
Solution:
(a) f x2dx = [~
3
T =i-1 _]_ - 3•
Area under y = x2 from x = 1 to x = 2 is equal to 21 square units.
(b) Area of trapezium= !(b- a){f(a) + f(b)} = !(2 - 1) {J(l) + /(2)} = !{12 + 22} =!x5 = 2! square units.
This approximation has an error = 2! - 21 _ _l - 6
which as a percentage of the exact area is 7·14%.
y
4
3
2
0 2 X
A closer approximation can be found by dividing the interval into two equal sub-intervals and then applying the trapezoidal rule to each sub-interval.
Divide the interval from x = 1 to x = 2 at x = 1!.
Thus the width of each sub-interval = 2
; 1
_J. - 2•
Required area =i= area A1 + area A2
=!. HJO) + !@} +!.!. UG> + /(2)} = HJC1) + 2 .f@ + /(2)} = H12 + 2@2 + 22} =HI+ 2.£ + 4} ={-X 9.5 = 2·375 square units.
This is a much closer approximation with an error of only 1·79%.
y
4
3
2
0
jl , I
I I I
A2 I I
2 X
In general, if y = f(x) is defined over the interval (a, b), the mid-point is x = a ; b then the width
of each sub-interval ish = b ; a.
The approximate area under the curve is
~. b; a {[f(a) +!(a; b) J +[!(a; b)+ f(b~} = ~h{f(a) + 2 .f(a; b)+ f(b)}.
y
0
I I 1 f(a)
I I
a
v=f(x)
I I I
I : I f(b)
I I f(a+b) I I 2 I I I I I
X
By increasing the number of sub-intervals n, the approximate area will come as close as required to the exact area.
In this case the width of each sub-interval is b-a
h=--11
and the x values for the ordinates are a, a + h, a + 2h, a + 3h, ... , a + (n - l)h, a + nh = b.
119
y v=f(x)
0 a .J: .J: .J: .J: :£. b X
+ N M o:t + + +
,.. =a+nh "' I
"' "' "' c:
+ "'
Area oftrapezia = th[{f(a) + f(a +h)}+ {!(a+ h)+ f(a + 2h)} +{!(a+ 2h) + f(a + 3h)} + ... +{!(a+ (n- 1)h) + f(b)}].
and
fb h
a f(x)dx ~ 2{/(a) + 2f(a +h)+ 2f(a + 2h) + 2f(a + 3h) + ... + 2f[a + (n- 1)h] + f(b)}
b-a whereh = --
n
Example 1:
Find an approximation for I2
x 2dx using ten equal sub-intervals.
Solution: 2 - 1
h = ----w- = 0·1 and the x values for the co-ordinates are 1, 1·1, 1·2, ... , 1·9, 2. r x 2dx ~ 0~1 {!(1) + 2/(1·1) + 2/(1·2) + 2j(1·3) + 2/(1 A) + 2/(1·5) + 2/(1·6) + 2/(1·7)
+ ¥(1·8) + 2/(1·9) + /(2)} = 0·05{1 2 + 2[1·1 2 + 1·22 + 1·32 + 1·42 + 1·52 + 1·62 + 1·72 + 1·82 + 1·92
] + 22}
= 2·335 by calculator.
Example 2:
Find an approximate value for the area bounded by y = .!, the x axis and the ordinates at x = 1, X
x = 2. Use ten strips. (equal sub-intervals).
Solution:
h = 2 ~ 1 = 0·1 with x values of the ordinates 1, 1·1, 1·2, 1·3, ... , 1·9, 2.
A _,__ Qi {..!_ _2_ _2_ _2_ _2_ _2_ _2_ 2 _2_: _2_ ..!_} -;- 2 1 + 1·1 + 1·2 + 1·3 + 1·4 + 1·5 + 1·6 + 1·7 + 1·8 + 1·9 + 2 = 0·6937713 by calculator and the area is approximately 0·694 square units.
Note: The exact area is 0·6931471 square units. -
120
Give all answers correct to three decimal places.
1. (a) Using the trapezoidal rule and one application find an approximation to the area under the curve y = x 2
, above the x axis and between x = 0, x = 1. (b) Find a better approximation to this area by usingfive equal sub-intervals.
2. Find an approximation to the definite integral f x 3dx, using the trapezoidal rule and:
(a) one application (b) five trapezia (c) ten trapezia. What is the exact value?
3. (a) By making five equal sub-intervals and applying the trapezoidal rule to each, find the approximate area bounded by y = tx2
, the x axis and the ordinates at x = 1, x = 2. (b) What is the exact area? (c) Express the error caused by the approximation as a percentage of the exact area.
4. Using the trapezoidal rule and five sub-intervals find approximations for the following definite integrals:
(a) J: (x2 + x)dx
(b) J3
-;dx 1 X
(c) J: 2xdx
f4·3 1
(d) r::. dx 3-8 -yx + 1
5. Use the trapezoidal rule, applied to four strips, to find the approximate area bounded by y = x - x 2 and the x axis.
6. By means of the trapezoidal rule, applied to five equal strips, evaluate the following definite integrals:
(a) r JXdx
(b) 12
sin xdx (make sure your calculator is set to radian measure)
(c) J1 1 dx
o )1 + xz
7. An important use of numerical integration is that it can be used to approximate the definite integral of a function that is given only in tabulated form. (a) In an experiment the following values of x andy = f(x) were recorded:
X 1·0 1·5 2·0 2·5 3·0 3·5 4·0
y = f(x) 3·1 4·0 4·2 3·8 2·9 2·8 2·7
Use the trapezoidal rule and six strips to approximate the definite integral of this function, 14
/(x)dx.
121
(b) During an experiment the following values of an unknown function f(t) were recorded:
t 1·0 1·2 1·4 1·6 1·8 2·0
f(t) 0·5283 0·7281 0·8762 0·9852 0·6829 0·5631
Use the trapezoidal rule to find the approximate value of f f(t)dt.
8. During a survey the area of an irregular headland was to be found. The surveyor used a base line divided into ten equal sub-intervals, each of width five metres. Offset measurements f(x) were taken at each value x across the base line and tabulated as below:
0 5 10 20 30 40 50 X
METRES
x (metres) 0 5 10 15 20 25 30 35 40 45 50
offset (metres) 0 10·2 13·2 16·3 13·8 16·0 17·0 18·6 17·4 10·8 0
Use the trapezoidal rule to find the approximate area of the headland.
SIMPSON'S RULE This rule is another formula for finding the approximate area under a curve. In Simpson's Rule, instead of using a straight line between two points on the curve an arc of a parabola is used and this gives a more accurate approximation.
If y = f(x) is defined on an interval a ::( x ::( b, then the parabola which has an equation of the form y = Ax2 + Bx + Cis made to pass through three points on the curve:
(a,J(a)), (b,f(b)) and the mid-interval value (a; b,J(a; b))
y
0
Now the area under the parabola is approximately equal to the area under the curve y = f(x).
122
The formula for this area is derived in APPENDIX IX and is stated as:
Note: Three function values are required for each application of Simpson's Rule.
Example 1:
Use Simpson's Rule to give an approximation for 12
x 2 dx.
Solution:
A= 12
x2dx
=i= b ~ a {r(a) + 4fe ; b) + f(b)} where a = 1, b = 2,j(x) = x 2
= 2 ~ 1 {f(l) + 4fc ; 2) + f(2)}
= t{12 + 4@2 + 22} = t{1 + 4(£) + 4} =!X 14 = 1 square units which is the exact value.
Simpson's Rule will always lead to an exact value when the given function represents a parabola.
Example 2:
Find the approximate area under the graph of y =_!_from x = 1 to x = 2 by applying Simpson's X
Rule to: (a) the whole interval. (b) two equal sub-intervals.
Solution:
. b-a{ (a+b) } (a) A =;= -6
- f(a) + 4 .f -2
- + f(b) where a = 1, y
1 b = 2;f(x) =-
X
= 2 ~ 1
{J(l) + 4 .f(~) + f(2)}
= 1H + 4 ·l +U = t{1 + i + 0·5} =i= 0·6944 square units by calculator.
0 2 X
123
(b) Take two sub-intervals and
1~x~~ ~~X~ 2.
mid-point of the first sub-interval 1 + ~ 2
_.2_ - 4•
mid-point of the second sub-interval = ~ ~ 2
_]. - 4•
In each case the sub-interval width = b - a =~-I _1 - 2•
Apply Simpson's Rule to each sub-interval:
y
0
A~ ~ {/(1) + 4 .j(~) +!G)}+ ~{!G)+ 4 .j(~) + /(2)}
1 { 1 1} 1 { 1 1 1} = 12 1 + 4 . t + ~ + 12 ~ + 4 . i + 2 = lz {1 + \6 + -t + -t + 176 + !} ~ 0·6932539 units, by calculator.
X
Compared with the results in example 2 in the previous section where the trapezoidal rule was used with ten strips resulting in 0·6937713 while the exact result was 0·6931471, two applications of Simpson's Rule gives a closer approximation.
For use with calculators and computers an extended version of Simpson's Rule is used. A full derivation is given in APPENDIX X. In this formula the interval (a, b) is divided into n equal sub-intervals where n must be even. Let
the points of division be a= x 1, x 2, x 3, x4, ... , xn, xn+l = b with corresponding ordinates
Y1,yz,y3,y4, · · · ,y,,Yn+l·
The width of each sub-interval ish = b - a. n
Simpson's Rule is now applied to each group of three division points, xl,Xz,X3;x3,X4,xs; ··· ;xn-l•Xn,Xn+l·
y
# I I I
I I
I Y1 I y2 Y3 y4 y5 : Yn-1 I Yn I I I
I I I I -----
0 x1 x2 x3 x4 x5 xn-1 x.
=a
124
v=t(x)
: Yn+1
xn+1 X
=b
and the formula used is
. h { A =;= 3 (Yl + Y,+l) + 2(Y3 + Ys + · · · + Y,-1) + 4(Yz + Y4 + + y,)}
This reads
A ~ ~{(sum of end ordinates) + 2(sum of other odd ordinates) + 4(sum of even ordinates)}
It is a very easy formula to apply and increased accuracy is achieved by increasing n, provided n remains even. This means that there is always an odd number of x values and the first and last ordinates are both odd.
Example3:
Use the extended form of Simpson's Rule to find an approximation to J2
.!_dx using: 1 X
(a) two sub-intervals (or three function values) (b) four sub-intervals (or five function values) (c) ten sub-intervals (or eleven function values).
Solution: (a) This corresponds to the previous example, part (a).
2 - 1 1 h=-
2-=0·5; f(x)=:x=y
x1 =1; x2 =1·5; x3 =2
A ~ o~5 { G + ~) + 4 (1 ~ s)} 0·6944444 by calculator.
(b) This is equivalent to the previous example, part (b).
(c)
2- 1 1 h=--=0·25· y=-
4 ' x'
x~: ~:25 x{2(~ 1·~)5; x;(=/\0; 4x(
4 7 1·75~)5}= 2. .. 3 + 2 + 1·50) + 1·25 + 1·75 ~ 0·6932539 by calculator.
2- 1 1 h = ---w = 0·1; y = :x·
x1 = 1; x 2 = 1·1; x 3 = 1·2; x4 = 1·3; x 5 = 1-4; x6 = 1·5; x7 = 1·6; X 8 = 1·7; X 9 = 1·8; X 10 = 1·9; X 11 = 2.
A~ 0~1 { G + D + 2(/2 + 1~4 + /6 + /s) + 4 (t\ + /3 + /5 + /1 + /9)}
~ 0·6931501 which is a very close approximation to the true value 0·6931471.
125
1. Use Simpson's Rule with three function values to find the approximate values of the following definite integrals:
(a) I: x 2dx I
2 1 (d) r-:;.dx
1 2-yx
(b) r (x 3 - x)dx (e) 15
log10 xdx
(c) J2
-4dx 1 X
2. Use Simpson's Rule with five function values (or four equal sub-intervals) to find an approximation to the area under the following curves, above the x axis and between the given ordinates:
(a) /(x) = x4; x = 0, x = 1
1 (c) f(x) =
4 + x 2 ; x = 0, x = 2
4 (b)f(x)=--;x=O,x= 1
x2 + 1 (d) f(x) = l ' X = 0 X = 0·6
~4- x 2' '
3. For the functionf(x) = JX, defined on 2 ~ x ~ 3 find:
(a) the exact value of the definite integral r JXdx. (b) the approximate value, using:
(i) the trapezoidal rule using four equal sub-intervals. (ii) Simpson's Rule with two equal sub-intervals.
(iii) Simpson's Rule withfour equal sub-intervals. and (iv) Simpson's Rule with ten equal sub-intervals. (Leave your answers as obtained from your calculator.)
4. Use Simpson's Rule with four sub-intervals to find an approximation for 11
-1
dx . Jo + x
5. (a) y = f(x) is known to be a continuous function, and experimentally the following table of results was recorded:
X II 20 125 I 30 I 35 I 40 I 45 I 5·0 I y = f(x) 3·1 3·8 I 4·0 3·6 2·6 2·5 I 2·4
By means of Simpson's Rule find an approximate value for J: f(x)dx.
(b) Assuming that v = f(t) is a continuous function of timet, with the following set of tabulated values:
t (sec) 0 0·5 1 1·5 2 2·5 3
v = f(t)(m sec) I 0 15 32 50 42 30 14
use Simpson's Rule to approximate 13
f(t)dt.
6. (a) A surveyor makes a straight line traverse across the bend of a creek and measures perpendicular offsets every two metres as shown. By application of Simpson's Rule, find the approximate area bounded by the traverse line and the creek.
126
E N
I I
I I
El El E El E I Nl "' Nl I ~I 6t ()()
c:i <t>l 001 .-! .... ail
0 2 4 6 8 10 12 14
(b) Consider the quadrant of a circle of radius 2 units. By dividing the radius OP into ten equal sub-intervals the following table of ordinates resulted:
16
X 0 0·2 0·4 0·6 0·8 1·0 1·2
y 2 1·99 1·96 1·91 1·83 1·73 1·60
E
18 20
y
2
p
0 2 X
1·4 1·6 1·8 2·0
1·43 1·20 0·87 0
Use these values and Simpson's Rule to find the approximate area of the quadrant and hence of the circle. Compare this with the area given by the usual formula A = nr2
•
If a plane area is rotated about an axis, in three dimensional space, it creates a solid. For example: 1) If a rectangle is rotated about an edge (as axis) it creates a cylinder.
2) A semi-circle rotated about a diameter creates a sphere.
127
/i / I I I I I I I \ \
3) A ray, through the origin, rotated about the x axis generates a cone.
y y
I I
====>-- I I
X 0 I X I I \
Such solids are called SOLIDS OF REVOLUTION. Consider any continuous function y = f(x) in the x-y plane. Let the area bounded by y = f(x),
the x axis and the ordinates x = a, x = b, rotate about the x axis.
y y
/I I I
I I I I f(b) I I I I
_!__ _ _J__
0 a b X 0 X I lb I I I I \ I \ I
\ I
Now a solid of revolution is formed which has the following properties: 1. Any cross section, perpendicular to the axis is a circle with its centre on the x axis and radius equal
to the ordinate at that point. 2. The radii of the end circles aref(a) andf(b) respectively. Thus the area of the plane section:
through x = a is A1 = n[f(a)] 2
and that through x = b is A2 = n[f(b)Y, Suppose now that the circular disc (or lamina) of width Ax is taken between x and x +Ax. This
section is enlarged below.
y
B i
I
D
x+L\x 1
--t---+-+--i- - - - - - - - .;- - j_ - +----l~ p I I 0 ~---L\x '
A
c
128
Now the area of the circle APB = n[f(x)]\ and the area of the circle CQD = n[f(x + Ax)JZ. Let the volume of the disc be AV, then
n{f(x)} 2• Ax< AV < n{f(x + Ax))Z. Ax
AV Hence n{f(x)} 2 < Ax < n{f(x + Ax))Z
as Ax -t 0, {f(x + Ax) }2
-t {f(x) }2
and hence AV Ax -t n{f(x) )2
lim AV = dV = n{f(x))Z Ax->O Ax dx
:. V = J n{f(x))Zdx.
Now if the volume of revolution from x = a to x = b is required, it is given by the definite integral
V = f n. {f(x)} 2dx or V = f ny2dx
Thus, the volume of the solid generated by the revolution about the x-axis of the area between y = f(x), the x axis and the ordinates at x = a and x = b is given by:
v = n r {f(x))Zdx
= n 1b y2dx.
Example (i): What is the volume of the solid generated when the area between y = x 2 + 1, the x axis and the ordinates x = - 1, x = 1 is rotated about the x axis?
Solution:
v = n r {f(x))Zdx
= n f1
(x2 + 1)2 dx
= n f1
(x4 + 2x2 + 1)dx
= n [xs + 2x3 + x] 1
5 3 -1
= n[(l + ~ + 1) - (-1- ~- 1)] = [3 + 10 + 15 + 3 + 10 + 15]
n 15
56n 15'
Th 1 f 1 . . 56n b. . e vo ume o revo utton ts 15 cu 1c umts.
129
y
2
-2
v=x2 +1
/I I I I 1 I I
- - - - _J - L -t------J~>-1 11 I I
\ : X
\ I
Note: From the symmetry of the diagram the same volume would be given by
V = 2n I (x 2 + 1) 2dx.
If a plane figure is revolved about the y axis, a similar formula exists.
y
b
0
X
-----a
I
V = nfb x2dy a
Example (ii): A mould for producing glasses is made by rotating the area bounded by y = x 3
, the y axis and the lines y = 1 andy = 8 about the y axis. What would be the volume of the glass in cubic centimetres?
Solution: Now y = x 3
,',X= {jY = y! v = n r x 2
dy
= n f {yt} 2 dy
= n 18
yfdy
= nDy!I = 35n [8!- HJ = 35n ~2- 1]
93n
5
~ 58·43. Thus the volume of the glass is 58-43 cubic centimetres.
130
y
0 X
Example (iii): A shell has a length of 20 em and its shape is generated by revolving y = JX about the x axis. Find its volume.
Solution:
V = n f y 2dx
= n ro (JX)2dx
= n l20
xdx
= n [~]:o = n [4~0] = 200n ~ 628·32.
The volume is 628·32 cubic centimetres.
y
0
This method of finding the volume enables the derivation of basic volume formulae.
Example (iv):
I I I
I I I I I J20
I I X
I I I \ I \ I
Find the volume of a sphere generated by rotating the semi-circle given by x 2 + y 2 = r2 about the x axis.
Solution: x2 + y2 = r2
y2 = r2 _ x2
As the circle is symmetric about each axis, the total volume is twice the volume of the hemisphere generated by rotating the quadrant OAB about the x axis. Thus the limits of integration are from x = 0 to x = r.
V = 2n Iy2 dx
= 2n L (r2 - x 2)dx
[ x3Jr = 2n r2x- 3 0
= 2n[(r3-
1:) _ o]
2r3
=2n. 3 = 4nr3
131
y
r B
A
0 lr x
Example (v): Find the volume of the cone generated when the line y = mx is rotated about the x axis over the interval from x = 0 to x = h, where h is the height of the cone.
Solution:
V= n ry2 dx
= n r (mx) 2 dx
= nm2 L'x2 dx
= nm2 [~
3
I 2
= nm (h3 _ O) 3
= tnm2h3 Now let the radius of the base at x = h be PQ = r. Thus m = gradient of OP
=tan e PQ OQ r h
V = tn(~:). h3
= tnr2 h
Example (vi): Find (a) the intersection of the curves y = x 2 and x = y2
;
(b) the area of the region between these curves; (c) the volume when this area is rotated about the x axis; (d) the volume when this area is rotated about the y axis.
Solution: (a) solving y = x2
X= y2
(b)
substitute from (1) into (2): x = x4
x4- x = 0
x(x3 - 1) = 0
giving x = 0 or x 3 = 1 Thus the curves intersect at x = 0 and x = 1 i.e. at (0, 0) and (1, 1).
A = f (.jX - x 2)dx
= f (x!- x 2)dx
=[~xi- ~3I = <t- t)- 0 -1 -3
.'. area = t square units.
132
y
0
y
0 \
p j
// I I I
I 1 I r I I I I 0
I \
lh I I I I I
X
X
(1) (2)
(c) The volume required = (volume generated by y = JX) - (volume generated by y = x 2
)
V = n f (.jX)2- n f (x2
)2dx
= n f (x - x4)dx
= n[~2- ~SJ:
= n[( ~- ~)- o] 3n 10
V 1 . 3n b' . o ume Is TO cu Ic umts
(d) V = n f (JY) 2dy- n L (y
2)
2dy
= n f (y - y4)dy
3n 10
V 1 . 3n b' . . () o ume Is TO cu Ic umts as m part c .
1. In each case leave your answer in terms of n. (a) Find the volume of the solid of revolution when the area under y = x 2
, above the x axis and between x = 0 and x = 2 is revolved around the x axis.
(b) The area bounded by y = 2x, the x axis and the ordinates x = 1 and x = 4 is rotated about the x axis. Find the volume of the frustrum of the cone generated.
(c) The area below the graph ofy =!,in the first quadrant and between x = 1, x = 4 is rotated X
about the x axis, what is the volume of the generated solid? (d) The arc of the parabola y = 2x2 between (0, 0) and (3, 18) is rotated:
(i) about the x axis (ii) about the y axis. Evaluate the volumes of the solids generated in each case.
(e) Find the volume of the hemisphere formed by the rotation of the quadrant of the circle x 2 + y 2 = 4, from x = 0 to x = 2, about the x axis.
2. Use calculators to give answers correct to three decimal places in each example. (a) Calculate the volume when the region bounded by the graph of y = x 2 - 4 and the x axis is
rotated about: (i) x axis
(ii) y axis. (b) Find the volume of the solid generated when the region in the first quadrant, bounded by the
x axis, the line x = 8 and the curve y = xi is rotated about: (i) the x axis
(ii) they axis. (c) The line y = x - 2 is rotated about the x axis. Find the volume of the solid of revolution
between the planes x = 2 and x = 3.
133
(d) The area between y = x 2 - 2x and the x axis is rotated about the x axis. What is the volume
of the solid so formed? (e) The area in the first quadrant bounded by y = 9 - x 2
, y = 0 andy = 9 is rotated about the y axis. Calculate the volume of the solid generated.
3. For the following sets of curves find their points of intersection and the volume when the area between the curves is rotated about the x axis: (a) y = x 2
; y = h (b) y 2 = 3x; x 2 = 3y.
4. (a) The area of the region bounded by the curves y = x 2 andy = x + 2, between x = 0 and x = 1 is rotated about the x axis. Find the volume of the generated solid.
(b) The region in the first quadrant bounded by the graphs of f(x) = ·h3 and g(x) = 2x is rotated about they axis. Find the volume of the solid so formed.
(c) The area under y2 = 3 - 2x - x 2, above the x axis and between x = -3, x = 1 is revolved
about the x axis. Calculate the volume of solid of revolution so generated. (d) The curve x 2 = 16 - 4y2 is rotated about they axis. Find the volume of the generated solid.
5. (a) Find the area bounded by the curve y = J9=X and the x andy axes. Find the volume of the solid generated when this area is rotated about: (i) the x axis
(ii) they axis. (b) Show that y = 2)1 - x 2 andy = )1 - x 2 intersect at (1, 0) and ( -1, 0). Find the volume
of the resultant solid when the enclosed region is rotated about: (i) the x axis
(ii) they axis. (c) Find the volume of the cone formed by the rotation, about the x axis, of the interval cut off
the line y = 2x - 1 between the two axes. (d) A parabolic mirror is made by rotating the area bounded by the parabola y = h 2
, they axis and the line y = 4 about they axis. What volume does it occupy?
6. (a) If the curve xy = 1 is rotated about the x axis, find the volume generated by the arc of the curve intercepted between the planes x = 1 and x = 4.
(b) Find the volume of the solid when the region between the curves y = x 2 and y = 2 - x 2
in the x-y plane is rotated about: (i) the x axis
(ii) they axis. (c) A petrol tank is designed by the rotation of the curve y = tx(x - 40) about the x axis between
the planes x = 0, x = 40. If units are in centimetres, how many litres would the tank hold? (d) Find the equation of the tangent to the parabola y = 2x2 at (1, 2). Calculate its point of
intersection with the x axis and the volume of the solid formed when the area between the parabola, the tangent line and the x axis is revolved about the x axis.
134
Englishman Charles Babbage's Difference Engine was based on the same principles as today's computers. This machine could generate logari!hmic and astronomical tables to six decimal places.
I • IC
Photograph courtesy of I.B.M.
s
CHAPTER 19
The word exponent is often used instead of the word index. Functions where the variable is in the index (e.g. 3X, 102x) are called exponential functions. One aim of this chapter is to learn how to use exponential functions and in particular to discover two very important functions.
Exponential graphs have a particular shape. Here is part of the graph of y = 2x and the table of values for it.
X -2 -1 0 1 2 3
2x 1 1 1 2 4 8 4 2
You can learn more about exponential graphs by thinking about the questions in the following exercise.
1. (a) What is the value of 2x if x = -3 and if x = -4? (b) What is the value of 2x if x = 4 and if x = 5? (c) What is the value of 2x if x = 10?
-3 -2 -1
(d) How would you describe the behaviour of the graph of y = 2x as x becomes larger and larger?
(e) Can you think of any value of x which would make 2x negative? (f) If x approaches negative infinity, what value does 2x approach?
2. Try these similar questions about y = 3x. (a) Do you expect 3x ever to the negative? (b) What is the behaviour of3x as x becomes smaller and smaller (x ---+ - oo)? (c) Where does the graph of y = 3x cut they-axis? (d) If xis any positive value, say x = 4, is 3x larger or smaller than 2x? (e) If a is negative which is larger 2a or 3a?
3. Complete the following table of values for y = Y and draw a sketch graph of the curve. For this graph the domain is limited to lxl ~ 3. If necessary use a calculator.
13:: r-3r-2r-1 I 0 It 11 5 I 2 I 25 I 3
136
4. On a different set of axes because of the scale difference sketchy = lOX and verify that the same fundamental properties hold, namely: (a) wx is never negative (b) wx -t 0 as x -t - oc;
(c) wx increases rapidly and without limit as x gets larger (d) the graph passes through the point (0, 1).
5. Without drawing up a table of values draw a rough sketch of y = 4x merely showing vital features of shape. What is one important point the graph passes through?
6. Draw your own graph or copy the graph of y = 2x and draw a tangent at the point (0, 1 ). Check if the gradient of this tangent is about 0·7.
7. Find rough values for the gradients (slopes) of the graphs y = 3X, y = 4x andy = wx at the point (0, 1) on each.
8. Verify from your answers and from the shape of your curves that the gradient of the tangent at (0, 1) on y = Y is greater than for y = 2x and similarly it increases for y = 4x and y = wx. Given that the gradients ofy = 2x andy = 3x at (0, 1) are approximately 0·7 and H respectively, guess what the gradient of y = 2·5x at (0, 1) might be.
9. Draw up a table of values and sketchy = 2·5x from x = -1 to x = 2 and estimate the gradient at (0, 1).
Consider any function y = ax where a is a constant greater than 1. Since a0 = 1 this graph passes through (0, 1) for any value of a. Let us try to find the gradient at this point in the way we have used in the calculus.
Take the point P(O, 1) and let Q be a point on the curve near P so that OM= Llx and let QR = Lly. Q, being on the curve y = aX, must have coordinates (Llx, a11x)
Lly QR a11x- 1 Then - = - = -,-------
Llx PR Llx The limit of this expression as Llx -t 0 is the gradient at (0, 1).
We will call this gradient m and use the ideas above to differentiate ax.
Since Q is (Llx, a11x) then QM = a11x
137
Let P(x, y) and Q(x + ilx, y + ily) be two points on the curvey =ax. Now y =ax and y + ily = ax+l1x so ily = ax+!1X - ax
thus
and
=ax. a/J.x- ax = aXCa/J.x- 1)
ily ax(a/J.x - 1)
ilx ilx
dy = lim Ay dx /J.x--+0 Ax
. ax(a/J.x - 1) = hm --'~---'-/J.x--+0 Ax
/J.x--+0 Ax = ax . lim (a/J.x --: 1)
But we have previously found that the limit lim (a/J.xA- 1) is the gradient of the curve at the point
/J.x--+0 tlX (0, 1) and which for any curve we call m.
So dy = max where m is the gradient at (0, 1). dx
A number of important results follow: 1. We can differentiate y = ax if we know its gradient at (0, 1). The gradient of y = 2x at (0, 1) is
0·69 approx. so for that curve :~ = 0·69 x 2x.
2. On any exponential curve the gradient at any point is proportional to the value of the function at
that point 0.e. dx = my) so when the function is small it increases slowly but when the function
is large its gradient is also large.
3. Looking at the expression dx =max it follows that if we could find a function ax for some a where
the gradient m at (0, 1) was 1 then we would have found a function which is unchanged upon differentiation. Such a function can be found, it is called the exponential function and it has great importance in mathematics.
The gradient of y = 2x at (0, 1) is approx 0·69. The gradient of y = 3x at (0, 1) is about 1·1.
Somewhere between 2 and 3 is a value e such that ex has a gradient of 1 at (0, 1). e turns out to be 2·71828 ... approximately.
y = ex is called the exponential function.
It has the property that if y = ex then dx = ex.
Calculators and books of tables have values of this function.
138
Example: Using tables or calculator find: (a) e2-6 3
Solution: (a) 13-87377
1. Evaluate: (a) eo·z
(b) el-675
(b) e-0·7
(b) 0-49659
(c) e-0·5
(d) e-3·7
(c) 0·01479
(e) 5e3
6 (f)~
6
2. Draw the graph of y = ex for -1 :::;; x :::;; 3 and check that the gradient of the tangent at (0, 1) is 1.
3. (a) Find approximately the value of the function y = ex where x = 1· 5. (b) Find the gradient of the graph y = ex drawn above when x = 1· 5. Is the gradient approx
imately equal to the value of the function at this point?
4. Draw a rough sketch of y = e2 x for -2 :::;; x :::;; 2.
5. Earlier we said that the gradient of y = ax at (0, 1) could be found as the limit of a,.,:: 1 as .ilx
becomes small. Using the calculator or tables for values of et.x complete this table:
LlX = 0·1 0·01 0·001
et.x = 1·10517
et.x- 1 = 0·10517
et.x- 1 1·0517
LlX
Example: Sketch the graphs of y = 2x andy = 2-x and compare.
Solution:
X -2 -1 0 1 2
2x 1 1 1 2 4 4 2 4
2-x 4 2 1 1 1 2 4
2 The graphs are reflections of each other in they-axis.
139
1. On the same axes draw the graphs of y = ex andy = e-x for -3 ~ x ~ 3.
2. By adding ordinates of the above graphs or otherwise draw graphs of y = ex + e-x and of
3. Sketch the graphs y = 3x andy = rx and compare them.
4. Draw a sketch of the graph y = e- 3x for -2 ~ x ~ 2. Where does this graph cut they-axis? On the same axes also draw y = 2e- 3x.
5. Draw graphs of y = -!-ex andy = ef on the same axes.
Remember
Examples: Differentiate with respect to x. (a) e6x+3
Solution: (a) y = e6x+3 (b) y = e2x3
Let y = e" where u = 6x + 3 dy = dy x du dx du dx
= e" x 6 = 6e6x+3
1. Differentiate with respect to x:
(a) e3x
(b) esx-6 (c) e?x+S
2. Differentiate with respect to x:
(d) e-4x
(e) e2 x + 9 (f) ex + 5x2 + 2
(a) 5e6x (c) x- e7 -x
(b) ex2 (d) ex3 + 2x
3. Find second derivatives of: (a) e2x (b) e3x+2
4. If f(x) = ~(e! + e-i), find f'(a).
Let y = e" where u = 2x3
dy=dyxdu dx du dx
= e" x 6x2
= 6x2e2x3
(g) 3x + _!_ + e-x X
(h) -!-(ex - e-x) (i) 2e 7-x
(e) e-4x2
(f) e~
(e) el-x
5. Find the gradient of the tangent to the curve y = ex at the point (1, e).
6. Show that the tangent to the curve y = 2eX, at the point where x = 2, is parallel to the tangent to the curve y = e2x at the point where x = 1.
140
7. Show that the curves y = e3x andy = 3ex each have the same gradient where they cross they-axis. Find the equation of the tangents to each graph at these points.
d2y 8. If y = ex + e-x, show that dx2 = y.
9. If y = 4e-x + 5e- 3X, show that~ + 4: + 3y = 0.
10. Show that the tangent to the graph y = ex at the point where it crosses they-axis is inclined to the x-axis at an angle of 45°. Hence find the angle between the tangents toy = ex andy = e-x at the point (0, 1) which is on both graphs.
11. Show that if y = e ?x- 2, then dy = 7y and also that dd2{' = 49y.
dx x
12. Find the equation of the tangent to the curve y = ex + 1 at the point (1, e + 1 ).
We consider now the use of the product and quotient rules and the application of the calculus to maxima and minima.
Example (i):
( ) D 'f't" . ex + 1 . h a 1 terentlate y = --- wtt respect to x. 2x
(b) Find first and second derivatives ofy = x 2 ex.
Solution:
(a)
Let
ex+ 1 y=--
2x u
y = - where u = ex + 1 v and v = 2x du dv
v--u-dy dx dx dx v2
_ 2xex - (ex + 1) X 2 - 4x2
2xex- 2ex- 2 4x2
xex- ex- 1
2x2
Example (ii):
(b) Let
and
y = x2ex
y = u x v where u = x 2
and v =ex
dy = vdu + udv dx dx dx
= ex X 2x + x 2ex = (x2 + 2x)ex
d2y -
2 = ex(2x + 2) + (x2 + 2x)ex
dx = ex(x2 + 4x + 2)
Find the maximum and minimum values of y = x 2 ex. (Note: Results above may be used).
Solution: if y = x 2 ex then from above y' = (x2 + 2x)ex as ex is never zero then y' = 0 if x 2 + 2x = 0 if x(x + 2) = 0 if x = 0 or x = - 2 Substituting x = 0 and x = -2 The turning points are (0, 0) and (- 2, 4e- 2
).
141
Given the turning points found opposite we use the second derivative to examine their nature From above y" = ex(x2 + 4x + 2) if x = 0, y" = e0 x 2 (positive) if x = -2, y" = e-2 (4 - 8 + 2)
= - 2e- 2 (negative) So the point (0, 0) is a minimum and the point ( -2, 4e-2
) is a maximum.
1. I;>ifferentiate with respect to x: i/
/((a) )xex \, ~J/
(b) 3xex i
/ i
2. Find second derivatives of: (a) ex2 (c) 2xe-x
3. Rewrite 5x in such a way that the quotient rule is not needed and differentiate with respect to x.
e
4. Ifj(x) = (1 + 2x)e3X, findj'(2) andf"(O).
5. Find the equations of the tangents to the curves y = e3x + 2, y = e2x + 1, andy = ex at the points where they cross the y-axis. Show that these tangents all pass through the same point on the x-axis.
6. Show that the tangent to the curve y = ex - 2x at the point (1, e - 2) passes through the origin.
7. Show that, if y = Ae-ct where A and care constants, then:;? - c2 y = 0.
8. (a) Find the stationary value for the curve y = t(ex + e-x) and show that it is a minimum. (b) What happens to the value of y when x becomes large? (c) Show that the values of y when x = 10 and when x = -10 are the same. Also show that for
any value a the values of y when x = a and when x = -a are the same. What does this tell you about the shape of the curve?
(d) Draw a rough sketch of the curve. Note: This curve is called a catenary. It is the shape of a chain or rope hanging freely.
9. A sketch of the curve y = xe-x is drawn opposite. Check these features: (a) Show that the curve has one stationary point
which is a maximum. Find it. (See A) (b) Show that the curve passes through the origin. (c) Find the point of inflexion where the curve
changes curvature. (See B) (d) Find the point C where x = -1. (e) Show that the curve approaches the x-axis as
x becomes large. (See point D)
10. Consider the curve y = xex. Its graph is similar to the above but with obvious differences due to the positive sign. (a) Find the first and second derivatives. (b) Find any maximum or minimum values. (c) Find any point of inflexion. (d) Check if the curve passes through the origin or not. (e) What happens to y if x becomes large? (f) Is y large or small, positive or negative when
(a) X = -3 (b) X= -10? (g) Draw a sketch of the curve.
142
A
D
2 3
c
We need to use a little ingenuity in integrating exponential functions. Study these examples.
Example (i):
Show that if y = e5x+l, then: = 5e5x+t.
Hence find J e5x+ 1dx.
Solution:
Ify = e5x+l then dy = 5e5x+l ' dx
Hence
So
f 5e5x+ldx = eSx+l + C
fesx+tdx = !esx+l + C 5 .
Example (ii):
Find 12
xex2
dx.
Thus 12
xex2
dx = ! l2
2xex2
dx
= 1[ex2Ji = !(e4
- e).
Example (iii): Calculate the area under the curve y = !(e2x + e- 2x) and between the lines x = -1 and x = 1.
Solution: The area is given by
fl !(e2x + e-2x)dx -1
As the graph is symmetrical about the y-axis this is
2 f !(e2x + e-2x)dx or f (e2x + e- 2x)dx.
Either this integral or the original one may be calculated.
Area = [ (e 2x + e- 2x)dx
= [!e2x _ !e-2xJ&
= (!e2 - !e-2
) - ct- !) = !e2 - !e-2 =i= 3·627 if simplified answer is required.
143
t
\I ~
-1
1. Integrate with respect to x: (a) e4x (d) e3x + e-x (g) 6e4-x (b) e-2x (e) e-f (h) e1-6x
(c) eax+b (f) 3x2 + e2x (i) e-2x
4
2. Integrate with respect to x: (a) 4xex2 (c) xe7x2+1 (e) xex2 + xe-x2
(b) ex - xex2 (d) x2e2x'+1 (f) 2x
x2 e
3. Evaluate these definite integrals to 3 decimal places:
(a) r e3xdx (b) r xe-x2dx (c) f t(ex + e-x)dx
4. Find the area bounded by the curve y = eX, the x and y axes, and the line x = 2.
5. Find the area under the curve y = t(ex +e-x) from x = -2 to x = 2.
6. The diagram shows the curves y = e2x and y = e-x. Calculate the area under each curve from x = 0 to x = 1 and hence find the area shaded between the curves.
7. Show that the area between the curves y = ex and y = e2x from x = 0 to x = 1 is given by (tez - e + t).
8. The diagram shows part of the graphs of y = ex and y = xex
2• Show that these graphs both pass through the
point (1, e) and that one crosses they-axis at y = 0 and the other at y = 1. Find the area shaded from x = 0 to x=l.
144
9. The shaded region in the diagram below the curve y = ex is rotated about the x-axis. Find the volume of the solid
of revolution formed. Hint: Remember V = nIb y2dx.
0
10. The region under the graph y = !(ex + e-x) and between the lines x = -2 and x = 2 is rotated about the x-axis. Find the volume of the solid of revolution formed.
11. Find the area shaded which is bounded by the curve y = eX, the line y = e and they-axis.
12. Sketch the curve y = e- 2x. Find the area between the curve, the x-axis, and the lines x = 1 and x = 2. If this region is rotated about the x-axis find the volume of the solid generated.
13. Differentiate xex and show that :X (xex) - ex = xex. Hence prove that f xex = xex - ex + c.
Connected with the exponential function is another very important function called the logarithm function. You have already noted that an index can be called an exponent. It can also be called a logarithm. More particularly, if y = ax then x is called the logarithm of y to the base a (written x=logay)
Thus if y = aX, then x = loga y of special interest, if y = eX, then x = loge y
You may have met logarithms before including the common logarithms to the base 10. Logarithms to base e are called natural logarithms or Napierian logarithms after a Scotsman John Napier who published an article on logarithms in 1614.
Follow these examples to investigate the theory behind logarithms. Remember, a logarithm is an index.
145
Example (i): Given 26 = 64 and 34 = 81 write down (a) log2 64 (b) log 3 81
Solution: (a) log2 64 = 6 (b) log3 81 = 4 (c) log10 100 = 2
because 26 = 64 because 34 = 81 because 102 = 100
Example (ii): Find (a) log2 16
Solution: (a) Since 16 = 24
then log2 16 = 4
Example (iii): Find (a) log3 J'YI Solution: (a) J27 = 3J3 = 31
.·. log3 J'i7 = 1·5
Example (iv): Prove that: (a) log" xy = log" x + log" y
X (b) log" = log" x - log" y y (c) log" x" = n loga x
Solution: Let ak = x and a"' = y (a) xy = ak x a"'
= ak+m
Hence log" xy = k + m
(b)
i.e.
(c)
Example (v):
= log" x + log" y. X ak k-m -=-=a y a"' X
log"- = k - m = log" x -log" y. y x" = (ak)" = ak"
log" x" = kn = n loga x.
(c) log10 100
(b) logs 125
(b) Since 125 = 53
then logs 125 = 3
(b) logvz 32
(b) 32 = 2s = (.j2)10
so logvz 32 = 10
Given loge 3 = 1·09861 and loge 2 = 0·69315, calculate loge 6 and loge 81 to 4 decimal places.
Solution: (a) loge 6 = loge (3 X 2)
= loge 3 + loge 2 = 1·09861 + 0·69315 = 1·79176 = 1·7918 (to 4 places)
146
(b) log3 81 = loge 34
= 4loge 3 = 1·09861 X 4 = 4·39444 = 4·3944 (to 4 places)
Example (vi): What is the natural logarithm of: (a) e3·57? (b) 1·7?
Solution: (a) By definition if y = e3'57, then loge y = 3·57.
Note: In general loge ea = a. (b) From tables or calculator loge 1·7 ~ 0·5306.
(c) 11o? (d) 0·17?
(c) From calculator loge 170 = 5·1358 direct. Tables usually only have logarithms of numbers from 1 to 10. Hence if using tables proceed as follows: loge 170 = loge (1·7 X 100)
= loge 1·7 + loge 100 = loge 1· 7 + 2 loge 10 ~ 0·5306 + 2 X 2·3026 ~ 5·1358
(d) loge 0·17 ~ -1·77196 from calculator direct 1·7
= loge lO if using tables
= loge 1·7 - loge 10 = 0·5306 - 2·3026 = -1·772.
1. Use calculator or tables to find to 4 places: (a) loge 5 (c) loge 1·384 (b) loge 7 (d) loge 10
(e) loge 1 (f) loge 2·7
2. Given loge 10 = 2·3026 and loge 7 = 1·9459, without tables or calculator find: (a) loge 1000 (c) loge 70 (e) loge 0·7 (b) loge 0·0 1 (d) loge 7000 (f) loge 0·007
3. Given loge 2 = 0·69315, and loge 3 = 1·09861, find without tables or calculator to four decimal places: (a) loge 4 (b) loge 27
(c) loge 1·5 (d) loge 12
4. Use your knowledge of indices (powers of 2 and 3) to find: (a) log2 128 (b) log3 9)3
5. Use the fact that eo = 1 to find loge 1.
6. What is loga 1 for any positive vah1e of a?
(e) loge 18 (f) loge 13·5
(c) log2 J2
7. Let a > 0. Can you think of any number c such that ac would be negative? (Think of the graph ofy = 2x.)
8. Is the statement loga( -4) meaningful if a is greater than zero?
9. Can you think of any number z such that ez = 0? Can there be any value for loge 0?
10. Draw the graph of y = loge x for values of x from 0·1 to 6. Use the same scale on the x andy axes. From your graph estimate the value of x for which loge x = 1.
11. On the same axes as for question 10 draw the graph of y = ex but for convenience of scale draw for -2 :;::; x :;::; 3. On your graph draw also the line y = x. What do you notice about your two
· graphs in relation to the line y = x? [Note: y = ex andy = loge x are called inverse functions.]
147
12. On the same set of axes in each case draw the following curves. (N.B. Unless otherwise stated log x means loge x). (a) log x and log 3x (N.B. log 3x = log 3 + log x). (b) log x, log (x + 2) and log x + log 2. (c) log x and log x 2
•
13. What is the maximum domain which can be assumed for log x? Assuming this domain, what is the range?
14. Simplify log a 3 - log a2
•
15. Simplify loge (e2 + e) - loge (e + 1).
16. Solve: (a) ex= 3 (b) e- 2x = 4.
17. (a) Whatislog2 64? (b) Whatis210g2 64 ?
18. (a) What is log3 81?
19. What do you think e1og,
2 is? Note: If y = aX. then x = loga y
So y =ax= a1og,y.
20. Evaluate: (a) e1og
3
(b) elog 7-5
21. Solve the equations
(c) elog 10
(d) elog 1
(a) elogx =log 2 (b) elog3x = 6
22 .. Show that 3 log x + 2log y -!log z = log (x)z2
}
dy 1 We now prove that dx = dx
dy
Proof: If y = f(x), then with the usual notation Ay _ 1
Taking limits as Ax -+ 0
Ax- Ax
Ay
1 .. Ay 1 tmtt-=---
t.x-->0 Ax 1. 't Ax tml-
t.x-->0 Ay 1
1. Ax tm
t.y->o Ay dy 1
Thus -=-dx dx
dy This result can also be arrived at geometrically.
148
(because if Ax -+ 0 then Ay -+ 0)
dy dx or- x- = 1
dx dy
Given y = loge x Then this means x = eY
from which
But
dx Y -=e. dy dy 1 dx = dx
dy 1
eY 1 X
This is a very special result.
(previous result)
dy 1 If y = log x then - = -
e ' dx X
Hence also f ~dx = loge x + C
Notice the importance of these results. Up to date we have had integrals for all other powers of x 1
except-. X
Example (i): Find the derivative of (a) loge (5x - 2)
Solution: (a) y = loge (5x - 2)
Let y = loge u where u = (5x - 2) dy _ dy du dx-du'dx
= .!_ 0 5 u
5 5x - 2
Note: Here is a quick method you may wish to use.
dy f'(x) If y = loge f(x) then dx = f(x) .
Check that this is so using the example above.
Notation: loge x may be written log x or In x .I 149
(b) loge (7x2 + 3x).
(b) y = loge (7x 2 + 3x) Let y = loge u u = 7x2 + 3x
dy _ dy du dx- du · dx
= .!_. (14x + 3) u 14x + 3
- 7x2 + 3x
1. Differentiate with respect to x: (a) log 3x (e) log (3x + 1) (b) log x + 2x (f) log (2x - 3) (c) log x 2 (g) log (x2
- 5) (d) e2 x + log 2x (h) 5x - log 2x3
(i) log x4
(j) 8 log 3x (k) 2 log x + 5 log 2x (1) ! log 3x
2. Show that log (x + 2)(x - 3) is the same as log (x + 2) + log (x - 3). Perform the differentiation of this function in two ways. Which is easier?
3. Differentiate y = log x + 1 with respect to x by the way you consider easier.
x+4
4. Show that log JX+9 = ! log (x + 9). Hence or otherwise differentiate this function with respect to x.
5. Differentiate with respect to x:
(a) ln (2x + 1)(x - 5) (c) ln (x + 6) 3 ( ) 1 (x + 1)(x - 1) e n (x + 2)
X (d) ln JX+4 (f) ln (3x + 2) 2 (b) ln--
1 x+
6. Differentiate with respect to x: 1 1
(c) 1
(a) ln- (b) ln (2x - 1) ln-X 2x3
7. Differentiate these products with respect to x: (a) x ln x (b) 2x3 ln (x - 2) (c) X ln X - X
8. Don't confuse y = log x 2 andy = (log x) 2• Differentiate each with respect to x.
9. Now think of y = log (log x). This can be written as log u where u = log x. Differentiate with respect to x.
10. Find second derivatives of: (a) log (3x + 1) (b) X log X - X
d2 (d )2 11. If y =log x, show that d; + Jx = 0.
12. The line x =a cuts the curves y = log x andy = log 2x at P and Q. Show that the tangents to the curves at P and Q are parallel. Also show that the distance PQ remains constant for all values of a.
13. Find the equation of the tangent to the curve y = ln x (a) when x = 1 (b) when x = 3.
14. At what point on the curve y = ln 2x is the gradient ! ? Find the equation of the tangent at this point.
15. Show that the minimum value of y = x log xis _!.Sketch the graph from x = ! to x = 5. e
16. Find the maxima, minima, or points of inflexion for the following functions and sketch a graph for each.
(a) y = X ln X - X (b) y = ln x X
150
1 (c) y = ln x +
X
I7. Differentiate with respect to x:
(a) In ~ ex
(b) e-x In x
(c) ex . In x
(d) In ex + I ex- 1
18. Remember that if x = -3, then lxl = 3 and draw the graph of log 14 I9. Write In (x 2
- 9) as the sum of two logarithms and hence differentiate with respect to x.
20. What is the maximum assumed domain of y = log (x + 2)? Where does the curve cross the x-axis? What is its gradient at this point? Sketch the curve.
21. For what value of x is the ratio of the natural logarithm of a number to the number itself the greatest.
Hint: This question really asks-when is the function log x a maximum? X
Remember d 1
dx (log x) = x' hence I ~dx = log x + C
d f'(x) dx log f(x) = f(x) ' hence I jf;j dx = log f(x) + C
The art of integrating some fractional expressions is to try to put them into the form f'(x) as shown below. f(x)
Example (i):
Find I_!__dx. 2x
Solution:
I I III I 2X dx = l Xdx = llog X + C
Example (ii):
Findi~dx. X + 7
Solution: Notice that if f(x) = x 2 + 7, then f'(x) = 2x.
So J x 2
6: 7dx =I ~2~2;dx = 3 I x 2
2: 7dx = 3log (x
2 + 7) +C.
Example (iii):
Find J 4x + 3 dx.
2x + I
Solution: . 4x + 2 We notice that = 2.
2x +I
151
Hence we break up the fraction into two parts as follows:
f 4x + 3 dx = f 4x + 2 + 1 dx 2x + 1 2x + 1
= J G~ ! ~ + 2x ~ 1) dx
= f2dx+1f2x~ 1dx
= 2x + t log (2x + 1) + C.
1. Integrate the following with respect to x: 1
(a)-4
(c) 4x - 3 (e) 2 2 5 3x + x--4x X
(b) _1_ 1 (f)
3 x+2 (d) 6x - 7 2x - 1
2. Integrate these expressions with respect to x: 2x (c)
x-2 (e)
ex (a) x2 + 1 x2 - 4x + 1 ex + 1
(b) X (d) 2x (f)~ x2 - 3 1 - x2 x4 + 3
3. Simplify, if necessary, and integrate with respect to x:
(a) 5x2 + ~x - 6
(b) (x + : 2 J (c) x2e2x2 - 1
X X
4. Evaluate:
fe-1 1
(a) --1
dx 1 X+ J
2 X
(c) - 2--1
dx 0 X +
(b) J" 2t + ?:_dt 1 t J
3 1 (d) 1 (2t- 1)dt
X + 7 8 Je+1 X + 7 5. Show that --
1 can be written as 1 + --
1 and hence show that --
1 = e + 7.
x- x-2
x-
6. Show that 11
+ x = -1 + - 2- and hence find f 11
+ x dx. -x 1-x -x
7. Find J2x + 5
dx. x+4
1 1 2 . J3 1 8. Show that --
1 - --
1 = - 2--
1. Hence or otherwise show that - 2 --
1 dx = t log 1·5.
x- x+ x- 2 x-
9. Find the area under the curve y =~(a) from x = 1 to x = 5 (b) from x = 1 to x = e.
10. Check the area under the curve y =~from x = 1 to x = e found above, by an approximate numerical method. ·
11. Find the area under the curve y = ~ from x = 1 to x = 2.
12. Sketch the curve y = 1 - ~· Where does it cross the x-axis? Find the area between the curve and the x-axis and bounded by the line x = 1.
152
13. Sketch the curve y = 2 - ~.Show that it cuts the x-axis where x = t and find the area between the curve and the x-axis and from x = t to the line x = 1.
14. The gradient at any point (x, y) on a curve y = f(x) is x ~ 1
. The curve crosses they-axis at the
point (0, 4). Find the equation of the curve.
15. The region under the curve y = Jx from x = 1 to x = e is rotated about the x-axis. Find the
volume of the solid of revolution.
16. Find the area under the curve y = ~2 from x = 1 to x = 3. X + -
17. The region under the curve y = J x 2
2:
1 and bounded by the lines x = 0 and x = 1 is rotated
about the x-axis. Find the volume of the solid of revolution.
18. The region between the curve y = 1 - 3 and the x-axis X
and from x = 1 to x = 2 is rotated about the x-axis. Show that the volume generated is n(3 - 4 log 2) cubic units.
19. Find the area of the region shaded between the curve y = ln x and they-axis and between the lines y = 0 and y = 1. Do it this way: (a) Rewrite y = ln x as x = eY. (b) Integrate eY with respect toy, from y = 0 toy = 1.
0
------ (e, 1) I I I I I I I
20. In a similar way (i.e. by first rewriting y = ln x as x = eY), find the volume of the solid of revolution formed when the region described in question 19 is rotated about they-axis.
21. Find the volume of the solid of revolution when the area between the curve y = In x and the y-axis and from y = 0 to y = 2 is rotated about they-axis.
22. Sketch the graph of y = In 2x. Rewrite this equation as x equal to some function of y. Find the volume formed by rotating the region between this curve and the y-axis and from y = 0 to y = 1, about the y-axis.
153
So far we have only differentiated exponential expressions such as ex or ef(xl. We look now at general expressions such as ax or afCxl where a is some constant. Remember that if a = eY then y = ln a by definition. This means that a = e1na (which is an important result)
Now ax = (e!nay = ex!na
So d(ax) = ln a ex!na dx ·
(as ln a is a constant)
Example:
Ify = a2x, find z.
Solution: y = a2x
Let y = a" where u = 2x
dy = dy. du dx du dx
= ln a. a". 2 = 2ln a. a2 x.
1. Find the slope of the tangents to the curves y = 2x and y = JX when x = 0 and compare with · the values of0·7 and 1·1 estimated from graphs drawn in exercise 19.1.
2. Find the slope of the curve y = 10x when x = 2.
3. Differentiate with respect to x. Leave logarithms unsimplified. (a) a4x (c) 5x . 4x
(b) (6x2 + 1)ax (d) 3x X
4. If y = sx find: 0 Hence find I sxdx.
5. Find I lOxdx.
6. Find the area under the curve y = 2x from x = 0 to x = 3 to 2 decimal places.
Before the advent of calculators, logarithms to base 10 were very useful. Logarithms are useful because being indices they can be added or subtracted when numbers are being multiplied or divided and most people find the first two operations easier. But why logarithms to base 10?
It is impossible to print a table with logarithms of all numbers shown separately because of space. Lists in tables are restricted usually to logarithms of numbers from 1 to 10. Logarithms of other numbers can then be calculated as for example.
log 17 =log (1·7 x 10) =log 1·7 +log 10 log 0·017 = log (1·7 + 100) = log 1·7 - 2log 10
154
If logarithms are to the base 10 these transpositions are easy because log10 10 = 1 and log10 100 = 2 etc.
This all raises two questions (a) How can logarithms be changed from one base to another? (b) How can logarithms to different bases be differentiated or integrated?
CHANGE OF BASE Let y =log" x Then x = aY Taking logarithms of both sides to base b
Iogb x = Iogb (aY) = y .Iogb a = Ioga x . Iogb a (by substitution)
This is usually written:
log x = Iogb x a Iogb a
and a special case is:
Example (i):
DIFFERENTIATION OF y = log8 x Let y = log11 x Before differentiating rewrite this as
loge X y = -- (see opposite) loge a
dy 1 .. dx = loge a. x (as loge a is a constant)
Remember:
d 1 1 -d (log11 X) = -
1-.-
x oge a x
and a special case is :
Use the fact that loge 8·2 = 2·104 and that loge 10 = 2·3026 to find log10 8·2 by change of base.
Solution: log 8·2
By change of base, 1og10 8·2 = log: 10
2·104 Hence log10 8·2 =
2.3026
= 0·914. Note: As a check log10 8·2 = 0·914 to 3 places from calculator direct or from common logarithm
tables.
Example (ii): Differentiate log10 (4x - 1) with respect to x.
Solution: y = log10 (4x - 1)
= log10 u where u = (4x - 1) dy dy du dx du'dx
=-1-! 4 loge 10 · u'
4 loge 10 . (4x - 1) ·
155
1. Use tables or calculator to find the following logarithms to base e: (a) log 1·95 (b) log 5·803 (c) log 9·61
2. Using the change of base rule, change each logarithm above to an equivalent logarithm with a base of 10.
3. Show that log10 x = 0·4343. . . x loge x.
4. Find the following logarithms to base e and then in each case change them to base 10. Answers to 3 decimal places. (a) log 84 (b) log 0·059 (c) log 10
5. Differentiate with respect to x (leave unsimplified).
(a) log10 3x (c) log10 x (e) log10 (5x2 - 7x) X
(b) 2x2 log10 x (d) log10 (x2
)
6 Sh th t _E_ (1 ) - 0·4343 ... . ow a d og10 x -X X
7. Prove that loge 10 = -1
-1-. Look up both loge 10 and log10 eon a calculator or in tables and
oglo e check that the result is correct.
Some interesting properties of the number e:
(a) The area under the curve y =_!'from 1 toe is 1 square unit. X
(b) The base for natural logarithms is e.
(c) If y = ex then 1x = ex (i.e. unchanged). Likewise I exdx = ex.
(d) At all points on the curve y = ex the gradient is equal to the value of the function at that point. . dy 1.e. dx = y.
(e) If y = ekx, then: = ky. (i.e. the gradient is k times the value of the function at any point).
(f) The curve y = ex has a gradient of 1 where it crosses they-axis.
(g) e = lim (1 +!)".An approximate value can be found fore by using this result which is proved ,1---..oo n
in the next section.
Take the point P(11.x, et.x) on the curve y = ex and join P to Q(O, 1).
PR Slope of PQ = RQ
et.x- 1
11.x But because the slope of the tangent at Q is 1
t.x 1 lime- =1
t.x->0 11.x
1. Llx 1 or 1m =
t.x->0 et.x - 1 (by reciprocal)
156
0
+ IZ>
/; ~
P (L1 x. e"•)
M
Now let -!z = e11x - 1 from which e11
x = (1 + -!z) and Llx = loge ( 1 + -!z} By substituting these
results for e11x and Llx in the line above we have:
loge(1 + !) lim
1 n=1
11x-+O
n
or lim n loge (1 + !) = 1 (because n --+ oo when Llx --+ 0) n-too n
( 1)" lim loge 1 + - = 1 r~--too n
( 1)" lim 1 + - = e1 = e. n-+co n
So
or
( 1)" use your calculator to find the value of 1 + n when
(a) n = 100 (b) n = 1000 (c) n = 10000
( 1)" Can you see that for n very large the value of 1 + n approaches 2·7182818 ... which is the
approximation given by the calculator for ex when x = 1?
1. Differentiate with respect to x:
(a) e-2x
(b) loge (1 + x) (d) (3x + 2)ex
(e) loge (5x + 2)
(f) lnx X
2. Find the equation of the tangent to the curve y = e3x at the point where x = 1. Find where this tangent cuts they-axis.
3. Find primitives for:
(a) e2x+3 1 (b) (5x- 4)
4. Find the area under the curve y = e2x from x = 1 to x = 3.
4x (c) xz + 1
5. Find I2
(ex - e-x)dx. What is the area under the curve y = ex - e-x from x = 1 to x = 2?
d2 d 6. If y =A + Be-41
, show that dJ. +4ft= 0.
d2 d 7. If y = e2
x + e4 X, show that d}. - 6dx + 8y = 0.
8. Differentiate with respect to x: (a) e.fi (c) 3x In x
(b) e~ (d) In (x - 1) 2
157
(e) ex'+1 X (f) In-2--
1 X +
9. Find these integrals (leave answers unsimplified)
(a) t1
x 2 : 3dx (b) t2
(5x + ~)dx 10. Show that the tangent to the curve y = ex at the point where x = 1 passes through the origin.
Find the area under the curve from x = 0 to x = 1 and also the area between the curve, the tangent mentioned above, and the y-axis.
11. Find the area under the curve y = f from x = 1 to x = 5. Also show that the area under this curve from x = 1 to x = e is 1 unit of area.
12. Find the area shaded in the diagram between the curve y = In x and the y-axis and bounded by the lines y = 0 (the x-axis) andy = 0·5.
Then find the volume of the solid of revolution formed if this region is rotated about they-axis. Hint: Note that y = In x can be rewritten in another form.
0.5
0
13. Find the area under the curve y =!(ex + e-x) from x = 0 to x = 3.
14. Calculate the volume of revolution if the region under the curve y = ex and between the lines x = 0 and x = 2 is rotated about the x-axis.
1 15. Prove that Iogb a = -
1 b. You may assume the change of base rule.
oga
16. Sum the series log10 2 + log10 4 + log10 8 + · · · to 10 terms.
17. Consider the function y = ~~ . X +
(a) Find any stationary values.
(b) By writing the curve as y = -1- 1 , show that as x becomes large y becomes very small.
x+-x What happens when x ~ - oo?
(c) Sketch the curve (it is called the serpentine curve). (d) Find the area under the curve from x = 0 to x = 2.
18. Sketch the graph of y = 1
. Find the area bounded by the curve, the x-axis and the lines 2x - 1
x = 1 andx = 3.
19. By noticing that-1- - __1_
2 = +-
4 find I 6
~4dx. (Leave answer without using x-2 x+ x-
3x-
tables or calculator).
20. Show that I~ : ~ dx = x + 6loge (x + 1) + c.
21. Use numerical integration to find an approximate value for the area under the curve y = f from x = 1 to x = 3 and check by formal integration.
22. Use Simpson's Rule to find the area under the curve y =__!_from x = 2 to x = 4 and check 2x
by formal integration.
158
1. Differentiate with respect to x: (a) 4x3 + JX (b) esx- 2
2. Write down the primitive function of:
(a) 3x2 + 2x + 1 (b) JX 1 (c)-
x+2
3. Find the values of kif x 2 + (k - l)x - (2k + 1) = 0 has equal roots.
4. Sketch the curve y = 4 - x 2 and find the area cut off between the curve and the x-axis.
5. Two standard dice, with numbers 1 to 6 on their faces, are thrown. What is the probability that: (a) they both show 6? (b) they show a 1 and a 6? (c) at least one of them shows a 6? (d) they show a total of six?
6. The lines AB and AC have equations 2x - 5y = 11 and 5x + 2y = 3. Show that they meet at right angles and find the coordinates of the point A.
1. Differentiate:
(a) x + e2x (b) (5x- 1)5
2. Evaluate:
(a) J: (3x2 + 2x)dx (b) r3
(4 - 2x)dx
3. A function is defined by the following rule:
l-2ifx~-5
f(x) = 0 if -5 < x < 2 xifx~2
Find: (a) f( -6) + f(l) + f(6) (b) f(a 2 + 2).
(c) 2x + 1 3x - 2
(c) J ~2 (6x2
- 4)dx
4. ABCD is a parallelogram in which DC = 8 em and BC = 4 em. E is a point on CB produced such that DE cuts AB at F, where AF = 5 em. (a) Calculate the length EB. (b) If angle DAB is 110 degrees find the area of triangle EBF.
5. Find the stationary points of the curve y = x3 - 7x2
- 5x and use the calculus to determine their nature.
6. Given the geometric sequence 2, -6, 18, ... find the smallest value ofn such that IT..I > 5000.
159
1. (a) Express 2x2 + 3x + 1 in the form A(x - 1)2 + B(x - 1) + C. (b) For what values of k is the quadratic expression x 2 - 6x - 2k positive definite?
2. Solve each of the following: (a) -5 ~ l - x < 2 (b) lx- 11 = 5
3. Find the first and second derivatives of xe2x.
4. For the curve y = ax2 + bx + c where a, b and c are constants, it is given that : = y when
x = 2. Show that b = -c.
5. Find the area between the curve y = x 2 - 4, the x-axis and the ordinates at x = l and x = 3.
6. (a) Find the centre and radius for the circle whose equation is x 2 + y2 - 4x + 6y = 3. (b) Draw the graph of the circle and shade the region where the following inequalities both
hold: x 2 + y2 - 4x + 6y < 3 and x > 0.
1. (a) Find the exact roots for the quadratic equation 2x2 + 2x - 3 = 0. (b) Solve the equation x4
- 5x2 + 4 = 0.
2. Find the point on the curve y = 3x2 + 5x - 2, which has a tangent parallel to the straight line y = 2- X.
3. Find: for each of the following:
(a) y = 3x2 - x (b) y = (x2 - 3)2 x3 (c) y = x 2 loge 2x
4. (a) What is the condition that the numbers a, b and c should be in geometric progression? (b) Show that the series 12 - 4 + lj- - · · · is geometric and find its limiting sum.
5. A point P(x, y) moves so that it is always equidistant from the points A(- 3, 4) and B(l, 2). Find the equation of its locus.
6. Find the indefinite integrals:
wf~+~~ ~JG+~~ (c) f 3 ~ 2xdx
d2 d 1. If y = (1 + 3x)e2
x, show that d {- 4dy + 4y = 0. X. X
2. In /:,ABC, LA = 80°, LB = 40°, a = 6 em. Find the length of side b.
3. What is the equation of the line passing through the point of intersection of the two lines x + 2y - 3 = 0 and 2x - y - l = 0, which also passes through the point (4, 5).
4. Find the volume of the cone formed when the area between the line y = 3x, the x-axis and the ordinate at x = 2 is rotated about the x-axis.
5. (a) Find all the roots of the equation x4 - 13x2 + 36 = 0.
(b) Find the values of k for which the quadratic equation x 2 - (k + 3)x + (k + 6) = 0 has real roots.
160
6. An integer is selected at random from the first 40 positive integers. What is the probability that it is: (a) divisible by 4 and 5? (b) divisible by 4 or 5? (c) not divisible by 6?
1. Find the derivative of each of these expressions:
(a) loge 2x (b) xex1 (c) (2xz + It X
2. Write down the primitives of the following functions:
(a) X - __!__ (b) fx - _l (c) (2x + 1)2
xz V-" JX 3. Find the equation of the parabola having:
(a) focus(0,2),directrixy = -2 (b) focus(O, l),vertex(O, -1)
4. For a certain curve ~~ = 6x - 10. Find the equation of the curve if it passes through the
point (1, 1) with gradient -1.
5. Find the volume enclosed by the surface generated when the curve x 2 + 4y 2 = 16 is rotated about the x-axis.
6. For what domain of x is the function x 3 - 9x2 + 24x + 2 increasing? Find the maximum and
minimum points on the curve and the point of inflexion. Sketch the curve.
161
In designing the Sydney Opera House, the engineers made the surface of every shell part of a sphere .
• r1
cti
Photograph courtesy of the Sydney Opera House Trust
tric
CHAPTER 20
In earlier work involving trigonometry the degree was used as a unit of measurement. It was a convenient unit for practical, numerical measurement but when dealing with trigonometric functions a new unit, called the radian is more appropriate. On your calculator there is a selector switch which may be set to radians or degrees.
In a unit circle, with centreD, two radii DP, DQ are drawn so that: the arc length PQ = radius DP = radius DQ
= 1 unit. The angle PDQ subtended at the centre by the unit arc PQ is defined to have a measure of one radian. p
Definition: A radian is the measure of the angle which an arc of unit length subtends at the centre of a unit circle.
If, in a circle of unit radius, an angle RDS has its vertex at the centre, its radian measure is equal to the length of arc on which it stands. If arc length RS = x units then L RDS = x radians.
Examples: (i) If the length of arc RS = 1· 3 units, then L RDS = 1· 3 radians.
(ii) If the arc RS measures 0·73 units, then L RDS = 0·73 radians. As in earlier trigonometry, if the rotation is anticlockwise the angle is taken as positive; if clockwise
as negative. Hence in the unit circles below: ·
a
L PDQ = ~radians. LPDQ = -2·25 radians.
164
In this way, any arc length, which may be measured by a real number, may be interpreted as an angle measured in radians. When theoretical work is being done with the trigonometrical functions, the radian measure is always used. The degree measure is still of vital importance in practical situations such as carpentry and surveying.
It is important to be able to convert between radians and degrees. This may be done by means of tables or on some scientific calculators there is a key to automatically carry out the conversion.
However, equivalent values may be found by means of a simple relationship. The circumference of a circle is given by
C = 2nr
h · h . . 1 b d fi d h . f circumference d . . 1 1 w ere n 1s t e trratwna num er e me as t e rat10 o d' an ts approxtmate y equa to
3·14159. In the case of the unit circle, the circumference is
C = 2n (as r = 1). Thus, the length of the semi-circular arc PQR = n units. If follows that the straight angle POR measures n radians.
tameter
R
c=2rr
n is the length of the semicircular arc in a unit circle and hence is the measure in radians of the straight angle PO R.
Now the number of degrees in a complete revolution is 360 degrees and this corresponds to an arc length of 2n units and so the radian measure of a complete revolution is 2n.
It follows that . 180°
1 radtan = -n
~ 57·2958°
and 1 degree = 1 ;O radians
~ 0·01745 radians.
Hence 2n radians = 360° and
The circular measure of angles involves the measurement of angles in radians.
165
Examples: (i) Express 90° in radians.
180° = n radians
1o n d' = 180
ra tans
. 90o = 90 x n . . 180
n d' = 2 ra mns.
(iii) Find the equivalent of~ radians in degrees.
n radians = 180° n . 180° - radmns = --3 3
= 60°.
(ii) Express -450° in circular measure. 180° = n radians
1 o = 1~0 radians
_ 4500 = _ 450n 180
5n d' = --ra tans 2
(on dividing numerator and denominator by 90)
1. Express the following angles in radians, leaving your answers in terms of n. (a) 30° (f) 210° (k) -60° (b) 45° (g) 300° (1) 54° (c) 60° (h) 420° (m) -270° (d) 120° (i) 22t0 (n) 480° (e) 135° (j) 18° (o) 262t0
2. State in degrees the following angles measured in radians. (a) -r (d) tz (g) 9"o (j) 5s" (b) 3n (e) 54' (h) I~ (k) ?2"
(c) 43 (f) 4n (i) I~ (1) i
(m) _z9, (n) s3" (o) - 7s"
3. Using a calculator, express in degrees (correct to four decimal places) the following angles which are given in radians. (a) 1·5 (b) 2·0
(c) 1·2 (d) 0·05
(e) 2·75 (f) 0·3
(g) -3·15 (h) 6·28
4. Give the circular measure of the following angles, correct to four decimal places. (a) 28° (c) 113° (e) 116·5° (g) 298°30' (b) 77° (d) 137° (f) 97·35° (h) -317·652°
5. For each of the following angles, give its radian measure in terms of n and write its sine and cosine as surds. (a) 60° (b) 225°
(c) 300° (d) 120°
(e) 135° (f) 420°
6. For each of the following angles, give its measure in degrees and evaluate its sine and tangent ratios, leaving the answers as surds. (a) 2
3" radians (c) 8:f radians (b) 3n radians (d) 3
2" radians (e) 1 ~" radians (f) 5
4" radians
7. With your calculator set to radian mode, find correct to 4 decimal places: (a) sin~ (d) sin 0·6 (g) cos i2 (j) tan (5n - 1) (b) cos !1: (e) cos 5·8 (h) tan 3·7 (k) sec 34 (c) tan 1·2 (f) tan 7
5" (i) sin fo (1) cot 3
166
8. (a) Two angles of a triangle are~ and 23" radians. What is the measure of the third angle in degrees?
(b) Three angles of a quadrilateral have as their measures-!,~ and 56" radians. Express the size
of the remaining angle in degrees. (c) What is the size of each interior angle of the following regular polygons? Give your answers
in radians. (i) hexagon (ii) octagon (iii) dodecagon (12 sides).
(d) The sum of two angles is 2330" radians and their difference is 1l0 radians. Find the size of each
angle in degrees.
In a circle of radius r units, an angle measuring e radians stands on an arc of length I units.
Now, as the length/, of the circular arc increases so does the angle at the centre.
Finally, the arc becomes the circumference of the circle and 8 becomes a complete revolution (2n radians).
Hence, length of arc, I units
circumference of the circle (in the same units) e radians
2n radians e
and 1 2nr 2n . 1 = 2nre
· · 2n and I= re.
In a circle of radius r units, the length I units of the arc AB subtending an angle of e radians at the centre is given by
l = re.
From this formula it follows that the measure of an angle e in radians is given by
e d. arc length
ra 1ans = . length of radms
Now consider the sector AOB (shaded in the above diagram). As the angle e increases, so does the area of the sector. Finally, when e = 2n radians the area of the sector becomes the area of the circle (nr2
).
H area of sector AOB e radians ence, .
area of the circle 2n radians· Now, let the area of the sector AOB be A square units.
Then, A e
167
In a given circle of radius r units, a sector contains an angle of 8 radians at the centre. The area of the sector is given by
Example (i): Find the length of the arc in a circle of radius 6 em, subtending an angle of 0·86 radians at the centre. Find also the area of the corresponding sector.
Solution: I= r8
= 6 X 0·86 = 5·16
length of arc PQ = 5·16 em A= !r28
= ! X 6 X 6 X 0·86 = 15-48
area of sector POQ = 15-48 cm2•
Example (ii): Find the angle, in radians and degrees, subtending an arc of length 9 em in a circle of radius 8 em.
Solution:
8=!_ I'
_)/_ -s = 1·125
.'. angle 8 = 1 · 125 radians
Example (iii):
1·125 X 180 d egrees n
~ 64·4578 degrees by calculator.
A sector of a circle, which contains an angle measuring~ radians at the centre has an area 4· 7059 cm2•
Find the radius of the circle correct to four decimal places.
Solution: A= !r28
4·7059 = !r2• ~
4·7059 X 2 X 6 2 ------=1'
n
,',I'= J12 X :·7059
. ·. radius ~ 4·2397 em by calculator.
Example (iv): In a circle of radius r, show that the minor segment (shaded), subtending an angle of 8 radians at the centre, has an area given by !r2(8- sin 8). Find (a) the area of the segment and (b) the length of the chord PQ when r = 10 em and 8 = 45°.
Solution: Area of minor segment = area of sector 0 PQ - area of triangle 0 PQ
= !r2 8 - !r2 sin 8 (area of triangle = !ab sin C) = !r2 (8 - sin 8).
168
p
/
r //
0 ~) ',
0
l area of minor segment = !r2 (8 - sin 8)
(a) 8 = 45° = *radians. . area of minor segment = ! x 102 (* - sin *)
= 50(* - 0·707107) ~ 3·91456 cm2 by calculator.
(b) Draw OR _l PQ. Then L ROQ = ! L POQ = -if radians
sin ROQ = ~~ RQ = 10 sin -if
chord PQ = 20 sin -if ~ 7·6537 em by calculator.
1. A circle has a radius of 3· 5 em. Find: (a) its area and circumference. (b) the length of arc subtending an angle of 1· 7 5 radians at the centre. (c) the area of the sector formed by the arms ofthis'angle.
p
2. For given circles, the following values of the radius are given together with vaiues of 8 for the angle subtended by an arc at the centre. In each case find the corresponding arc length. (a) r = 8 em 8 = 1 radian (d) r = 10 em 8 = 3 radians (b) r = 5 em 8 = 2·5 radians (e) r = 5·8 em 8 = 135° (c) r = 6 em 8 = 1·76 radians
3. Copy and complete the following table giving answers correct to three decimal places:
ANGLE AT CENTRE (8) ARC LENGTH (/)
3 radians
2 radians 3·6 em
8·21 em
60 degrees 4·82 em
34' radians 15·86 em ------------------------------------
4. The circumference of a circle is 198 em. Calculate the length of arc which subtends an angle of 120 degrees at the centre.
5. (a) Taking the radius of the earth as 6341 km, what angle would be subtended at the earth's centre by an Equatorial arc of length 9000 km?
(b) What arc length would be subtended by an angle of one minute at the earth's centre?
169
6. Find the area of the sectors in a circle of radius 10 em where the angles at the centre are; (a) 0·78 radians (d) 150 degrees (g) 34' radians (b) 1·04 radians (e) -if radians (h) 37 degrees (c) 60 degrees (f) 2
{ radians
7. A sector of a circle has an area of 28·86 cm2 and contains a central angle of -if radians. Find: (a) the radius of the circle (b) the arc length of the sector.
8. The area of a sector of a circle of radius 12 em is 188·5 cm2• Find the central angle contained in
the sector in : (a) radians (b) degrees.
9. Complete the following table for a given sector of a circle:
RADIUS ANGLE AT CENTRE AREA
(a) 15 em 90 degrees --------- ----------------- ----------
(b) 5·6cm !f radians --------- ----------------- ----------
(c) 6·4cm radians 47·9 cm2
--------- ----------------- ----------
(d) 10m radians 26·18 cm2
1--------- ----------------- ----------
(e) 56" radians 61·14 cm2
1--------- ----------------- ----------
(f) 1·762 radians 88·10 cm2
1--------- ----------------- ----------(g) 2·8m 2
3" radians 1--------- ----------------- ----------
(h) 63 degrees 48·06 cm2
r--------- ----------------- ----------
(i) 7·14 em degrees 100 cm2
10. A pendulum of length 1·2 m swings through an angle of~ radians. What is: (a) the length of arc traced out by the end of the pendulum? (b) the area of the sector formed by the swing of the pendulum?
11. What is the area of the sector of a circle of radius 5 em, if the length of arc bounding the sector is: (a) 5 em? (b) 3 em? (c) 9 em?
12. Find the area of the minor segment in a circle with: (a) radius 4 em and subtending an angle of 1· 32 radians at the centre (b) radius 5 em and subtending an angle of~ radians at the centre (c) radius 6·2 em and subtending an angle of 135 degrees at the centre.
13. (a) What is the area of the major segment of a circle in which the minor segment subtends a central angle of 56 radians and the radius is 7 em?
(b) What is the area of the minor segment cut off a circle of radius 10 em by a chord of length 12cm? Check your answer by the approximate formula
area of minor segment ~ Ubase) x height.
14. A circular piece of wire has a radius of 12 em. This is cut and bent to form an arc of a circle whose radius is 60 em. Find the angle subtended at the centre by this arc. Give your answer to the nearest degree.
170
15. A circular pond has an area of 615·75 m2• It is divided into sectors, one of which contains a
central angle of 23". Find: (a) the length of the arc in this sector (b) its area.
16. Find the shaded areas in the following figures where the arcs are quadrants of circles. (a) l (b)
17. In an equilateral triangle with each side 8 em, circular arcs, each of radius 2 em are drawn as shown below. Calculate the total arc length of the curve.
l E 0
"'"
.______.__.j f-0------4 em PI
If an angle is expressed in radian measure then the trigonometric ratios may be defined as functions of a real variable. For example, sin xis defined as the sine of an angle of size x radians and for every value of x there is one and only one value of sin x.
(a) The graph ofy =sin xfor -2n:::;; x:::;; 2n.
171
(i) The graph of y = sin x is continuous and periodic resembling a wave motion. The period or wave length is 2n, the distance between two successive crests or troughs on the curve. It can be seen that the graph repeats at intervals of 2n.
sin x = sin (x + 2n) (ii) The amplitude of the function is half the difference between its maximum and minimum
values. Hence the amplitude of y = sin x is 1. (iii) y = sin x is an odd function, sin (- x) = -sin x, and its graph has point symmetry about
the origin. (iv) For the given domain:
-the solution to sin x = 1 is x = - 32, ~· -the solution to sin x = 0 is x = - 2n, - n, 0, n, 2n.
(b) The graph ofy = 2 sin xfor 0 ~ x ~ 2n. For y = 2 sin x the period is 2n, that is, it is the same as for y = sin x. However the amplitude is doubled. What is the amplitude for
~~J•-----~~~~~t- y = 3 sin x?
(c) The graph of y = sin 2x for 0 ~ x ~ 2n . For y = sin 2x the amplitude remains the same as for y = sin x. However the period for this graph is n, that is half the period of y = sin x. We notice that the graph shows two full wave motions from 0 to 2n. For the given domain, the solution to sin 2x = 1 is x = !f, 54'.
Note: In this section of work it is sufficien.t to draw neat sketches of the functions which illustrate their essential characteristics. It is not necessary to plot a great many points.
1. On the same axes draw sketch graphs of: (a) y = sin x (b) y = 2 sin x (c) y = 3 sin x (d) y = 4 sin x and give the amplitude and period of each function. Note: For functions of the form y = a sin x, a gives the amplitude of the graph.
2. What is the amplitude of the graph of: (a) y = t sin x? (b) y = 5 sin x?
172
3. Draw sketch graphs of the following functions and give the amplitude and period for each. (a) y = sinf (b) y = sin4x (c) y = sin3x Note: For functions of the form y = sin bx, the value of b affects the period but not the amplitude of the graph. The period of y = sin x is 2n.
The period of y = sin bx is 2bn.
4. What is the period of the graph of: (a) y = sin 5x? (b) y = sin .p (c) y = sin nx?
5. Draw freehand sketch graphs of the following functions and give the amplitude and period of each. (a) y = -sin x (b) y = 2 + sin x (c) y = 2 sin 2x (d) y = sin (x + ~)
6. Use the graphs drawn in question 3 to solve the following equations. - (a) sin f = 0 for the domain 0 ::::; x ::::; 2n.
(b) sin 4x = 1 for the domain 0 ::::; x ::::; n. (c) sin 3x = 0 for the domain 0 ::::; x ::::; n.
The graph ofy =cos xfor -2n ::::; x::::; 2n.
-rr
-1
Period
2rr
(i) The graph of y = cos x has the same wave shape as the graph of y = sin x. In fact if the sine curve is translated a distance of~ to the left it falls on the cosine curve.
(ii) The period of y = cos x is 2n and the graph repeats at intervals of 2n. cos x = cos (x + 2n)
(iii) y = cos xis an even function, cos (- x) = cos x, and its graph is symmetrical about they-axis.
1. What is the amplitude of the curve y = cos x?
2. Over the given domain, use the graph of y = cos x to write down the solutions to the equations: (a) cosx = 1 (b) cosx = -1 (c) cosx = 0
3. Make freehand sketches of the following graphs. Give the period and amplitude of each. (a) y = 2 cos x (d) y = cos 3x (g) y = -cos x (b) y = 5 cos x (e) y = 2 + cos x (h) y = cos (x + i) (c) y = cos 2x (f) y = cos f (i) y = 3 cos 2x
173
4. On the same axes sketch the graphs of y = cos x andy = sin x. By using the addition of ordinates method, sketch the graph y = cos x + sin x.
., 5. On the same axes sketch the graphs of y = 4 cos x andy = 3 sin x from 0 :::;; x :::;; 2n. By adding
ordinates sketch the graph of y = 4 cos x + 3 sin x. What is the period and amplitude of the new graph?
The graph ofy =tan xfor -2n :::;; x :::;; 2n.
I I I I I
··----·+r----~-- -~~-~jl> I ~n 2n I I I I I I
Since tan x = sin x, tan xis not defined for cos x = 0. Hence the graph of y = tan xis disceintinuous COS X ,
in this domain for x = - 32", -~, ~' 32.
The period of y = tan x is n and it can be seen that the graph repeats at intervals of n. tan x = tan (x + n).
Draw sketch graphs of the following functions and give the period of each.
1. y = -tanx 2. y =tan 2x 3. y =tan i
Here is the graph of y = sec x for - 2n :::;; x :::;; 2n. The graph of y = cos x is shown dotted.
')1 --- j '
_3n 2 I
~-(\-' ~/:-~ I · I I I I I I I I l
I I
-1 r/\:, 1/ \
174
Since sec x = -1-, the graph of y = sec x can be sketched by using reciprocal values of cos x. We
COS X .
note' also that sec x'is not defined and its graph is discontinuous when cos x = 0. That is, when x = - 3
2", --I, 'f, 3z" the graph of y = sec x has vertical asymptotes. The range of values for y = cos x is - 1 ~ y ~ 1. The range of values for y = sec xis y ~ -1 andy ~ 1. The period of y = sec x is 2rc.
1. (a) Sketch the graph of y = sin x f.or - 2rc ~ x ~ 2rc. (b) For what values of x is cosec x not defined? (c) Sketch the graph of y = cosec x on the same axes as used for y = sin x.
2. (a) For what values of xis cot x not defined? (b) Draw in the asymptotes and sketch the graph of y = cot x for - 2rc ~ x ~ 2rc.
3. What is the period of the curve y = sec 2x?
Example: Find the graphically the number of solutions of the equation sin x = 1-. Solution: The diagram shows the graphs y = sin x and y = 1- drawn on the same axes.
The graph of y = 1- was drawn through (-f, !f) and (0, 0).
The two graphs intersect in three points showing that the equation sin x = 1- has three solutions.
1. Show graphically that the equation (a) cos x = x has one solution (b) cos x = i has three solutions.
2. Sketch the graphs y = sin 2x and y = 1- and find how many solutions there are to the equation sin 2x = 1'·
3. The graphs of y = sin x, y = j, y = j andy= i v are given. Read off approximate solutions to the following equations in the domain 0 ~ x ~ rc: (a) sin x = 1-(b) sin x = -f (c) sin x = i
175
0 2 3 X
4. (a) On the same axes, draw accurate graphs of y = 2 sin x, y = x, y = 1, y = 1 in the domain 0 ~ x ~ n.
(b) From the graphs find approximate solution to the equations (i) 2 sin x = x (ii) 2 sin x = 1 (iii) 2 sin x = 1
5. Solve the equation sin 2x = 1 in the domain 0 ~ x ~ n.
6. In the domain 0 ~ x ~ n draw the graphs of y = sin x and y = cos x. Solve the equation sin x =cos x.
1. (a) If 3 degrees equals x radians, find x, sin x and tan x to four decimal places. What is the value of cos x?
(b) For small x, what can you say about sin x, x and tan x? (c) Is it true to say for small x, that cos x ~ 1?
2. In the diagram, 0 is the centre of the unit circle and PT is the tangent at P. LPOA = x radians. It can be seen that: 60PA < sector OPA < 60PT. Find the areas: (a) 60PA (b) sector OPA (c) !::,.OPT and show that
sin x < x < tan x. If xis small
60PA ~sector OPA ~ 60PT.
Thus sin x ~ x ~ tan x for small x.
As seen above sin x < x < tan x.
Dividing through by sin x we have 1 < ~ < - 1-.
Sill X COS X
Now as x--+ 0, cos x--+ 1, hence lim~= 1.
1. DERIVATIVE OF sin x In Appendix XI we prove that
x-+0 Sln X
If follows that lim sin x = 1. x-+0 X
; (sin x) = cos x
176
2. DERIVATIVE OF cos x Using the above result we now find the derivative of cos x.
Since cos x = sin ( ~ - x) !!__ (cos x) = !!__ [sin (!!_ - x)] dx dx 2 .
= cos ( ~ - x) x (- 1) Using the function of a function rule.
=sin x x ( -1) = -sin x.
3. DERIVATIVE OF tan x Using both the above results and the quotient rule we find the derivative of tan x.
. sin x Smce tan x = -cos X
-anx----dx dx cos x d (t ) _ d (sin x)
_ cos x . cos x - sin x . (-sin x)
Example (i):
- cos2 x cos2 x + sin2 x
cos2 x 1
cos2 x =sec 2 x
Differentiate sin 3x.
Solution:
d dx (tan x) = sec2 x
Let y = sin 3x, then by the function of a function rule. dy d dx = cos 3x x dx (3x)
= 3 cos 3x.
Example (ii):
If y = cos (2x + 5) find dx. Solution: Using the function of a function rule.
dx = -sin (2x + 5) x :X (2x + 5)
= -2 sin (2x + 5).
177
Example (iii): Differentiate tan (3 - 5x).
Solution: Let y = tan (3 - 5x), then by the function of a function rule.
dy = sec2 (3 - 5x) x !!.__(3 - 5x) dx dx
= -5 sec2 (3- 5x).
Example (iv):
If y = ex tan x, find ddy . X
Solution: Using the product rule
dy =tan x. !!_(ex) + e·~. !!_(tan x) dx dx dx
= tan X • ex + ex . sec2 X
= ex(tan x + sec2 x).
Example (v): Differentiate cos2 4x.
Solution: Let y = cos2 4x = (cos 4x)2
, then by the function of a function rule d d Jx = 2 cos 4x x dx (cos 4x)
= 2 cos 4x [-sin 4x x ~ (4x)J
= 2 cos 4x x ( -4 sin 4x) = -8 cos 4x sin 4x.
Example (vi):
Find ~(log sin x).
Solution: Let y = log sin x, then by the function of a function rule
dy = -.-1- x !!_(sin x) dx sm x dx
1 =-.-X COSX
Slll X
=cot x.
Example (vii): 3
Differentiate ___:;__with respect to x. Slll X
Solution: x3
Let y = -.-, then by the quotient rule Slll X
. d ( 3) 3 d ( . ) dy = Slll X • Jx X - X • Jx Slll X
dx sin2 x sin x x 3x2
- x 3 x cos x sin2 x
x2 (3 sin x - x cos x) sin2 x
178
Differentiate the following with respect to x.
1. sin 6x 17. cos 3x - sin 2x ex
33. -.-Slll X
2. 3 tan x 18. tan 2x + sin 4x 34. sin3 x
3. cos 4x 19. 2 sin (3x - 4) 35. cos2 2x
4. sin !x 20. 4 cos(~- x) 36. log sin 2x
5. sin nx 21. x sin x 37.~ COS X
6. cos (2x + 5) 22. sin x cos x 38. log cos x '
7. pin (3x - 1) 23. x 2 tan x 39. e 2x sin 2x
8. sin 2x 24. ex sin X 40. e-x tan x
9. cos (3 - 2x) 25. x 3 cos 2x 41. cos (2x + ~) 10. 4 tan 2x 26. esinx 42. x 2 sin 2x
11. sin bx 27. 2x sin "t 43. 1- sinx 1 + sin x
12. cos (bx + c) 28. ex COS X e4x
44.--tan x
13. sin x - cos x 29. tan 2x cos x 45. sin~ X
14. 2 cos x + 3 tan x 30. sin x
46. )cos 2x
15. sin 3x + 2 cos x 31. _1_ 47 . sin 2x COS X cos 3x
16. tan 1- x 32. 48. log tan 3x tan x
The following primitives or indefinite integrals are inverse results of the derivatives in the previous section.
Since~ (sin ax) = a cos ax
then f cos axdx =~sin ax+ C.
f cos x dx = sin x + C
f sin x dx = -cos x + C
f sec2 x dx = tan x + C
179
This leads to the following more general results.
Example (i):
f cos ax dx = ~ sin ax + C
f sin ax dx = -~cos ax + C
f sec2 ax dx = ~ tan ax + C
Find the primitive function of cos 3x.
Solution: f cos 3x dx = t sin 3x + C
Example (ii):
Find f sec2 (4x + 3)dx.
Solution:
f sec2 (4x + 3)dx = t tan (4x + 3) + C
Example (iii):
Evaluate 1~(1 + sin 2x)dx.
Solution:
J:(l + sin2x)dx = [x- ~cos2x]~ = [~- ~(-1)]- [o- ~(l)J
1t 1 1 =2+2+2
n =2+1.
Example (iv): Find the area bounded by the curve y = cos x and the x-axis between x = 0 and x = ~·
·Solution:
A= 1~cos xdx
n
y
= [sinxF 0
= (1) - (0) =1.
The area is 1 square unit.
180
X
1. Find the primitives (indefinite integrals) of: (a) sin 2x (f) sin (2x + 3) (b) cos 5x (g) cos (1- - x) (c) sec2 x (h) cos x - sin x (d) sin 3x (i) sec2 (4x + 3) (e) sec2 1 (j) cos 3x + sin 2x
2. Evaluate the following definite integrals.
(a) I*sec2 xdx (d) I' (1 + cos x)dx
(b) L-I sin x dx (e) J; cos x dx
(c) I-I sin 2x dx (f) t-I (sin x + cos x)dx
3. Find the area under the curve: (a) y = sin x between x = 0 and x = n (b) y = cos x between x = 3
2" and x = 2n (c) y = sin 2x between x = 0 and x = !f (d) y = 2 cos 3x between x = 0 and x = ~ (e) y = 3 cos x + 4 sin x between x = 0 and x = £.
(k) 1 -cos 3x (1) sin (n - x) (m) x - sin (2x - 3) (n) 1 + sin (-I - x) (o) sec2 2x - sin 4x
(g) I1
sin nx dx
(h) t1
(cos -Ix)dx
(i) f-~ 2(x + sec2
x)dx
4. (a) Show that the volume of the solid formed, when the area under the curve y = sec x between x = 0 and x = £,is rotated about the x-axis is given by
V = n J:sec2 xdx.
(b) Find this volume.
5. Differentiate sin3 x and use the result to evaluate
t-I sin2 x cos x dx.
1. Differentiate : nx (a) cos 3 (c) tan nx
(b) sin2 x (d) e2 x sin 2x
2. Find the second derivative of: (a) sin 5x (b) cos 4x
d2y 3. If y = 3 sin x- 4 cos x show that dx 2 + y = 0.
4. Iff(x) =sin 2x, findf'(i) andf"(O).
5. If y = sin X0
, express X0 in radians and hence find 1x.
181
(e) log cos 3x
(f) sin 3x 3x
6. Find:
(a) f sin 3x dx (b) f sec2 4xdx (c) f cos nxdx
7. Evaluate :
(a) J:cos 2xdx (b) J:(l + sec2 x)dx (c) J: (sin 3x - x)dx
8. If y = tan 2x0, find Jx. First express 2x0 in radian measure.
9. Use the identity sec2 e = 1 + tan 2 e to show that
f tan 2 e de = tan e - e + c.
10. Differentiate esinx and hence find f: cos X esinx dx.
11. Differentiate tan 3 x and hence find f: tan2 x sec2 x dx.
12. Differentiate loge sin x and hence find f cot x dx.
13. Sketch each of the following curves and hence find the area under the curve from x = 0 to x = *· (a) y = cos x (b) y = sin 2x (c) y = 3 cos x (d) y = 2 sin h.
14. On the same diagram sketch the curves: (a) y =sin x andy= sin2 x (b) y = cos x andy = cos2 x.
15. Find the gradient of the curve y = sin x at the point where (a) X = 0 (b) X = ~ (c) x = -I
16. Sketch the graph of y = 2 cos 3x for - n ~ x ~ n. What is the period of the function 2 cos 3x?
17. If:= cos2xandy = 1 whenx = !,findywhenx =-f.
18. Sketch the curve y = 2 sin -fx from x = -1 to x = 1. Find the area bounded by this curve, the x-axis and the ordinates at x = - 1 and x = 1.
19. Find the maximum and minimum values for y = sin 2x in the domain 0 ~ x ~ n.
20. The range of a shell fired from a gun having an elevation of e radians is given by R = V2
sin 28 g
where V and g are constants. For what angle of elevation will the range be a maximum?
21. Use Simpson's Rule with four equal subintervals to find an approximation for t1
tan x dx.
182
Air pressure at an altitude of x metres is given by the exponential formula P = P0 ekx where k = -0.00012.
Photograph courtesy of Qantas
I I
CHAPTER 21
s ~I
-~---~~-~~~ -~~~~~~~--'
0 p
A particle P moves in a straight line. Suppose its distance, s metres, from a fixed point 0 after t seconds is given by s = t2
•
The distance-time graph would appear as shown with distance plotted on the vertical axis and time on the horizontal axis.
VELOCITY Now velocity is the rate of change of distance with respect to time. ·
Suppose we wish to find the average velocity of the particle between the second and fourth seconds.
A 1 't distance travelled verage ve oct y = . k
ttme ta en The average velocity is represented graphically by the
average gradient of the graph between the points P(2, 4) and Q(4, 16)
. 16- 4 12 i.e. Average veloctty =
4 _
2 = 2 = 6.
s
16
14 T
12
10
8
6
4 p
2
2 3 4
The instantaneous velocity at timet, is given graphically by the gradient of the tangent to the curve at timet.
Thus the instantaneous velocity at any point is given by the value of ~ at that point.
Now
thus
s = t2
ds = 2t dt .
ds Hence at P where t = 2, v = dt = 4.
The velocity after 2 seconds is 4 metres per second.
The symbol~; is sometimes written s, the dot denoting differentiation with respect to t.
If a particle moves in a straight line and its distances from a fixed point in the line is expressed as a function of time t, then its instantaneous velocity v at any time t is given by
ds v = dt'
184
Example: A particle moves on a line so that its distance, s metres, from the origin at time t seconds is given by s = 4 - 3t + t2
• Find: (a) the distance of the particle from the origin at the start and after 2 seconds. (b) the velocity of the particle when t = 2 (c) at what instant the particle comes to rest. (d) the closest distance of the particle to the origin.
Solution: (a) s = 4 - 3t + t2
At the start t = 0 If t = 0, s = 4 . ·. at the start the particle is 4 metres from the origin. If t = 2, s = 4 - 6 + 4 = 2 . ·. after 2 seconds the particle is 2 metres from the origin.
ds (b) v = dt = - 3 + 2t
when t = 2, v = - 3 + 4 = 1 . ·. after 2 seconds the velocity is 1 metre per second.
(c) The particle is at rest when v = 0, i.e. when - 3 + 2t = 0 giving t = Ii
.'. The particle is at rest after I! seconds. (d) The closest distance to the origin occurs when sis a minimum
i.e. when: = 0 and this occurs when t = Ii If t = I!, s = 4 - 4! + 2t = It .·. The closest the particle comes to the origin is Ii metres.
ACCELERATION When velocity is not constant and changes with time the rate of change of velocity with respect to time is called acceleration. Acceleration must therefore be measured in units of velocity per second. For example in (metres per second) per second.
If the velocity v of a particle is expressed in terms of timet, then the instantaneous acceleration · a at time t is given by:
dv d2s ( . ds)
a = dt or a = dt 2 smce v = dt
Acceleration is sometimes written s where the two dots indicate double differentiation with respect to t.
Example (i): A particle moves in a ~tiaight line such that its distance, s metres, from a fixed point 0, in the line, at the end oft seconds is given by s = 12 + 3t + 2t3
• At the end of 4 seconds find: (a) its distance from 0. (b) its velocity. (c) its acceleration.
185
Solution: s = 12 + 3t + 2t3
v = ds = 3 + 6t2
dt dv
a=-= 12t dt
(a) Using s = 12 + 3t + 2t3
when t = 4, s = 12 + 3 x 4 + 2 x 43 = 152 . ·. After 4 seconds the particle is 152 metres from 0.
(b) Using v = 3 + 6t2
when t = 4, v = 3 + 6 x 42 = 99 . ·. After 4 seconds the velocity of the particle is 99 metres per second.
(c) Using a = 12t when t = 4, a = 12 x 4 = 48 . ·. After 4 seconds the acceleration of the particle is 48 metres per second per second.
Example (ii): A point moves in a straight line so that its distance s metres from a fixed point 0 at time t seconds is given by s = 8 + 12t- 4t2 + lt3
.
(a) Find the position of the point when t = 0. (b) Find the equations for velocity and acceleration. (c) At what times is the velocity zero and what is the acceleration at these times?
Solution: (a) s = 8 + 12t - 4t2 + lt3
when t = 0, s = 8 . ·. The point is 8 metres from 0 when t = 0.
(b) s = 12 - 8t + t2
§ = -8 + 2t (c) If the velocity is zero 12 - 8t + t2 = 0
(t - 2) (t - 6) = 0 t = 2 or 6
When t = 2, § = - 8 + 2 x 2 = -4 . ·. When t = 2 the acceleration is -4 metres per second per second.
When t = 6, § = - 8 + 2 x 6 = 4 . ·. When t = 6 the acceleration is 4 metres per second per second.
Note: A negative velocity means that the point is moving in the direction of s decreasing. A negative acceleration means that the velocity is decreasing. For example a negative acceleration could cause a velocity of 8 metres per second to decrease to 6 metres per second or a velocity of - 8 metres per second to decrease to -10 metres per second.
1. A particle is moving in a straight line so that at time t its distance from a fixed point is given by s = 2t3
'- t2• At what time is its velocity zero?
2. An object travels in a straight line with velocity given by v = 4t5 • Find the acceleration at time t = 2 if t is in seconds and v in metres per second.
186
3. A particle is moving in a straight line so that at time t seconds, the distance from a fixed point is given by s = t3
- 2t + 1. At what time is the acceleration equal to 12 units per second per second?
4. A particle moves in a straight line so that its distance, s metres, from a fixed point 0, on the line, at time t seconds is given by s = t2
- 6t + s: Find: (a) the initial position of the particle. (b) the position of the particle when t = 3 and t = 4. (c) the distance travelled by the particle in the fourth second. (d) the initial velocity and the velocity when t = 4. (e) at what time the velocity is zero. (f) the acceleration of the particle.
5. A body falling under gravity falls s metres in t seconds, where s = 4·9t2.
(a) Find the distance the body falls in the third second. (b) Find the velocity after 3 seconds. (c) Find the acceleration of the falling body. (d) If the body hits the ground after4 seconds from what height did it fall and what was the velocity
at impact?
6. A body moves in a straight line; and s, the number of metres from a fixed point 0 after t seconds, is given by the formula s = 3 - 4t + t2
•
(a) How far is the body from the fixed point at zero time? (b) What is the velocity equation? Sketch its graph. (c) What is the velocity when t = 1 and when t = 3. What interpretation do you give to the
different signs of these two velocities? (d) When the velocity is positives is increasing. When does this first take place? (e) When does the body pass through the point 0? What interpretation do you give to the two
answers?
7. If the velocity of a particle, after t seconds, moving in a straight line is given by v = 10 + 6t - t2.
Find the acceleration when t = 2 and the greatest positive velocity attained by the particle.
8. ThedistancexmetresofabodyfromafixedpointOaftertsecondsisgivenbyx = 2t- 3t2 + 2t3•
Find the velocity and acceleration after 3 seconds.
9. A particle moves in a straight line in such a way that its distance, x metres, from a fixed point 0, in the line, after t seconds is given by x = 4t - 5t2 + 2t3
• Find: (a) the position of the particle when t = 0, t = 2, t = 3. (b) the difference in position from when t = 2 to when t = 3. (c) the velocity when t = 2. (d) the time when v = 0. (e) the value of x when v = 0. (f) the acceleration equation and calculate the velocity when the acceleration is zero.
10. A particle is moving in a straight line so that its distance, x metres, from a fixed origin 0, after t seconds, is given by x = l2t - t 3
•
(a) Show that the particle is at the origin initially and find its initial velocity. (b) Show that the particle moves away from the origin 0 for 2 seconds and that when it returns
to 0 its velocity is 24 metres per second in a negative direction.
11. A body rolls down an inclined plane so that after t seconds its distance x metres from the top, measured along the incline, is given by x = t + 4t2
•
(a) Find the velocity of the body when t = 2. (b) Show that the acceleration is constant and find its value. (c) If the velocity of the body is 25 metres per second at the bottom of the inclined plane, find the
time taken for the body to reach the bottom and the length of the inclined plane.
187
Example (iii): The distance x metres at time t seconds of a particle moving in a straight line, from a fixed point 0 is given by x = 2 cos 2t. (a) Sketch the graph of this equation. (b) Find the times when the particle is at rest and where the particle is at those times. (c) What is the maximum distance of the particle from 0?
Solution:
(a) X
( 27!
-2
(b) The particle is at rest when ~~ = 0.
X= 2 COS 2t dx
4 .
2 v =- =- sm t dt
v = 0 when sin 2t = 0 i.e. when t = 0, ~, n, 3
2", 2n. At these times x = 2, -2, 2, -2, 2.
(c) The maximum distance from 0 is 2 metres.
1. A particle moves along a straight line, its distance from the origin being given as a function of time by x = sin 2!, where xis measured in metres and tin seconds. (a) Sketch the graph of this distance equation. (b) Find the velocity of the particle when t = 0. (c) Show that the particle is initially at 0 and find when next it returns to 0. (d) Find the first two times when the particle changes direction and find its acceleration at these
times.
2. The distance x metres at time t seconds of a particle moving in a straight line from a fixed point 0 is given by x = 2 - 2 cos 2t. (a) Sketch the graph of this equation for 0 ~ t ~ 2n. (b) Find the times when the particle is at rest and where the particle is at these times. (c) What is the maximum distance of the particle from 0?
3. The displacement of a particle in a straight line is given by x = cost + sin t, where x is the displacement in metres after t seconds. Find: (a) the initial displacement.
(b) the displacement after J!: seconds. (c) the velocity when t = 0, t = Jl:, t = ~· (d) the acceleration when t = '3-, t = 34'. (e) sketch the graph of the displacement equation for 0 ~ t ~ 2n.
4. The displacement of a particle moving in a straight line is x metres after t seconds where x = e' - t. (a) Show the particle is at rest when t = 0. (b) What is the velocity of the particle after 1 second?
Let us now consider the problem of finding the distance equation of a point as a function of time, when given its velocity or acceleration equation as a function of time. This problem is solved by using primitive functions.
188
Since dv
v = f adt a= dt'
and since ds
s = f vdt v = dt'
Example (i): A point initially at the origin, at rest, has an acceleration which is given by a = 2 + St, where tis in seconds and a is in metres per second per second. Find its velocity and distance equations as functions oft.
Solution: dv
a=-= 2 + St dt v = 2t + ~t 2 + c
but v = 0 when t = 0, hence C = 0
thus v = ds = 2t + 512
dt 2 giving
but s = 0 when t = 0, hence C = 0 Thus s = t 2 + it 3
Example (ii):
Alternative set out using integral notation a= 2 + St
v = f (2 + St)dt
= 2t + ~t 2 + c but v = 0 when t = 0, hence C = 0 thus v = 2t + 4t 2
s = J (2t + ~t 2)dt ~~2 + it3 + c ~
but s = 0 when t = 0, hence c = 0 Thus s = t 2 + it 3
A particle, which was initially at rest at the origin 0, has an acceleration after t seconds of (6 + 2t) metres per second per second. What is its velocity at the end of 3 seconds, and what is the distance travelled in this time?
Solution:
d 2 s a=-= 6 + 2t
dt2
u v = ~; = 6t + t 2 + c Since the particle was initially at rest v = 0 when t = 0, thus C = 0.
ds 2 Hence v = - = 6t + t r-- - dt
Wh\~n t = 3~o= 6 X 3 + 32 = 27. That is, after 3 seconds the velocity of the particle was 27 metres per second.
Since ds = 6t + t 2
dt 2 t3
s = 3t + 3 +c. As the particle was initially at the origins = 0 when t = 0, thus C = 0
3
therefore s = 3t 2 + 1
3
33 When t = 3, s = 3 x 32 x 3 = 36.
That is after 3 seconds the particle has travelled 36 metres.
189
1. The velocity of a particle is given by v = 6t 2• If initially the body starts from the origin, find:
(a) the velocity after 3 seconds. (b) the distance equation as a function of time. (c) the displacement after 3 seconds. (d) when the particle is 16 metres from the origin.
2. A particle moves along the x-axis with an acceleration given by a = 6t i;f3. When t = 0, x = 5 and the velocity of the particle is 1 unit per second. Find the position of the particle after 3 seconds.
3. A particle starts from rest and moves so that its acceleration at time t is given by a = -6. If the particle is initially 27 metres from the origin, find: (a) velocity v in terms oft. (b) distances in terms oft. (c) the time at which the particle reaches the origin and its velocity at that time.
2 ,' ;
4. Given that ~t~ = l2t - 2, that~; = 0 when t = 0 and that s = 4 when t = 1, finds when t = 5.
d2 s 5. Given that dt 2 = 2t, v = 0 when t = 0 and s = 10 when t = 1, finds when t = 2.
6. Given that § = 12 - 2t and s = 16 when t = 0, find: (a) the velocity of the particle after 3 seconds. (b) the distance travelled by the particle in the first 3 seconds.
7. A particle starts from rest and moves so that its acceleration at timet is given by§= -4. If it is initially 18 metres from the origin find : (a) sin terms oft (b) sin terms oft (c) the time at which the particle reaches the origin and its velocity at that time.
8. A point, with initial velocity of 20 metres per second, starts from the origin and has an acceleration given by a = 8 - 2t. When does it first come to rest? How far has it moved in this time.
EXPONENTIAL GRAPHS Exponential functions are readily graphed using theBkey on a calculator to find ordered pairs on the graph.
Example: Draw the graphs of the following exponential functions which are of the form y = ekx. (a) y =ex (b) y = etx (c) y = e2x (d) y = e-zx
Solution: TABLE OF VALUES
X -2 -1 0 0·5 1 2 3
y =ex 0·14 0·37 1 1·65 2·72 7·38 20·08
Y = etx 0·37 0·61 1 1·28 1·65 2·72 4-48
y = ezx 0·02 0·14 1 2·72 7·38 54·60 403·4
y = e-2x 54·60 7·38 1 0·37 0·14 0·02 0·002
190
Note: 1. Ask increases in value the curve steepens. 2. y = e2
x andy = e- 2x are reflections in they-axis. 3. All these curves pass through the point (0, 1).
1. (a) Copy and complete the given table of values for functions of the type y = Aex.
(b) Sketch the curves. (c) What is the effect of
changing the value of A iny = Aex?
2. (a) Copy and complete the given table of values for functions of the type y = Aekx.
(b) Sketch the graph of each function.
y
y
y
y
X -2
y =ex
y = 2ex
y = 1·5ex
y= -2ex
X -2
= 2etx
= !etx
= -!etx
= -!e-tx
191
-1
-1
-0·5 0 0·5 1 2
-0·5 0 0·5 1 2
THE GROWTH AND DECAY FUNCTION The exponential function has many practical applications. It is involved in natural examples of growth and decay such as: population growth, growth of bacteria in a culture, the rate of cooling of a body, the rate of decay of radioactive substances and the rate of change of pressure with altitude, to name a few.
The exponential function y = Aekt is known as the growth function. Note that A and k are constants andy is a function oft. The independent variable is given as t since the growth function usually varies with time.
Note that if y = Aekt then ~ = kAekt
= ky.
Thus the growth function y = Aekt satisfies the equation dt = ky.
This equation states that the rate of change of y is proportional toy. For any practical situation for which this is true the growth function will be of the form y = Aekt.
If k is negative the y is decreasing continuously and the equation represents the decay function. Substituting t = 0 in the formula y = Aekt gives y = Ae0 i.e. y = A. Thus A represents the initial value of y and the formula is often given as y = y 0 ekt or N = N0 ek', etc.
The exponential growth and decay formula is y = Yoekt
where (i) y0 is the initial value of y when t = 0. (ii) k is the growth or decay constant for a particular population.
Example (i): The number of bacteria found in a certain culture is given by N = N0 e0
'12
', where tis in hours. Initially the culture population was 240 000 bacteria. How many were present after 5 hours?
Solution: The growth formula is N = N0 e0 ·12t.
From the data N0 = 240 000 .'. N = 240 000e0
'12
'
when t = 5, N = 240000e0 ' 12 x 5
= 240 000e0'60
=i= 437 308-48 (by calculator) After 5 hours there are approximately 437 000 bacteria present.
1. The number of people in a town is given by N = 12000e0'09
', where tis the number of years since 1980. (a) What was the initial population in 1980, i.e. when t = 0? (b) What would be the expected population in 1990, i.e. when t = 10?
2. The number of bacteria found in a certain culture is given by N = N0 e0 '241
, where tis in hours. Initially the culture population was 50 000 bacteria. How many bacteria would be present in the culture after: (a) 2 hours? (b) 3! hours?
3. The number of people in a town is given by N = 2000e0'3', where tis the number of years since
the end of 1970. (a) Draw the graph of the number of people in the town against time for t = 1 to t = 10. Use
your graph to find the year in which the population first exceeded 15 000. (b) Use your calculator to find the expected population of the town at the end of 1990.
192
4. In a certain chemical reaction the amount P grams of a substance which remains unchanged after t minutes is given by the equation P = P0 e- 0
"151
• If the initial mass of the substance is 200 grams, find the amount remaining after (a) 1·5 minutes (b) 4·5 minutes (c) 6 minutes.
5. The temperature K (in degrees Kelvin) of a certain star in t million years time is given by the equation K = 9000e-0 "1381. (a) What is the present temperature of the star? (b) Find what the temperature of the star will be in 2 million years time.
6. A piece of radioactive material having a mass of 500 grams is decaying according to the formula M = 500e-0
"251
, where Mis the mass in grams of material remaining after t years. (a) Draw a graph of this equation for values oft from 0 to 4. (b) Use your graph to find:
(i) what time elapsed before there was only 400 grams of material remaining. How long after this before there was only 300 grams remaining?
(ii) how much material remained after 15 months? Check this answer using your calculator.
Example (ii): The city of Lismore had a population of 20 904 in 1971 and this had grown to 21 650 in 1975. Find the annual growth rate as a percentage, correct to two decimal places, assuming that the growth rate is proportional to the population.
If the growth rate is assumed constant, what would be the population 1987?
Solution:
Since the rate of change of population is proportional to the population (i.e. ~ = kP) we can use
the exponential growth equation P = P0 ek1•
From the data P0 = 20 904 . ·. P = 20 904ekt
When t = 4 (1975) 21650 = 20 904e4 k . 21650 4k I.e. 20 904 = e
Taking logarithms of both sides to the base e
loge ~~ ~~~ = 4k
. ·. k ::\= 0·0087662 (by calculator) ::\= 0·88%
Thus the annual growth rate is 0·88%. In 1987, t = 16. Using this value fort and taking k = 0·0088 the population is given by
p = 20904eo·oossx16
~ 24064·567 (by calculator) Thus population in 1987 would be approximately 24100.
Example (iii): A ship is travelling at 6 metres per second when its motors are stopped. Its deceleration is proportional
to its velocity (i.e. ~ = kv) and after 100 seconds its velocity is 3 metres per second. Find when its
velocity will be 1 metre per second.
Solution: We can use the decay equation Substituting v0 = 6, v = 3 and t = 100
193
v = Voekt 3 = 6e1ook
0.5 = e1ook
Taking logarithms of both sides to the base e loge 0·5 = lOOk k ~ -0·0069314
When v = 1, 1 = 6e-o·o069314t
f, = e-0·0069314t
Taking logarithms of both sides loge f, = -0·0069314t .'. t ~ 258-49893 (by calculator)
Thus the speed will be I metre per second after approximately 258 seconds.
1. The mass of an organism growing in a pond after t days is given by M = M 0ek1• If the initial
mass is 1· 5 kg and k = 0·05, what would be the mass of organism after: (a) 5 days (b) 10 days
and (c) how long would it take the mass to double?
2. The population of a town has its rate of growth proportional to the population. If the annual growth rate is 0·02 and the original population is 2400, what would be the population in: (a) 4 years? (b) 10 years? (c) How many years would it take for the population to reach 3400?
3. A radioactive material decays according to the exponential function M = M 0 ek1• If the initial
mass is 50 grams and the mass after 10 years is 40 grams, find: (a) k (b) the amount present after 15 years. (c) the half-life of the material. (The half-life of a radioactive material is' the time taken for the
material to halve its mass.)
4. (a) Show that the half-life of any radioactive element is f(loge !) and so is independent of the initial mass.
(b) The rate of decay of radium 226 is -0·0004278. (i) Find its half-life to the nearest year.
(ii) How long would it take fort of the original to decay?
5. If money is invested to earn interest that is compounded continuously, the amount after t years is given by A = Pek1
, where Pis the principal invested, k is the interest rate expressed as a decimal and t the number of years. If $2000 is invested for 5 years at 10% interest compounding continuously, what amount would be due. Note: Because interest is not usually compounded continuously the formula generally used is
A = P 1 + - , where P 1s the pnnc1palmvested, r = , N = number ( r)"N . . . . percentage interest rate n 100
of years and n =number of times paid each year.
6. An object travelling through a gas is subjected to a retardation, proportional to its velocity. If initially the object has a velocity of 100 mfs, while after 4 seconds the velocity is 20 mfs, find its velocity after: (a) 8 seconds. (b) 10 seconds. (c) 20 seconds.
7. In a colony of bacteria, the number present initially is 8000. If the number doubles in 5 days, how many would be present after 12 days?
8. In a sample culture at zero time there were 4·6 x 106 bacteria. At the end of 5 days there were 7·2 x 106 bacteria. Find: (a) the growth rate. (b) the number expected to be present at the end of 12 days.
194
9. Air pressure at an altitude of a metres is given in kilograms per square metre by the formula P = P0 eka, where k = -0·00012. If the pressure at the earth's surface is 10000 kg/m2
, what is the pressure at an altitude of: (a) 5000 metres? (b) 15 kilometres?
10. The chemical nitrogen pentoxide is a solid which has a decay rate of -0·0005 when time is measured in seconds. If the initial mass is 10 grams: (a) how much remains after 500 seconds? (b) what is its half-life?
11. In 1961 the aborigine population in Australia was 7 5 309 and in the 1966 census was 80 207. Assuming that the rate of population growth is proportional to the population, find: (a) the growth rate. (b) the estimated aborigine population 1981.
12. To historically date objects less than 50000 years old, carbon 14 dating is used. Carbon 14 has a half-life of 5730 years. Before death animals and vegetation have a reading for carbon 14 of 12·5 counts per minute on a Geiger counter. Show that the decay rate for carbon 14 is -1·21 x 10-4
.
If a piece of wood from an excavation site has a reading of 7 counts per minute, show that the wood's age is approximately 4800 years.
195
1. Differentiate the following with respect to x. (a) (2x + 1)5 (b) J2x- 5 (c) ecosx (d) loge (x 2 + 1)
2. Find the value of:
(a) L (x - x 2)dx (c) r: (2 sin X + 3 COS X)dx
3. a, b, 8 are in arithmetic progression while b, a, 21~ are in geometric progression. Find a and b.
4. The adjacent sides of a parallelogram are 10 em and 15 em respectively. If the shorter diagonal is 16 em, find the angles of the parallelogram and its area.
5. Find the stationary values and point of inflexion for the function y = 2x3 + 3x2 - 12x + 8.
Sketch the corresponding curve and show that it cuts the x-axis at just one point.
6. ABCD is a quadrilateral and X is the mid-point of the diagonal BD. From X parallels are drawn to BA and BC, meeting AD and DC in Yand Z respectively. Prove that YZ is parallel to AC.
1 F . d dy 'f· · m dx 1
·
(a) y = sin3 x (b) y = (x2 + 1)ex 1 (c) y = (3x - 1)4 (d) y = sin x cos x
d2y dy 2. Given that - 2 = 2 - 3x2
, and when x = 0, y = 10 and -d = -5, find y when x = 3. ~ X
3. If A(x - 1)(x - 2) + B(x - 1) + C = 2x2 - 3x + 5, find A, Band C.
4. Find the equation of the tangent to the curve y = x 3 + 22 at the point where x = 1.
X
5. Find the area enclosed between the curve y = 1 - x 2 and the x-axis. This area is rotated about the y-axis. Find the volume of the solid of revolution.
6. A particle is moving in a straight line, its displacement from a fixed origin 0 after t seconds is given, in metres, by x = t 3
- 3t2 + 4t - 7. What is the velocity when the acceleration of the particle is zero?
1. The gradient at any point (x, y) on a curve is given by 1 + 3x2. If the curve passes through the
point (1, 4), what is its equation?
2. Evaluate: 30 12
(a) I 3n + 5 (b) I 3(2)11-\ leave your answer in index form.
11=1 11=1
3. If y = 8 - 3
6x, find ddy and dd
2
{. Find the value of x for which each of the functions y, ddy and X X X X
~;;.is separately zero. Find a minimum value of y for positive values of x. 196
4. In a circle of radius 12 em an arc subtends an angle of 40 degrees at the centre. Find: (a) the length of the arc. (b) the area of the sector bounded by the arc and the radii to its end-points. (c) the area of the segment bounded by the arc and the chord joining its end-points.
5. In the quadrilateral ABCD, AD = BC and X, Y, Z are the mid-points of AB, AC, DC respectively. Prove that the triangle XYZ is isosceles.
6. The number of bacteria in a culture is given by the equation N = N0 ek1, where N0 is the initial
number of bacteria, k is the growth rate per hour of the bacteria population and t is the time in hours. If the number of bacteria in a culture was initially 50 000 and the growth rate is 15% per hour, find the number of bacteria present after 5 hours.
1. Find the derivatives of the following functions.
(a) (x2 - 1)(x + 5) (b) x2
-1
2x + 3
2. Find primitives for:
(a) 3x2 - 2x + t (b) JX + xJX
. 1 (c) -.-3-
sm x
2 (c) 5x- 3
3. (a) Complete the table of values for the function y = )2 + x 2.
(b) Use Simpson's Rule to evaluate r )2 + x2 dx.
4. If a and f3 are the roots of x 2 - 2x - 5 = 0, find the value of:
(a) a+ f3 (b) af3 (c) _!_ + _!_ a f3
(d) log e2 x
5. The acceleration of a body after t seconds is 2t + 2 metres per second per second. If initially the body was at rest 8 metres from the origin, find the displacement after 3 seconds.
6. With one throw of a die, what is the probability of: (a) throwing a 3 or a 6? (b) throwing an even number or a number less than 4?
1. The first 3 terms of an arithmetic sequence are -2, 4, 10. Find the twentieth term and the sum of twenty terms.
2. Sketch the graph of y = sin 2x for 0 ~ x ~ n. Find the area enclosed by the curve and the x-axis from x = 0 to x = -r.
3. (a) Find the solution to the equation 2x4 - 5x2
- 7 = 0. (b) If 3x2 + 5x + 4 = (x + 1)(ax +b) + c(x2 + 1), find a, band c.
4. Two straight roads XY and XZ diverge from X at an angle of 37 degrees. Car A travels at 60 km/h along road XY and car B travels at 72 km/h along road XZ. If both cars leave X at the same time how far apart are they after 2 hours 20 minutes?
197
5. Evaluate:
(a) rz X- l dx Jl X
(b) t2
(e2x - 1)dx
6. In a certain chemical reaction the amount P grams of a substance which remains unchanged after t minutes is given by the equation P = 20e-0
'15r. Find the number of grams of the substance
(a) at the beginning of the reaction. (b) remaining after 4·5 minutes.
1. (a) Sketch the curve y = ex and find the area of the region enclosed by the curve the x-axis and the ordinates at x = 0 and x = 2.
(b) If this area is rotated about the x-axis through 2rc radians, find the volume of the generated solid.
2. Evaluate:
n=l
00 ( 1)"-1 (b) I 5 -- . n=l 2
50
(a) I 5n- 3
3. Find the equation of the tangent to the curve y = x3 + 3x2 at the point of inflexion.
4. A window consists of a rectangle surmounted by a semi-circle having as its diameter the width of the rectangle. If the perimeter of window is 10 metres find the radius of the semi-circle which will give the greatest possible area for the window.
5. The area under the curve y = loge X from X = 1 to X = 5 is given by r loge X dx. Since we do
not know the integral of loge x use Simpson's Rule to find an approximation to this area. Use two sub-intervals 1 :::::; x :::::; 3 and 3 < x :::::; 5.
6. ABCD is a parallelogram in which the diagonals AC and DB are drawn. LABD = L CBD = 30°. Show that: (a) LADE= LCDB (b) L.BAD is isosceles (c) AC bisects LDAB.
198
1. Simplify (x + 3)(x - 4) - x(x - 1).
2. Find the factors: (a) x 2
- x- 2
3. Simplify 4x22 - 4x + 1. 2x - 4x + 2
4. If PV = pv(l + 1 ~0). find t if P = 7,
v = 140, p = 6, v = 130.
5 Sl x-5+x+5_ 5· . o ve 10 5
- .
6. If9x = 2\, find x.
. . x 2 - x x 2
- 5x 7. Stmphfy -;-. ---
x-3 3-x
8. Find the factors: (a) 10 + 3x - x 2
(b) x 2 + 2x - ax - 2a.
9. Simplify 3J2 + 4J8- J32. 10. Express Pi
2 with a rational denom-
...;3- 1 inator.
11. If 2 = m0'623 and 3 = m0
'847
, express as a power ofm the number (a) 6 (b) 9.
12. If (2.j5 - 3.jf0)2 = a + b.J2, find the value of a and b.
13. Solve 3x + 7y = 27} 5x + 2y = 16
14. Solve l3x - II = 5.
15. Given m = (J3 - .J2)2 + J24, find m in its simplest form.
b2 16. Use the formula e2 = 1 - 2 to find a
a when e = 0·29 and b = 5.
17. Solve 13 - 2xl < 6.
18. Solve (2x - l)(x + 4) < 0.
19. If tan 8 = f and 8 is acute, find cos 8.
20. In !::::.ABC; a = 7, b = 6, c = 5, find the largest angle of the triangle.
199
21. ABC is a triangle in which B = 135°, A = 30° and b = 10 em; find a in surd form.
22. Find the equation of a straight line which: (a) passes through (- 3, 5) and has a
gradient ofi. (b) is perpendicular to 3x - 2y - 7 = 0
and passes through (2, -1).
23. In a throw with two dice, find the probability of scoring (a) either 7 or 11
(b) more than 6.
24. In a family of three children, what is the probability of there being (a) 3 boys? (b) at least 2 boys?
25 S. l'f X + 2 X + 3 . tmpty-----. x+3 x+2
26. Simplify (J7 - 2J3)(J7 + 2)3).
27. Simplify a +~when a = .j5- 1.
28. Solve 3x - y = - 5 } y = x 2
- 2x- 1
29. Solve lx + 121 = l2x + 61.
1
2x- 1 I 30. Solve 3
< 4.
31. In !::::.ABC, AB = 17 em, AD = 15 em and LDAB = 132°. Find the area of f:::.ABD.
32. Find the 30th term of the sequence 6, 9, 12, 15, ...
33. In an arithmetic sequence T7 = 23 and T13 = 41, find the first term and the common difference.
34. For the arithmetic sequence 5, 12, 19, 26, ... , find T50 and S50 •
35. Which term of the geometric sequence 3, 6, 12, ... is equal to 768?
36. If x, x + 4, x + 16, are in geometric sequence, find x and the common ratio.
30
37. Evaluate L: (4n - 1). n=1
12
38. Evaluate L 5 x 2"-1•
n=l
39. Find the limiting sum of the geometric series in which a = 25 and r = 0·2.
40. Solve 2x2 + 3x - 14 = 0.
41. Find the roots in surd form for x 2 + 1 = 6x.
42. If 3 is a root of the equation x 2 + kx - 6 = 0, find the value of k.
43. The roots of the equation 2x2
- 7x + 12 = 0 are IX and f3. Find (a) IX + f3 (b) IX/3 (c) ~ + * (d) IX2 + /32
44. Solve for x if 2x4 - 26x2 + 72 = 0.
45. Find the value of k for which 3x2
- 5x + k = 0 has equal roots.
46. Show that 2x2 - 3x + 6 is positive for
all values of x.
47. For what values of x is (3x - 1)(2 - x) positive?
48. If 2x2 - 9x + 1 = Ax2 + Bx + C, find
A, Band C.
49. Express 4x2 - 6x + 5 in the form
A(x - 1)2 + B(x - 1) + C.
50. Find the value of k if the roots of the equation 3x2
- 15x + k = 0 differ by 3.
51. Find t if:
3 (a) y = 2x + 2 (b) y = (x2 + 3)4
X
3x (c) y = 2x - 1
52. Differentiate: (a) (x - 3)(2x2
- 2x + 4)
(b) .jx2 - 6
53. Findf'(x) ifj(x) = 5x3
; x. X
54. Find the gradient of the curve y = x 2 (x - 2) at the points where x = 1 and where x = - 2.
55. The curve y = 2x2 + bx + 8 has a gradient of 3 when x = 2. Find b.
56. Find the equation of the tangent to the curve y = x 3
- 4x2 at the point (1, - 3).
200
57. Find the equation of the normal to the curve y = 4x - x 2 at the point where x=l.
58. If x 2 + 4x = ax(x + 1) + b(x + 1) + c, find a, b and c.
59. Find the value of k for which x 2 + (k + l)x + k = 0 has equal roots.
60. The sum of a number and its square is 132. Find the number.
61. Find the limiting sum of the series 16 + 8 + 4 + 2 + 0 0 0
62. Express o-37 as a geometric series and find its limiting sum.
63. At what points on the curve y = ~1 x+ is the tangent parallel to the line y = x?
64. If f(x) = 2x3 - 4x + 5, find f(2), !'(2),
f"(2).
65. What is the second derivative of (3x - 2)4 ?
66. For what values of x is f(x) increasing if f(x) = 2x3 + 6x2
- 18x + 3?
67. Find the centre and radius of the circle x2 + y 2
- 4x + 6y + 9 = 0.
68. Find the primitive of x 2 - 3x + 1.
69. Ifj'(x) = 3x2 - 2x + 1, findf(x).
dy 1 70. - = 2 + 3, findy.
dx x
71. The gradient function of a curve is 7 - 4x and the curve passes through the point (1, 10). Find its equation.
72. What is the focal length of the parabola x2 = 20y?
73. Find the equation of the parabola with focus at (0, 3) and directrixy = -3.
74. Find the minimum value of2x2 - 6x + 7.
75. Evaluate I3
(2x2 - x + 1)dx.
76. Evaluate t4
(2x - 1)2 dx.
77. Find the area under the cu:r:ve y = x2 + 2 from x = 1 to x = 3.
78. Find the point of inflexion on the curve y = 3x3
- 3x2 + 3x - 6.
79. Find the domain for which the curve y = 4x2
- x3 is concave upwards.
80. For the curve y = 2x3 - 12x2
- Sx - 3, find the equation of the inflexional tangent.
81. Find the stationary points on the curve y = x3 + x2 + 1 and distinguish between them.
82. Given that ~: = 3t2 - t and that s = 11
when t = 2; express sin terms oft.
83. The gradient of a curve is given by dy = dx
1 + 4x - x 2• What is the equation of the
curve if it passes through the point (3, 0)?
84. Find the equation of the curve given d 2 d ______,[ = 3x - 1 and when x = 2 __Z = 7 ~2 '~
andy= 3.
85. Evaluate:
(a) f1
5
3dx fl 3
(b) 0
x'dx
86. Find the area enclosed by the curve y = 6x - x 2 and the x-axis.
87. Find the indefinite integrals
(a) f ( 6x2
- ~3) dx (b) f (3x + 2)2
dx
88. Find f .)6x + 1 dx.
89. Find the area enclosed by the curve y = x(x + 2), the x-axis and the ordinates at x = - 1 and x = 1.
90. Find the area bounded by the curves y = 3x - x 2 andy = 3 - x.
91. Using the trapezoidal rule and three sub-intervals, find an approximation for
{
4
2xdx.'
92. Using Simpson's Rule with two equal sub-intervals find an approximation for 13
logexdx.
93. Find f"(4) if f(x) = logex.
94. Find the volume of revolution if the area below the graph of y = f between x = 1 and x = 3 is rotated about the x-axis.
201
95. Differentiate with respect to x: (a) 3x2
- 4e2x (b) loge 5x (c) exlogex
96. Find the length of the arc of a circle of radius 15 em which subtends an angle 2
9"
radians at the centre.
97. Find the area of the sector in a circle of radius 20 em where the angle at the centre is 1·86 radians.
98. A segment of a circle has an area of 50 cm2
and subtends a central angle of~ radians. Find the radius of the circle correct to one decimal place.
99. Draw sketch graphs of: (a) y = 4 cos x (b) y = 2 sin 2x
100. Differentiate: (a) sin 3x (b) tan(2x - 1) (c) 3 cos (I - x)
101. Find: if y = ecosx.
102. Find the derivative of 2x tan x.
103. Find the primitive of sin(4x + 2).
104. Evaluate {1
(2 - cos x) dx.
105. Find the area under the curve y = 2 sin x between x = 0 and x = ~·
106. Show that the derivative of loge cos x is -tan x.
A particle moves in a straight line and its displacement from the origin is given by s = 5 - 6t + t2
, find :
107. the distance of the particle from the origin after 2 seconds.
108. the times when the particle is at the origin.
109. the velocity of the particle when t = 5.
110. at what instant the velocity is zero.
111. the acceleration of the particle.
112. The velocity of a particle is given by 3t2 + 2. If the initial displacement is 10 metres from the origin, find the displacement after 4 seconds.
113. If .X= -2 and .X= 10 when t = 1, find the velocity function.
114.
115.
116.
117.
118.
119.
If d2
s = 3t - 4 find s given that when dt2
'
ds t = 0 s = 0 and -d = 0.
' t
The population of a town grows according to the formula N = N0 ek
1• If the popula
tion grows from 40 000 to 50 000 in 2 years, find k and the population after a further 6 years. Give answer to nearest thousand.
In a certain vehicle the cost $C of making 392
a specified trip is given by C = 2 V + V, where V is the average velocity in metres per second. Find the most economical average velocity.
Find the gradient of the tangent to the
curve y = x - 3 at the point (1, -1). X+ 1
Investigate the nature of the turning points at each of the stationary values for the
3 3 2 curve y = x - x .
Find values of k for which 5x2 - 3x + k
is positive definite.
120. If3x 2 + x + 1 and a(x- 1)(x + 2) + b(x + 1) +care equal for all values of x, find a, band c.
121. Find the point on the curve y = 3 + 4x - x 2 where the tangent is perpendicular to y = ~ + 1.
122. Find the value of k in the equation kx2
- 3x + 2k = 0 if twice the sum of its roots is equal to their product.
123. Show that the expression 2x - 5 - 3x2 is negative definite.
124.
125.
If the minimum value of the expression 3x2 + 6x + k is -2, find the value of k.
The parabola y = 7 - kx - 2x2 is sym-metrical about the line x = -2. Find the value of k and the maximum value of the expression.
126. Solve 4x - 18(2x) + 32 = 0.
127. For what range of values for y will the series y + y 2 + y3 + · · · + y" have a limiting sum? Find the limiting sum if y=;i.
128. If the limiting sum of the geometric series 1 + x + x2 + x 3 + · · · is 20, find x.
202
20
129. Evaluate L (10 - 2n). n=3
130. How many terms of the geometric series 1 + 3 + 9 + 27 + . . . are needed to give a sum of 3280?
131. For a geometric sequence T3 = 18 and T6 = 144. Find the first term and the common ratio.
132. Insert 3 numbers between 21 and 45 so that the five numbers form an arithmetic sequence.
133. Find the equation of the straight line which passes through the point (1, 3) and through the point of intersection of the lines x - 4y + 2 = 0 and 3x + y + 6 = 0.
134. Prove that the line 3x + 4y = 30 is a tangent to the circle with centre at the origin and radius 6 units.
135. Find the perpendicular distance from the point (2, -1) to the straight line 5x - 12y + 4 = 0.
136. Find the locus of a point P(x, y) which remains equidistant from A(3, 0) and B( -1, 4).
137. Find the equation of the parabola with focus (0, 6) and vertex (0, 0).
138. Find the coordinates of the focus of the parabola x 2 = -12y + 24.
139. Find the locus of a point P(x, y) which moves so that its distance from the fixed point (0, 4) is always equal to its perpendicular distance from the fixed line y = -4.
140. Differentiate: (a) y = e4x-1 (b) y = x 2 tan 2x
141.
142.
Find the indefinite integrals:
(a) f__!f:!_ (b) f ~xdx 3x - 2 e
Sketchy = cos 2x for 0 ~ x ~ nand find the area under the curve between 0 and *.
143. Evaluate t1
sin nxdx.
144. In l:c,ABC, D and E are point on the sides AB and A C such that ED = CE and LB = LC = 70°. (a) Find LA. (b) Find LADE. (c) Show that DE is parallel to BC.
145. ABC is a triangle right-angled at A. AD is perpendicular to BC. P is a point on interval BD such that CP = CA. If LAPC = 80° show that PA bisects LEAD.
146. ~ABC is right-angled at C and AB = 8 em, AC = 3 em. Find the length of the median AD.
147. Find the number of degrees in each angle of a regular decagon (10 sides).
203
148. How many sides has the regular polygon each of whose angles is 156 degrees?
149. Show that the angle formed by the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the remaining angles.
150. In the quadrilateral ABCD, ADIIBC, the diagonal AC bisects LA and BD bisects LD. Prove that AB = BC = CD.
Consider y = f(x) on the interval a ::::;; x ::::;; b.
Let the parabola y = Ax2 + Bx + C pass through the points P(a,J(a));
Q(a; b,J(a; b)} R(b,J(b)).
Thus each of these points must satisfy y = Ax2 + Bx + C, giving on substitution,
f(a) = Aa2 + Ba + C
t(a;b)=A(a;by +B(a;b)+c
y
I 1 f(a) I I
a
I
: f(~) I
I
: f(b) I
b
f(b) = Ab2 + Bb + C From these equations:
f(a) + 4 . t(a ; b) + f(b) = A { a2 + 4 . (a ; by + b2} + B {a + 4 . (a ; b) + b}
+C + 4C + C
(1)
(2)
(3)
= A{a2 + (a2 + 2ab + b2) + b2
} + B{a + 2(a +b)+ b} + 6C = 2A(a2 + ab + b2
) + 3B(a + b) +6C (a)
Consider now the area under the curve y = Ax2 + Bx + C from x =a to x =b.
A = Lb (Ax2 + Bx + C)dx
[Ax
3 Bx
2 Jb = -3- +2+ Cx
a
= A (b3 - a 3
) + B (b 2 - a2 ) + C(b - a)
3 2 A B
= 3 (b - a)(b2 + ab + a2) + l(b - a)(b + a) + C(b - a)
= !{2A(b - a)(b2 + ab + a2) + 3B(b - a)(b + a) + 6C(b - a)}
= b ~ a {2A(a2 + ab + b2) + 3B(a + b) + 6C} which from equation (a) gives
A = b ~ a{f(a) + 4f(a ; b)+ f(b)} which approximates the area under the given curve y = f(x) between x =a and x =b.
An alternate form is A ~ ~{Y 1 + 4y2 + y 3}
where a= x 1 ,J(a) = y 1 ; a; b = x 2 ;f(a; b) = y 2 ; b = x3 ;f(b) = Y3 and
h = b ~ a = width of each sub-interval.
205
Consider y = f(x) on the interval a :::;; x :::;; b. Divide the interval a :::;; x :::;; b into n equal sub-divisions where n is even. This results in an odd number of ordinates, namely (n + 1).
Let the points of division be a= xt> x 2 , x 3 , ••• , x,_1 , x,, x,+l with corresponding ordinates y1, y 2 , y 3, ... , Y,-1, y,, Y11 +1·
The width of each sub-interval ish = b - a. n
Apply the alternate form of Simpson's rule from the previous appendix to:
X1, Xz, X3 : A1 ~ ~(Y1 + 4yz + Y3)
Xs : Az ~ ~(Y3 + 4y4 + Ys)
X7 : A3 ~ ~(Ys + 4y6 + Y?)
Adding these results and noting that: (a) end ordinates y1, y,+1, occur once only. (b) even ordinates y2 , y4 , y6, ... , y, occur once only but are multiplied by 4. (c) the other odd ordinates YJ, y 5 , y 7 , ••• , y,_1 each occur twice.
h Hence, A~ 3{(y1 + y,+l) + 2(Y3 + Ys + Y7 + · · · + Y,-1) + 4(Yz + Y4 + Y6 + · · · + y,)}.
Before proceeding to find the derivative of sin x we note that
(a) 1. sinh
1 Im--= l!-+0 h
(already proved in Chapter 20)
We will now establish two further results. sin2 h 1 - cos2 h --,;r- h2 (b) (from the Pythagorean result sin2 h + cos2 h = 1)
sin2 h (1 - cos h)(1 + cos h) --,;r- = h2
sin2 h (1 - cos h) 1 + cos h . --,;r- = h2 (dividing each side by (1 + cos h))
206
i.e. I - cosh = I (sin h)2
h2 I +cosh h
lim(l -cos h) _ _!_ h-+0 h2 - 2 (
. sinh ) Smce ash ~ 0, cosh ~ I and -h- ~ I
(c) We need a formula for sin(x + h). A simple derivation of this formula follows. A Area !:::,ABC = icb sin A
= icb sin(x + h)
c k
b
also area !:::,ABC = ikp + ikq Hence icb sin(x + h) = ikp + ikq
. . k p k q .. sm(x + h) = b · ~ + c · b
that is sin(x + h) = sinh cos x + sin x cosh
We now proceed to find the derivative of sin x. d ( . ) 1. sin(x + h) - sin x -dSinX=lm h
X h-+0
I. sinh cos x + sin x cosh - sin x = lm -------'-----::-------
11-+0 h
I. (sin h) . (cosh - I) = 1m -h- cos x + sm x h IJ-+0
. (sin h) . (cos h - I) = ~~~~ -h- cos x + sm x . h h2
(. h O sin h I cos h - I I)
= cos x as ~ , -h- ~ , h2 ~ - 2
Therefore :X (sin x) = cos x.
207
1. ! 2. ~ 3. (a) ! (b) t 4. (a) t (b) ! (c) ~ 5. (a) !- (b) t 6. 20 7. yellow
1. M 2. M 3. # 4. +r 5. f.r 6. (a) 17 (b) -f.r (c) # 7. -1-r 8. ~ 9. 1-r 10. ~
1. (a) ! (b) t (c) !-3. (a) ¥o- (b) tt 5. (a) +,; (b) -b {c) if; 6. (a) !- (b) ~ (c) !
2. io 4. i
(d) A- (e) !- (f) if;
7. (a) t (b) t (c) t (d) -fo (e) to (f) # 8. (a) A- (b) t (c) ~ 9. (a) -% (b) t
10. (a) t (b) ! (c) ~ (d) ~ (e) zero 11. ~ 12. (a) ! (b) !- (c) ~ (d) 1\r 13. (a) -fs (b) !- (c) -h
1. (a) ! (b) 16 (c) ! (d) fr (f) rt (g) t (h)~ (i) *
2. (a) one die (b) one die
1. 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 5 4 3
3. no 4. win 2. 1 in 9 5. t 6. t 7. -tr
(e) ~ (j}!-!
3. fa
11 12
2 1
8. 6 or 8 9. 6, 8, 5, 9, 4, 10, (2, 3, 12)
1. (a) 6 (b) 12, 13, 21, 23, 31, 32 (c) t 2. (a) !- (b) t 3. (a) i (b) ! (c) t 4. (a) there will be 12 possibilities (b) t 5. (a) ! (b) i 6. ! 7. (a) ! (b) t 8. (a) fr (b) ! (c) ! 9. !
10. (a) 16 (b) ! (c) ! 11. (a) i (b) i (c) t (d) t 12. t 13. (a) twenty numbers can be formed (b) ! 14. (a) ! (b) ~ (c) ! (d) ! (e) t 15.! 16.t 17.~ 18. (a) 160000 (b) 4o1oo 19. (a) 24 (b) 1000, r:h
209
I. (a) t CbH 2. i9
1. (a) ! (b)! (c) !-2. {a) /o 3. (a) (i) 2 l6
(iv) N6 = f.r
(b) !o cu) m (b) no (b}t (b)!
4. (a) i 5. (a) t 6. ~ 9. to
10. (a) H 11. (a) 0·73 13. (a) ! 14. 0·729
1. t 4. (a) N.f6
6. * 8. m
1. (a) t
7. tl96
(b) H = ~ (c) /g (b) 0·53 (b)*
15. (a) 1 ~0
2. H (b) **
7. (a) fr 9. (a) a~o
(b}t
(d) -fr (c) ! (iii) N6 = .fr
(c) i
8. :~i6 = 0·15
(d) A-12. 2lo4
(c) 2l6 (b) To
3. * 5. n-
(b) rt (b) ~ri~
2. (a) -!,;- (b) iJ- (c) A- (d)# 3. (a) fa (b) fa 5. (a) -is (b) Js 7. (a) t (b) is
(d)!- (e) n 8. (a) t (b)t 9. -h
1. 23, 27, 31 2. 24, 31, 39 4. 96, 192, 384 5. 36, 49, 64 7. 3, -4, -11 8. 19, 28, 37
10. 125,216, 343 11. 21, 34, 55
4. ~ 6. t
(c) j (f) ¥o 10. i
3. 4, 1, -2 6. 486, 1458, 4374 9. 64, -128, 256
12. 16, 15, 19
1. (a) 2, 4, 6, 8 (b) 0, 3, 6, 9 (c) -3, -2, -1,0 (d) 2, 6, 10, 14 (e) 7, 8, 9, 10 (f) t, !, !, t (g) t, i, t, ~ (h) 1, 8, 27, 64 (i) 9, 6, 3, 0 (j) 1' 2, 4, 8 (k) f, t, ~' f, (1) 3, 9, 27, 81
2. 1, 3, 6, 10, 15, 21. Triangular numbers. 3. (a) even (b) odd (c) T, = 2n - I 4. (a) + 1 (b) -1 5. (a) -5, 7, -9
(c) 3, -6, 9 6. (a) 4 7. T9
8. (a) 11
(b) -1,t,-!(d) -1, 4, -9
(b) yes, T11 (c) 5
(b) no (c) 32, subtract 3
I. (a) d = 2; 11, 13 (c) d = -2!; 2, -! (e) d = 3; X+ 12, X+ 15 (g) d = 4; 11, 15 (i) d = )3; 4)3, 5)3
2. (a) yes, d = 8 (c) no
3. (a) 6, 13, 20 (c) -4, -2!,
5. X 21
I. (a) 43 (e) 34
2. (a) 211 - I 3. 7n 4. 23- 311
(b) -27 (f) 113 (b) 2n
5. 12, 16, 20, 24; d = 4 6. I, 3, 5; T60 = 119 7. T74
8. no 9. 63
(b) d = ---'2; 10, 8 (d) d = 10; 100, 110 (f) d = 7; 33, 40 (h) d= i; 7, 7t (j) d = 3x; 13x, 16x (b) no (d) yes, d = 0·4 (b) 8, 5, 2
(c) 31 (d) 106 (g) -160 (h) 5 + 23b
10. 11,9,7;d= -2;T100 = -187 11. d = 6 12. a= 2, d = 3 13. -58, -51, -44 14. 148 15. 20 16. (a) 89 (b) 176 17. T51 = 203 18. 24,29, 34,39,44 19. 43, 36, 29, 22, 15, 8, I, -6 20. (a) T,, = 55 - 711 21. $59 22. 23 23. a= 5, d = 3
1. (a) 365 (d) 3405
2. 5050 3. 930 4. 1080 5. 735 6. 4134 7. d = 7, 1390 8. 2500 9. 79799
10. 2500
(b) 414 (e) 450
(b) 12
11. T1 = 25, T40 = -92, S40 = 1340 12. 8 13. 10 14. 1683 15. $295 16. 1024 em 17. $11250 18. $51 000 19. $4500000 20. a= 15, d = -3 21. 610
(c) 7
210
22. 10 23. 4, 6, 8; d = 2 24. (a) d = 6 (b) 119 25. d = 4; 1890 26. I, 3, 5, 7, ... 27. 550 m 28. 12000 m 29. 12 em
I. 378 2. 730 4. 3625 5. 504 7. 416 8. -105
10. 360 II. -II
I. (a) r = 4, 64 (c) r= -3,-54 (e) r = 11-, 13! (g) r = It, 3Jt (i) r = x, ax3
2. (a) 4, 12, 36, 108 3. (a) yes, r = 2 (b) no
(c) 1240
3. 208. 6. -105 9. 306
12. 9090
(b) r = t, It (d) I'= t. 6t (f) I'=!. i (h) I'= -t, -4
(b) 8,4, 2, l,t (c) no
(d) yes, r = ! (e) yes, r = 2 (f) yes, r = a
4. c:_ = ~ b a
5. (a) 245 (d) ± 15
6. X= t 7. y = 4
I. (a) 160 2. (a) !z
(e) ab18
(i) 11114
3. 216
(b) 6t (e) ±3
(b) 28~ (b) 3Qi (f) 8 (j) 1116114
4. (a) 2, 4, 8; yes, r = 2 (b) 3,6,12;yesr=2
(c) 7! (c) 2m (g) 2m
(c) 5t (f) 16
(d) 2·7 (d) 121! (h) -9)3
5. (a) r = 2 (b) r = t (c) r = 3 (d) r = -i 6. (a) 7, 14, 28, ... or 7, -14, 28, ... where r = ±2
(b) ;i, f, 5, ... or;i, -f. 5, ... where r = ±2 (c) i, 2, 6, ... or -i, 2, -6, ... where r = ±3 (d) 216, 108, 54, ... or 216, -108, 54, ...
where r = ±t (e) 2, 3, 41-, ... or 2, -3, 4!, ... where r = ±It
7. T8
8. T4
9. 11 = 6 II. 54, 18, 6, 2 12. I'= 3 13. X= 2, I'= 2 14. X= 4t;y = IQi 15. p = 16; q = 6t 16. 6~5 17. 5, 10, 20, 40, 80 or 5, 10, 20, -40, 80 18. r = 4, a= 1 19. T7 = 364t 20. T8 = 3! 21. (a) x = ±,fo, r = ±,fo
(b) x = ±b, r = ±~ a
I. (a) 3280 (b) 31·75 (c) -2555 (d) 109·282 or 40(..j3 + I)
(e) 4-926 (f) x(y14 - I) y2- I
2. -10·5 3. 31·875 or 31~ 4. (a) -t (b) --h (c) 0·221 5. 162, I 08, 72, 48 6. 160, 80, 40, 20, 10, 5 7. $81·92, $163-83 8. 6 9. 7
10. 8 II. (a) 1456
(c) 906·641 (b) 13 021 (d) 57·593
I. $3387-80 2. $2107·39 3. $47234-62 4. (a) $8139·27 (b) $8577-02
(c) (i) $11446·60 (ii) $11566·04 (iii) $11594·36
I. $8870 2. $32003 3. $39477 4. $269 546
I. $72-30 2. $301·39 3. (a) $2586·84 4. $194-82
I. (a) r" --> 0 2. (a) (i) yes
(b) (i) 12 3. lxl < I;~ 4. (a) 1111·1
(d) 40 (g) 18
5. (a) 6 6. 4 7. 20 8. I'= t 9. I'= -i
10. X= 1 II. y= -t 12. (a) ~ 13. -b 14. 0·001 15. 84 metres 16. 16 metres 17. 12·5 18. (a) no
(b) 9·06%
(b) r" --> oc;
(ii) no (iii) no (iv) It
(b) 121·5 (e) 80 (h) 6·828 (b) 1 = 0·8
(b)~ (c) ~
(c) r" = I (iv) yes
(c) 80 (f) 156·25 (i) o· is (c) 4~ = 4·4
( (d) fS:.
(b) yes, r = i
211
19. (a) 61 (b) 7 20. X= I
21. (a) 4 x ( -ir1
(c) H = 2-462
(b) 32[1 - ( -i)"] 2-696 . 13 '
I. (a) 5;11;14 (b) 8; 10; 14 (c) 32; 37; 42 (d) 241; I9t; 81; 3t
2. no 3. T59 = 355 4. a= 3, d = 2 5. 8 6. 32 7. X= /o 8. a= 4t, r = 2 9. d = 10
10. a= 4, b = 6 or a= i, b = -It II. IYI < I; 3
12. 27_ + 27_ + 27_ + " .. 57 102 104 106 '99
13. (a) 55 (b) 315 (c) 40t 14. (a) -I~,~, 21, 4~, 6!, 9
(b) 1~, 1~, 21, 4, 6, 9 15. -9, -4, I 16. (a) 64, 511, 39t, 27 (b) 64, 48, 36, 27 17. $94504 18. (a) $441·90 (b) 4-46%
l.(a)x<2orx>6 (b)-l<x<2 (c) x = 0 or 4
2. (a) i (b) t (c) ! 3. 2-3 ha 4. (a) 3x2
, gradient = 3 (b) 3x - y = 0 5. 58; 590 6. 53°8'
I. (a) 4x3 - 4x
2. a= 7, b = -5 3. (a) t 4. (a) 17 em
5. (a) x- I X
6. 4 units; same side
1. 336·3
(b) 4x + 10
(b) i.; (b) 104·9 cm2
(b) X - 25 x2
- 25
(c) 6x2 - 3
(c) 164·9 cm2
2. (a) 9-6 em (b) 73°15' (c) 46 cm2
3. (a) 5620 (b) 27 4. (a) 150° (b) 47°16' or 227°16' 5. X + y - 2 = 0; (0, 2)
6. *
1. (a) x=torx= -4 (b) x= -2orx=7 (c) x = 1, y = -1 or x~ 3, y = 3
2. (a) 1 (b) ttan A 3. (a) 1770 (b) 3 4. (a) 108°12' (b) 2·5 (c) 4·7 square units 5. (0, 0), maximum; (f, --/r), minimum
1. Teacher 2. a= 6, d = 3 3. 7·1 km 4. n-5. 6903002 6. t
1. (a) x = 0 or x = f (b) 2 .;;; x .;;; 5 (c) x = 2, y = -1
2. (a) 126·9 square units (b) 27· 1 units 3. ( -t, H), maximum
(5, -75), minimum 4. (a) 1680 (b) 2 5. 3·75 6. -2
1. (a) f (b) 2 2. 79·0 m 3. (a) ~ (b) H (c) ~ 4. a= -3, b = 2 5. (a) t (b) t 6. a= It
1. (a) x = -1 or x = 2t (c) -1 <x<;; It
2. (a) W = 30 - x (c) 225 cm2
3. (a) r = ! 4. 39x- 16y = 0 5. area is lOf square units 6. 62·4 cm2
1. {x: x < -2} u {x: x > 2} 2. {x: 1 < x < 3} 3. {x: -4 < x < 2}
(b) x = 1 or x = -4
(b) A = 30x - x 2
(b)~
4. {x: x < -5} u {x: x > -2}
212
5. {x: x < -3} u {x: x > 7} 6. {x: -4 < x < t} 7. {x: -4 < x < 3} 8. {x: x < 2t} u {x: x > 5} 9. {x: 1 < x < 4}
10. {x: x < -2} u {x: x > 7} 11. {x: -7 < x < 3} 12. {x: -5 < x < -2} 13. {x: x < -lt} u {x: x > 4} 14. {x: -1 < x < f} 15. {x: -2 < x < 3t}
1. (a) -1 ± y'IO, irrational -5 + ,jT'J ' '
(b) 4 , mal!onal
(c) t, -4, rational I + y'I3
(d) -2
, irrational
(e) t, -1, rational (f) -t, -3, rational
-3 + .[5 (g) ~ , irrational
-l+y'41' ' (h) IO , matwnal
(i) -t, -1, rational 2. The number under the square root sign is negative. 3. t, b2
- 4ac = 0 4. t, 1 t; b2
- 4acis a perfect square. 5. b2
- 4ac > 0 and is not a perfect square.
1. real, rational, equal. 2. unreal. ' 3. real, rational, unequal. 4. unreal. 5. real, irrational, unequal. 6. real, rational, equal. 7. real, rational, unequal. 8. real, irrational, unequal. 9. real, irrational, unequal.
1. k = 5t 3. k = 9
2. a< -k
5. (a) k = -2! (c)k<-2!
4. k = 3 or -1 (b)k;;>-2!~ (d) k > -2!
6, k .;;; -12 or k ;;. 12 7. k= -lor-5
10. k .;;; -2 or k ;;. 2
1. (a) 2, 5 2
(c) k' 1
(e) 1!,! 2. (a) 5
(d) 21 (g) 5!
(b) -t, -It (d)'£k-l
4' 4 (f) -3t, -It (b) 2 (e) -4 (h) lOt
(c) 2t (f) 10 (i) 17
3. (a) -It (b) -2 (d) t (e) 2
4. (a) x 2 - 8x + 15 = 0
(b) x 2 - 2x - 8 = 0 (c) x2 + 6x - 7 = 0 (d) 4x2 8x + 3 = 0 (e) 2x2 + 9x + 4 = 0 (f) x2
- 3 = 0 (g) x 2
- 4x + I = 0 (h) x 2
- !Ox + 23 = 0 5. (a) 2 + k, 3k
(c) 4± (f) -6
(b) (i) k = 3 (ii) k = 4 (iii) k = -8 6. II= -3 7. (a) k = 4 (b) k = It (c) k = 5 8. k = 4
10. (a) -m
II. 3 14. (a) k = ~h2
I. 2
I (b) --
11
9. -2-4
12. 21 (b) h = -3k
2. (a) coefficient of x 2 is negative. b
(b) x = - 2a = 2 (c) II
4 (c) 11
2 + -II
3. (a) 5, min. (c) -9!, min. (e) -9, min. (g) -15·5, min. (i) 7-h, max.
(b) -6, min. (d) 3!, max. (f) 4, max. (h) 0, min. (j) -t, min.
4. (It, 2t), 2t 5. 9 7. 508
I. a> 0,11 < 0 2. a< 0, /':,. < 0 3. (a) indefinite
(c) positive definite (e) indefinite (g) positive definite
6. k = 3, Ii 8. 6 and 6
(b) negative definite (d) negative definite (f) indefinite (h) indefinite.
4. (a) f(x) < 0 for I < x < 5 f(x) = 0 for x = I or x = 5 f(x) > 0 for x < I or x > 5
(b) f(x) > 0 for all values of x. (c) f(x) < 0 for all values of x. (d) f(x) < 0 for -3 < x < I
f(x) = Oforx = -3 orx =I f(x) > 0 for x < -3 or x > I
(e) f(x) < 0 for -4 < x < 3 f(x) = 0 for x = -4 or x = 3 f(x) > 0 for x < -4 or x > 3
(f) f(x) < 0 for 0 < x < t f(x) = 0 for x = 0 or x = t f(x) > 0 for x < 0 or x > t
(g) f(x) > 0 for 0 < x < 2 f(x) = 0 for x = 0 or x = 2 f(x) < 0 for x < 0 or x > 2
(h) f(x) > 0 for -2 < x <! f(x) = 0 for x = -2 or x =!
. f(x) < 0 for x < -2 or x > !
213
5. (a) x < - 3 or x > 7 (c) x < I or x > 6 (e) t < x < 3 (g) X < 2 or X > 2t
(b) x < 0 or x > 4 (d) x < 3 or x > 3 (f) is never positive. (h)-l<x<3
6. (a) -4 < x < I (c) -6 < x < 2
(b) x < 0 or x > 5 (d) -2 <X < 8
7. x < -3 or x > 5 8. (a) k > t
(c) k > 4t (b) k < -I! (d) -8 < k < 8
(e) I < k < 5
I. a = 5, b = - 2, c = 9 2. A = 6, B = 4, C = 13 3. a = 2, b = 3, c = 8 4. 3(x - 1)2 + IO(x - I) + 12 5. 3(x + 1)2
- ll(x + I) + 10 6. 5x(x - I) + 3(x - 2) + 12 7. a= I, b = I 8. (a) a = I, b = -2, c = -I
(b) a= I, b = -3, c = 3 (c) a = 4, b = -8, c = 3 (d) a= I, b = -4, c = 4 (e) a = 4, b = 6, c = 5
9. (a) A= 2,B = -3orA = -2,B = 3 (b) A = 3, B = -12
I. x = ± I or x = ± 3 2. x =I or x = 2 3. (a) x = ±I or x = ±4
(b) X= ±2 (c) x= ±2orx= ±3 (d) X= ±-ft (e) x = 3 or x = -I (f) x = ±I or x = ±2
4. x = Oor x = 2 5. x = I or -3 or x = -I ± Ji 6. (a) x = I or x = 2
(b) x = I or x = 2
(c) x = 0 or x = 2 (d) x = 4 or -2 or x = I ± Ji (e) x =tor -It (f) x = 7 or - I or x = 3 ± Ji
(b)12x2 -4'
(e) 6x- 4
-2 (g) 48(2x + 3)2 (h) (x + V
2. 12 3. -18 5. 4 7. X=~ 9. X = t IQ. X = t
(c) -12x
(f) 6x + 4
(i) -4 --./(4x + 2)3
4. f"(x) = 2a 8. y =X+ I
3. X > 2~ 4. X > 1! 5. (a) concave upwards and
gradient positive from x = I to x = 3
(b) concave downwards and gradient positive from X= 1 to X= 4
(c) concave upwards and gradient positive from X= 2 to X= 3
(d) concave downwards and gradient negative from X= 1 to X= 2
6. (i) (ii)
/
) 2 3
I. (I, 2) 2. (2, -45) 3. ( -t -f.r) d2y
4. dx2 # 0 for any x 5. (2, -77), (-2, - 85)
6. (a) 3x2 - 2 (b) 6x (c) 2x + y - I = 0
7. X + y - 4 = 0 10. (t, 0)
I. minimum at (1, I) 2. maximum at (- I, 7) 3. maximum at (0, 2); minimum at (2, - 2) 4. minimum at ct, -I H); maximum at ( -1, I) 5. minimum at (2, 0) 6. minimum at(!, 16);maximumat(-3, 16) 7. minimum at (1, 3) 8. maximum at (0, 2); minimum at (I, I) and at (-I, I)
214
I. 18 cm2
3. r = 5 em; h = 10 em 5. 18 em by 9 em 7. 50
x7 I. (a) y = T + C
(c) y = ~x3 + C
(e) y = ~x2 + 2x + C
xs (g)y=TS+C
x3 2. (a) 3 + C
(c) 8x + C xs xz
(e)---+ C 5 2
2. 8 em by 8 em 4. 48 cm2
6. 20 knots 8. (b) 10
x6 (b) y =- + c
2
(d) y = 5x + C x3 x2
(f) y = - - - + 5x + C 3 2
3x4
(h) y = 2x - - + C 4
(b) 7:4 + c
(d) x 2 + 6x + C x4
(f) !6 + 2x + C
(g) x2
- x + C (h) 2x2 - ~x4 + C 3 2 16
3 3. (a) f(x) = 2x 2
- 5x + C
2 x 2
(b) f(x) = -x 3 - - + 2x + C
3 2 x4
(c) f(x) = !Ox - 2x2 - 4 + C
(d) f(x) = ~x5 - ~x3 + ~x2 + C
5 3 2 5. v = t/2
- 3t + c 6. V = %m·3 + C
9. (a) y = -x-1 + C x-4
(b) y = -4 + c
(c) y = ~xt + C 3
(e) y = -x- 3 + C
(g) y =~xi+ C 5
(i) y = -x-5 + C
(k) y =~xi+ C 5
I. y = x 2 + 5x + 2 2. y = - x 2 + 3x - 1 3. y = 3x2
- 4x - 2 4. s = ~13
- I - 5 x2
5. y = x 3 - T + 5x + 4
6. y = x 3 - 2x2 + x + 4
:\'3 7. (a) y = 3x + x 2
- ~
(d) y = 2x! + C
(f) y = 2x-2 + C
(h) y = -6x-1 + C
(j) y = ~x'+ C 2
(I) y = -~xt+ C 3
x3 (b) y = 3x + x 2
- - - 9 3 x 3 7
(c) y = 3x + x 2 - 3 - J
2 2 8. y = -xt + 2x +-
3 3
9. y = -± + 6 10. v = 3t - 2t2 + 24
s = ft 2 it' + 241 + 4 x 3 5x
ll.(a)y= +-+3 2 2 x3 3x2 7 (b) y =- + - + 2x +-3 2 6
12. X= 228
I. 591 2. maximum height = 225; range = 300. 3. (a) 2t (b) 2 (c) It (d) 2! 4. 111· 3 square units 5. (a) -9x2 + lOx- 6
(b) 2(x2 + 3x - 3) (2x + W
(c) 20(5x - 2)3
6. (a) ! (b) t
I. 0·2 2. A = 3, B\= -4, C = 5 3. (a) 'x = 3 or x = -2
(b) -2 < X < 3 (c) -6!
4. 18 5. 2x + y- 5 = 0 6. (a) 8·2 m (b) 95°58' (c) 42·2 m 2
I. (a) x = 0 or x = t (b) x = 0 or x = 2 2. x = t i.e. gap setting is 0·25 mm. 3. (a) 4880 (b) 10
4. (a) 7(1 - 2x3
)
(x3 + 1)2
(b) -24x(5 - 2x2)
5
1 5 6 (c)-+----
)X 2.;-;? x3
5. (a) 3 units (b) 3x + 4y + 1 = 0 6. (a) 77-8 m, 152·1 m (b) 135° (c) 214·3 m
I. (a) roots are real, rational and uneql,lal. (b) -2 <X <t
2. (2, -16), minimum; ( -2, 16), maximum 3. 3y = x 3
- 6x2 + 3x + 16 4. (a) lines intersect at (0, I 0).
(b) X+ 2y = 0 (c) 4,[5 units
5. (a) -h (b) !! (c) .ft 6. m=6orm= -2
215
I. (a) 107 (b) 310 2. (a) t\ (b) /., 3. (a) 50° (b) 600
(c) AC = 78·3 m, BC = 69·3 m 4. (a) x = -1 or x = t
(b) -1 <X< t (c) minimum value is -It.
(c) 6255 (c) t!
5. (a) 5 (b) A = I, B = 5, C = 7 x'
6. y =-+X- 6 3
2. (a) T,, = 47 - 7n (b) n = 12 (c) S,, = 18 3. y = 3x - 3 andy = - 3x - 6 4. (a) BC = 9·2 em (b) 34·6 cm 2
5. 30 6. (a) (1, 0) minimum, ( -1, 4) maximum
(b) (0, 2)
I. y = 2x 4. (a) y = x 2
(d) y =X+ 3
2. y =X+ 2 (b) X= 3y
3. x 2 + y 2 = 25 (c) y = x + 2
5. y = 5 6. x 2 + y 2 = 4 7. circle, centre (0, 0), radius 3 units.
I. y + 3x- 15 = 0 2. 5y- 3x + 1 = 0 3. (x - 1)2 + (y - 6) 2 = 9 or
x 2 + y2 2x - 12y + 28 = 0 4. x 2 + (y - 2)2 = 4, yes 7. 2x2 + 2y2
- 19x + 3y + 35 = 0 9. y2
- 6y - 4x + 13 = 0 10. x 2 = 8y
X- Y Pi II. (a) .j2 (b) (y2 + l)y- x = 0
12. ( -1, 5); (x + 1)2 + (y- W = 26
13. (a) y - 4, _Y_ (b) y - 4 X _Y_ = -I x x-5 x x-5
14. 16x- lOy> 17
I. x 2 - 6x + y 2 + lOy + 25 = 0
2. (a) (2,5);3 (b) (-1, -3);4 (c) ( -4, 1); 2 (d) (0, 6); 6
3. (a) x 2 + y2 = 9 (b) (x - 1)2 + (y - 4)2 = 25 (c) (x + 3)2 + (y + 2)2 = 1 (d) (x - 4)2 + y 2 = 16
4. x 2 + y2 + 2x - 12y + 12 = 0 7. (a) (1, -3); 3 (b) ( -2, -4); 3
1. (O,l);y= -1 2. (a) (0, 2);y = -2
(c) (0, t); Y = -t (e) (0, fr,); y = --h-
3. a = 1; (0, 1) 5. focal length; x 2 = 2y 6. (a) x 2 = 8y
(c) x 2 = -8y
(b) (0, -4); y = 4 (d) (O,t);y = -t (f) (0, -t); y = t
4. x 2 = -16y
(b) x 2 = 4y (d) x 2 = -I6y
l. 2; y = -1; x 2- 6x + 17 = 8y
2. (0, -l);x2 = 8(y + 1) 3. (a) (x + 1)2 = 8(y - 4) (b) x 2 = 8(y - I)
(c) (x- 1)2 = -8(y- 6) 4. 3;x=3 5. 16;x= -2 7. 2;(2,0) 8. y2 = 4x;x = -1
1. (iii) 0·2525 sq. unit
l. (a) 0·25 sq. unit 2. 4t sq. units 3. (a) 21 sq. units;
(d) I sq. unit; (g) 1 sq. unit; (j) 3~ sq. units.
4. (a) st (b) 7 (f) 1 (g) i
5. 9 sq. units 6. 18 sq. units 7. t sq. unit 8. (a) 31t sq. units;
10. (a) 4;
(b) I sq. unit;--- (c)-8sq. units; (e) 32t sq. units; (f) 36 sq. units; (h) 17t sq. units; (i) 5~ sq. units;
(c) 10 (d) 9 (e) ~ (h) 12~ (i) 9
(b) 2 sq. units. (b) 16
xs l. (a) S + C (b) 4x + C (c) x 9 + C
x3 (d) 3 - 2x + C
x4 (e)-- x 3 + C
2
(f)~xt+C 4
(g) -x-2 + C (h) ~x> + C 3
ax11 (i) -+ c
11
1·2
(') ~+ c J 1·2
2 (k) -xi+ C
5 x2
(I) - + 2xi + C 2 4x3
2. (a) - - x2 + C 3 3
(c) £ + x-1 + C 3
(e) x3 - x 2 + x + C
(b) 2xs + 3x4 + C . 5 4
(d) 4~
3
+ 2x2 + x + C
x6 x3 x2 (f) - - x4 + - - - + x + C
6 3 2
216
x3 (g)--x+C
3
(i) ~xi+ C 7
(k) ~xi- 2x-t + C 5
3. (a) 4x(x2 - 1) + C
3
(h) ~ + x 2 + x + C
(j) x3 - _!_ + C x2 2 -0·3
(I) ___ £__+ c jX 0·3
(b) x 2- 3x + C
12 1 (c) t3 - t2 + 5t + C (d) - - - + C
2 t 4u6 6u5 2u3 413 1
(e) -3- + S - J + C (f) 3 + 4t - f + C
x 5 4 x 2 x3 1 (g) -- -x> +- + C (h) - + + C
5 7 2 3 X
(i) 2u + C
(k) x 3- .!_ + C
X
v2 (j) 2 + 2v + C
4 (I) u2 + 3u - 2u>- -ui + C
5 4. y = x3
- 3x + C; C = 3 5. (a) y = x2 - 3x (b) y = x 2
- 3x (c) y = x 2
- 3x + 4 6. -4t
l. (5x - 1)4 C
20 + 1
3. 3 6(2x + 5)
5. (4x + J)! + C 6
7. /'~22x)6 + C
9 1 c
· 4(1 - 2x)2 +
l. (a) 15 (b) 1 (e) 21 (f) 0 (i) to
2. (a) 0 (b) -h (e) 251 (f) 2ot (i) 57t (j) 8t
4. (a) 4t (b) 4t 7. (a) a + f + t + t
2. (2x + 4)6 + C 12
4. I + C 6(3 + 6x)
6.t~+C
8. -~(3 - 3x)i + C
(c) 2H (d) It (g) 2 (h) 8/s
(c) 81 (d) -4 (g) lli (h) 4t (k)i (I) 11t
(b) a+ f + t + t
1. (a) st sq. units 2. (a) sst sq. units
(c) i sq. unit
(b) It sq. units (b) 101 sq. units (d) 4t sq. units
3. (a) 2t sq. units (d) 81 sq. units
4. (a) 3! 5. (a) 71 sq. units
(c) t sq. unit 6. st sq units 8. 1661 sq. units
10. t sq. unit
(b) 3! sq. units (c) 41 sq. units (e) 3 sq. units (f) t sq. unit (b) 4isq. units (b) 2Jt sq. units (d) 2! sq. units
12. 1 sq. unit, 1 sq. unit 14. fz sq. unit
7. sot sq. units 9. 8 sq. units
11. 1 sq. unit 13. It sq. units 15. 97t sq. units
1. It sq. units 3. 2t sq. units S. It sq. units 7. 21 sq. units
2. st sq. units 4. 2(.j2 - 1) sq. units 6. 4i sq. units; 2t sq. units 8. I8i sq. units
1. i sq. unit 2. 4 sq. units 3. n sq. unit 4. ! sq. unit S. ( -J, 2), (2, 8); 9 sq. units 6. (a) 2It sq. units (b) IQt sq. units 7. IQt sq. units 8. t sq. unit 9. i sq. unit 10. st sq. units
11. 4t sq. units 12. n sq. unit 13. +.,sq. unit 14. It sq. units IS. IQt sq. units
1. (a) O·S sq. unit (b) 0·34 sq. unit 2. (a) O·S (b) 0·26 3. (a) 1·17sq.units (b) I!sq.units 4. (a) 4·72 (b) 0·692
(d) 0·166 S. 0·1S6sq.unit 6. (a) 1·218 7. (a) 10·3 8. 666·S m2
(b) 1·397 (b) 0·764
1. (a) 6t (b) 2·2S (d) 0·4144 (e) 1·1948
(c) 0·2S3; 0·2S (c) 0·29% (c) 1·44S
(c) 0·880
(c) O·S046
2. (a) 0·200S sq. unit (b) 3·1416 sq. units (c) 0·3927 sq. unit (d) 0·3047 sq. unit
3. (a) l·S78483S (b) (i) l·S7814S7
(iii) 1· S78482S 4. 0·6933 S. (a) 9·7167 6. (a) 137·04 m2
(ii) 1· S784699 (iv) 1· S784834
(b) 9ot
(b) 3·1267 sq. units; 12·S067 sq. units; 12·S664 sq. units
1. (a) 3;" c. units (b) 84n c. units (c) ¥c. units
(d) (i) 97l" c. units; (ii) 81n c. units (e) 1 g" c. units 2. (a) (i) 107-233 c. units (ii) 2S·133 c. units
(b) (i) 60·319 c. units (ii) S?-446 c. units (c) 1·047 c. units (d) BSI c. units (e) 127·23S c. units
3. (a) (0, O),(t, t); 0·0 13 c. units (b) (0, 0), (3, 3); 25-447 c. units
4. (a) 19·268 c. units (b) 107·233 c. units (c) 3 ~" c. units (d) 12r c. units
S. (a) 18 sq. units; (i) 81" c. units (ii) 64r c. units (b) (i) 4n c. units (ii) ~c. units (c) if c. units (d) 16n c. units
6. (a) ¥c. units (b) (i) 1 g" c. units (ii) n c. units (c) 428·932 Jitres (d) y = 4x- 2; (t, 0); ~c. units
217
1. (a) t, -h (c) 1024 (e) No
2. (a) No (c) (0, 1) (e) 2"
s. (0, 1) 8. 0·9
1. (a) 1·2214 (c) 0·606S (e) 100·4277
3. (a) 4-48
(b) 16, 32 (d) increases rapidly (f) 0 (b) approaches 0 (d) larger
7. 1·1, l·S, 2·6
(b) S·3388 (d) 0·0247 (f) 67-2381 (b) Yes
1. (a) 3e3x (b) Sesx- 6 (c) 7e1x+s (d) -4e-4x (e) 2e2x (f) ex+ lOx (g) 3 - x-2 - e-x (h) !(ex + e-x) (i) -2e1 -x
2. (a) 30e6 x (b) 2xex'
(e) - 8xe-4 x'
(c) 1 + e1 -x
(f) -1-e.rx
2,JX 3. (a) 4e2x (b) 9e3x+ 2 (c) el-x
4. !Ce e-1) s. e
7. y- 3x- 1 = 0; y-! 3x- 3 = 0 10. 90° ' 12. y = ex + 1
1. (a) ex(l + x) (c) (2 - x)e-·' (e) (2x + l)e2x
(g) ex(x ; 1) X
(i) 2 - ex + e-x (e x - 1)2
2. (a) ( 4x2 + 2)ex' 3. se-x; -se-x
4. 17e6, 21
(b) 3ex(l + x) (d) (3x2 + 6x )ex (f) (1 - x)e-x
-ex (h) (e·' + 1)2
(b) (2 + x)ex (c) (2x - 4)e-x
S. y = 3x + 3; y = 2x + 2; y = x + 1 meet at ( -1, 0). 8. (a) Minimum (0, 1) (b) y---> oo 9. (a) A = (1, e-1
) B = (2, 2e-2) C = ( -1, -e)
10. (a) ex(l + x); (2 + x)eX (b) ( -1, -e-1) is minimum. (c) ( -2, -2e-2) (d) Yes (e) very large (f) small, negative
1. (a) ie4x + C (c) te•x+b + C (e) -2e-l + C (g) -6e4-x + C (i) -ke-2x + C
2. (a) 2~' + C (c) ne1x'+1 + C (e) !ex' - !e-x' + C
(b) -!e-2x + C (d) te3x - e-x + C (f) x 3 + !e2
x + C (h) -!e1-6x + C
(b) ex -!ex' + C (d) !e2x':1 + C (f) -e-x + C
3. (a) 6·362 (b) 0·175 (c) 3-627 4. e2
- I ~ 6·389 square units 5. 7·254 square units 6. 2·562 square units 8. 0·859 square units 9. 10·036 cubic units
10. 49·150 cubic units II. 1 square unit 12. 0·059 units2
; 0·0141 units 3
1. (a) 1·6094 (d) 2-3026
2. (a) 6·9078 (d) 8·8537
3. (a) 1·3863 (d) 2-4849
4. (a) 7 5. 0
(b) 1·9459 (e) 0 (b) -4·6052 (e) -0·3567 (b) 3·2958 (e) 2·8904 (b}t
6. 0 8. No 9. No, No
II. Symmetrical about y = x. 13. 0 < x <co; -co< y <co
(c) 0·3250 (f) 0·9933 (c) 4·2485 (f) -4-9619 (c) 0·4055 (f) 2-6027 (c}!-7. No
10. 2·7
14. log a 15. log. e = 1 16. (a) 1-0986 (b) -0·693 17. (a) 6 (b) 64 18. (a) 4 (b) 81 19. 2 20. (a) 3 21. (a) 0·693
1 1. (a) -
X
(b) 7·5 (c) 10 (b) 2
(d) 2e2x + _! ( ) 3 x e 3x + 1
2x 3 (g) x2 - 5 (h) 5 - ~
(j) ~ (k) 2 X X
(d) 1
(c) 3 X
(f) _2_ 2x- 3
(i) i X
1 (1) 2x
3. _1 ___ 1_ 4. __ I_ X + 1 X + 4 2(x + 9)
5. (a) _2_+ _1_ (b)_!- _I_ 2x+i x-5 x x+i
(c) _3_ (d) .. 1 x + 6 2(x + 4)
1 1 1 6 (e)--+----- (f)
X + 1 X - J X + 2 3x + 2 1 -2 3
6. (a) -- (b) -- (c) -=--x (2x- 1) x
7. (a) I +In x
(c) In x
8. 3,2lnx X X
9 10. (a)
(3x + ljl
2x3
(b) 6x2 In (x - 2) + -(x- 2)
9. _I_ xinx
(b) _! X
13. (a) y = x - I (b) 3y - x = 3 log 3 - 3 14. X= 2; 2y- X= 2Jn 4- 2 16. (a) min (1, -1) (b) max (e, e-1); inf. (et, !e-!)
(c) min (1, 1); inf (2, In 2 + t)
17. (a) 1 - x In x (b) -e-x(x In x + 1) xex x(in x)2
218
ex -2ex (c) ~ + ex in X (d) (ex + l)(ex - 1)
(e) 3 I 1 2x
19. -- +--or -2--
x+3 x-3 x -9 20. -2 < X < co; X = -1; I 21. x = e
I. (a) tinx + C (c) In (4x - 3) + C (e) x 3 + x 2
- 5 In x + C
(b) In (x + 2) + C (d) iin(6x -7) + C (f) ! In (211 - 1) + C (b) tIn (x 2
- 3) + C (d) -In (I - x 2
) + C (f) tIn (x4 + 3) + C
2. (a) In (x 2 + 1) + C (c) tIn (x2
- 4x + I) + C (e) In (ex + 1) + C
6 3. (a) 5x + 7 In x + - + C
X
b 1 3 1 ( ) -x + 2ln x-- + C 3 3x3
(c) ie2x
2
- In x + C 4. (a) 1 - log 2
(c) tIn 5 6. -x- 2ln (1 - x) + C 7. 2x - 3 In (x + 4) + C 9. (a) In 5
11.4ln2 13. 1 +log!= I -log 2 15. n cubic units 17. nln2 20. ~(e2 - I)
eY 22. x = -;jf(e2 - I)
2
1. 0·693, 1·099
(b) e2 + 1 (d) tIn 5
(b) 1 12. (3, 0); 12 - 3ln 31 14. y = In (x + I) + 4 16. tin 11- tin 3 19. (e - I) 21. j(e4 1)
2. 100 In 10 = 230·26 3. (a) 4log a . a4x
(c) 4x(5x In 4 + 5)
(b) ax[12x + (6x2 + !)log a]
(d) 3x(x In 3 - 1) x2
4. In 5. (SX);-1-CSX) + c
In 5
5. -1-(rox) + c 6. _]__ = 10·10
ln10 ln2
l. (a) 0·6678 2. (a) 0·2900
(b) 1·7584 (c) 2·2628 (b) 0·7637 (c) 0·9827
4. (a) 1·924 (c) 1
(b) -1·229 (may be written 2·771)
I 5. (a)-
x log 10 (c) 1 - In 10 log10 x
x 2 in 10
(e) lOx - 7 (5x2
- 7x)log 10
(a) 2·70481 (b) 2·71692
2x (b) -- + 4x log10 x
In 10
(d) _2_ x log 10
(c) 2·71815
1. (a) - 2e- 2x
(d) (3x + 5)e
(b) _I_ I+ X
5 (e) 5x + 2
2. y = 3e3x- 2e3; -2e3
ex ex (c)---
x x 2
(f) I -In x
3. (a) !e2x+ 3 + C (b) ! In (5x - 4) + C (c) 21n(x2 +I)+ C e6- e2
4.--2
5. e2 + e- 2 - e - e-1; area = e2 + e-2 - e - e-1
square units. e.fX
8. (a) r;; 2yx
(c) 31n x + 3
I (d) (x - I)
I - x2
(f) x(x 2 + I)
9. (a) !(In 4 - In 3) (b) 7! + 4ln 2 I I
(c)~--3 3e
10. (e - 1), (~ I) 12. (e - 1), ~(e - I) 14 n(e4 - I)
11. In 5 square units 13. !(e3 - e-3)
. 2 16. 55 log1o 2
17. (a) max(l,!)min(-1, -!) (b) approaches 0 (d) ! In 5
18. !In 5 19. i (In 4 + In 5 - In 8) = ! In 2·5
1. (a) 12x2 + 1r;;
2yx (c) x(1 + 2loge x)
2. (a) x 3 + x2 + x + C (c) loge (x + 2) + C
3. k =-I ork = -5 4. I 0~ square units 5. (a) if, (b) fa 6. (1/g-, -1~)
(b) 5e5x-2
(b) ix! + C
(c)* (d) -ft
1. (a) 1 + 2e2x (b) 25(5x - 1)4 (c) (
3x -_
7 2)2
2. (a) 12 (b) 0 (c) 16 3. (a) 4 (b) a2 + 2 4. (a) EB = 2-4 em (b) H cm2
5. ( -t, #),maximum; (5, -75), minimum 6. 9
1. (a) 2(x - 1)2 + 7(x - I) + 6 (b) k < -4!
2.(a)-1<x~6 (b)x=6orx=-4 3. e2x(l + 2x); 4e2 x(l + x) 5. 4 square units 6. (a) (2, - 3), radius 4
219
-1+,[7 1. (a) ~ (b) x = ± 1 or ± 2
2.(-1,-4) -3 2
3. (a) -2 + 3 (b) 4x(x2 - 3)
X X
(c) x(l + 2log. 2x) 4. (a) b2 = ac (b) 9 5. 2x- y + 5 = 0
6 (a) (x + 2)3 + C . 3
1 (b) log. x - x + C
(c) -2log(3- 2x) + C
2. 3·9 em 3. 4x - 3y - I = 0 4. 24n cubic units 5. (a) x= ±2orx= ±3 6. (a) io (b) t
(b) k ~ -5 or k ~ 3 (c) -M
1. (a) I - l~ge 2x (b) ex'(l + 2x2) (c) 16x(2x2 + 1) 3
X
x2 I 2. (a) :._ + ~ + C (b) ~xt- 2xt + C
2 X
(c) (2x ; 1)3 + C
3. (a) x 2 = 8y (b) x 2 = 8(y + I) 4. y = x 3
- 5x2 + 6x - 1
5 64n b' . , 3 cu !CUllltS
6. x < 2 and x > 4; (2, 22) maximum, (4, 18) minimum; (3, 20) point of inflexion.
1. (a) if radians (b) ;i radians (d) -\"-radians (e) ¥radians (g) ¥ radians (h) If radians (j) Tb- radians (k) -j' radians (m) -¥-radians (n) .!If radians
2. (a) 90° (b) 540° (d) 15° (e) 225° (g) 2° (h) 105° (j) 112·5° (k) 255° (m) -40° (n) 300°
3. (a) 85·9437° (b) 114·5916" (d) 2·8648° (e) 157·5634° (g) -180-4817° (h) 359·8175°
(c) j' radians (f) 1f radians (i) i radians (I) N radians ( o) 3
254 radians
(c) 240° (f) 720° (i) 70° (I) 45° (o) -157·5° (c) 68·7549° (f) 17-1887°
4. (a) 0·4887 radians · (b) 1·3439 radians (c) 1·9722 radians (d) 2·3911 radians (e) 2·0333 radians (f) 1-6991 radians (g) 5·2098 radians (h) -5·5441 radians
n ,j3 1 5n I 1 5' (a) 3' 2' 2 (b) 4' - .[2' - .j2
5n -,ft 1 2n ,j3 1 (c) 3' -2-,2 (d) 3' 2' -2
(e) 3n _!__ _ _!__ 4 ',fi' ,j2
ovG 6. (a) 120 , -, -yG 2
(b) 540°, 0, 0
(c) 480°, JJ, -..(3 2
(d) 270°, -1,doesnotexist
(e) 5WO ! -=-!_ (f) 225° -=! 1 '2' ..(3 ',fi'
7. (a) 0·8660 (b) 0·7071 (c) 2·5722 (d) 0·5646 (e) 0·8855 (f) 3·0777 (g) 0·2588 (h) 0·6247 (i) 0·8090 (j) -1·5574 (k) -1·4142 (I) -7·0153
8. (a) 30° (b) W
(c) (i) 2n d' (") 3n d' ("') 5n d' 3 ra 1ans; u 4 ra Ians; 111 6 ra 1ans
(d) 30°, 108°
I. (a) 38·4845 sq. em; 21·9911 em (b) 6·125 em (c) 10·7188 sq. em
2. (a) 8 em (b) 12·5 em (c) 10·56 em (d) 30 em (e) 13-666 em
3. (a) 18·6 em (b) 1·8 em (d) 4·603 em (e) 6·731 em
4. 66cm
(c) 2·932 radians (f) 49·114 em
5. (a) 1·4193 radians (b) 1·8445 km 6. (a) 39 cm2 (b) 52 cm2 (c) 52·36 cm2
(d) 130·90 cm2 (e) 26·18 cm2 (f) 104·72 cm2
(g) 117-81 cm2 (h) 32·29 cm2
7. (a) 10·5 em (b) 5-498 em 8. (a) 2·618 radians (b) 150 degrees 9. (a) 176·715 cm2 (b) 12·315 cm2 (c) 2·34 radians
(d) 0·524 radians (e) 6·83 em (f) 10 em (g) 8·210 m 2 (h) 9·35 em (i) 224·8 degrees
10. (a) 0·4712 m (b) 0·2827 m2
11. (a) 12·5 cm2 (b) 7·5 cm2 (c) 22·5 cm2
12. (a) 2·810 cm2 (b) 2·265 cm2 (c) 31·695 cm2
13. (a) 102·047 cm2 (b) 16·350 cm2; 16 cm2
14. 72° 15. (a) 29·3215 m 16. (a) 3·4336 cm2
(b) 205·25 m2
(b) 3-4336 cm2
17. 8n = 25·13274 em
I. (a) 1; 2n 2. (a) 1 3. (a) 1; 4n 4. (a) ¥
(b) 2; 2n (b) 5
(b) 1; ~ (b) 6n
(c) 3; 2n (d) 4; 2n
(c) 1;-¥ (c) 2
5. (a) 1; 2n (b) 1; 2n (c) 2; n (d) 1; 2n 6. (a) x = 0, 2n (b) X = f, -1f
(c) x = 0, lj, ~' n
I. 1 2. (a) x = 0, 2n 3. (a) 2; 2n
(d) 1;-¥ (g) 1; 2n
5. 2n; 5
(b) x = n (b) 5; 2n (e) 1; 2n (h) 1; 2n
(c) x = ~,¥ (c) 1; n (f) 1; 4n (i) 3; n
220
I. n 2. ~ 3. 2n
I. (b) -2n, -n, 0, n, 2n 2. (a) -2n, -n, 0, n, 2n 3. n
2. 3 3. (a) X= 0, 1·9 (b) X= 0, 2·28 (c) X= 0, 2-48 4. (b) (i) X = 0, 1·9 (ii) X = 0, 2-48
(iii) X = 0·52, 2-62 5, X= 0, 1·24 6. X= 0·79
I. 6 cos 6x 2. 3 sec2 x 3. -4 sin 4x 4. 1 cos lx 5. n cos nx 6. -2 sin (2x + 5) 7. 3 sec2 (3x - 1) 8. 2cos 2x 9. 2 sin(3 - 2x) 10. 8 sec2 2x
11. b cos bx 12. -b sin(bx + c) 13. cosx+sinx 14. -2sinx+3sec2 x 15. 3 cos 3x - 2 sin x 16. t sec2 ~ - 1 17. -3 sin 3x- 2 cos 2x 18. 2 sec2 2x + 4 cos 4x 19. 6 cos(3x - 4) 20. 4 sin(~ - x) 21. sin x + x cos x 22. cos2 x - sin2 x 23. x(2 tan x + x sec2 x) 24. ex(sin x + cos x) 25. x2(3 cos 2x - 2x sin 2x) 26. cos x. e•inx 27. 2 sin f + x cos f 28. ex( cos x - sin x) 29. 2 cos x sec2 2x - tan 2x sin x
COS X 30. --.-
2- = -cosec x cot x
sm x sinx
31. --2
- = secx tan x COS X
33 ex( sin x - cos x)
· sin2 x
35. -4 sin 2x cos 2x
sec2 x 32. --2 - = cosec2 x
tan x
34. 3 sin2 x cos x
36 2 cos 2x
2 2 . -.--= cot x sm 2x
37. 2(cos x + x sin x) 38. _sin x = _ tan x cos2 x cos x
39. 2e2x(sin 2x + cos 2x) 40. e-x(sec2 x - tan x) 41. -2 sin (2x + ~) 42. 2x(sin 2x + x cos 2x)
43. -2 cos x e4x(tan x - sec2 x)
(1 + sin x)2 44' tan2 x
45 x cos x - 2 sin x - sin 2x
• 3 46.---X ,jcos 2x
47. 2 cos 2x cos 3x + 3 sin 2x sin 3x
cos2 3x
48. 3 sec2 3x = 3 tan 3x sin 3x . cos 3x
I. (a) -lcos 2x + C (c) tan x + C (e) 2 tan f + C (g) -sin(~ - x) + C
(b) t sin 5x + C (d) -!cos 3x + C (f) -1 cos (2x + 3) + C (h) sin x + cos x + C
(i) t tan (4x + 3) + C (j) t sin 3x - t cos 2x + C (k) x - t sin 3x + C (1) (m) -f.- + t cos (2x - 3) + C (n) x + cos (t - x) + C
cos (n- x)
(o) !tan2x + icos4x + C 2. (a) 1 (b) 1
(d) n (e) -1 (g) ~ (h) ~
3. (a) 2 (b) 1
(c) 1 (f) 2 (i) 4 (c) t
(d) t (e) 4 - t,fi 4. (b) n cubic units 5. t
1. (a) --J sin¥ (c) n sec2 nx
(e) -3 tan 3x
2. (a) -25 sin 5x (c) 2(cos2 x - sin2 x)
4. 0; 0
(b) 2sinxcosx (d) 2e2x(sin 2x + cos 2x)
(f) 3x cos 3x - sin 3x 3x2
(b) -16 cos 4x
5. X0 = ts'o radians; rlo cos No
6. (a) -t cos 3x + C (b) t tan 4x + C (c) f. sin nx + C
7. (a) t (c) t- Fa
8. -fa- sec2 % 10. e- 1 1t'. t 12. loge sin x + C 13. (a) 0·7071
(c) 2·1213 15. (a) 1 16. '! 17. t 18. ~ ~ 2·5465 square units 19. 1, -1 20. t 21. 0·6165 square units
1. I= 0, I= t 2. 320 m/s2
3. I= 2
(b) 1 + t
(b)t (d) 0·9239
(c) 0
4. (a) s = 8 (b) s = -1, s = 0 (c) 1m (d) -6 mjs, 2 mfs (e) I = 3 (f) 2 mfs2
5. (a) 24·5 m (b) 29·4 mjs (c) 9·8 mfs2 (d) 78·4 m, 39·2 mfs
6. (a) 3m (b) v = 21 - 4 (c) -2 mjs, 2 mjs, opposite directions (d) after 2 seconds (e) at 1 = 1 and 1 = 3, the body ,passes through the
origin in opposite directions at the two times. 7. 2 units per second per second, 19 units per second 8. 38 mjs, 30 mfs2
9. (a) X = 0, X = 4, X = 21 (b)17m (c)8m/s
221
(d) I = 1, I = 2 (e) X = 1, X = 4 (f) a= -10 + 121, 0·16 mfs
10. (a) 12 mfs 11. (a) 17 mfs (b) 8 mfs2
(c) 3 s, 39m
1. (b) 2 mfs (c) returns when I = ~ (d) I = t, a = -4; I = ¥,a = 4
2. I= 2l 6", 1 = ¥;x = 3·73,x = 1·5 3. (a) 1 m (b) 1-41 m
(c) 1 mjs, 0 mjs, -1 mfs (d) -1·37 mjs2, 0 mfs2
4. (b) 1·7 mfs
1. (a) 54 units per second (b) s = 213
(c) 54 units (d) after 2 seconds 2. x=21·5 3.(a)v=-61 (b)s=-312 +27
(c) after 3 seconds, -18 metres per second 4. s = 228 5. s = 12t 6. (a) 43 units per second (b) 93 units 7. (a) s = -41 (b) s = -212 + 18
(c) after 3 seconds, v = -12 8. after 10 seconds, 266t m
1. (a) 12000 2. (a) 81 000 (approx) 3. (a) 1977 4. (a) 160 g (approx)
(c) 81 g (approx)
(b) 29 500 (approx) (b) 116000 (approx) (b) 807 000 (approx) (b) 102 g (approx)
5. (a) 9000°K (b) 6800°K (approx) 6. (b) (i) 0·9 years (approx); 1·1 years (approx)
(ii) 366 g (approx)
1. (a) 1·926 kg (b) 2-473 kg (c) 13-86 days 2. (a) 2600 (b) 2931 (c) 1Hyears 3. (a) -0·0223143 (b) 35·8 g (c) 31·06 years 4. (b) (i) 1620 years (ii) 948 years 5. $3297-44 6. (a) 4 mjs 7. 42000 8. (a) 0·0896048 9. (a) 8488 kgfm2
10. (a) 7-79 g 11. (a) 0·0126022
1. (a) 10(2x + 1)4
(c) -sin xe'ou
(b) 1·79 mfs (c) 0·03 mfs
(b) 1·3 X 107
(b) 1653 kgfm2
(b) 1386 seconds (b) 97 000 (approx)
(b)~ (d)~
x2 + 1
2. (a) f, (b) 290 (c) 5 3. a= 16,b= 12ora= -5t,b= It
· 4. 76°42', 76°42', 103°18', 103°18'; 146 cm2
5. (1, 1), minimum; (-2, 28), maximum; ( -!, 14!), point of inflexion.
I. (a) 3 sin2 x cos x (b) ex(! + x)2
-12 2 (c) (
3x _ l) 5 (d) cos
2 x - sin x
2. -16! 3. A = 2, B = 3, C = 4 4. 2x + y- 5 = 0 5. It square units; f cubic units 6. I metre/second
I. y = x 3 + x + 2 2. (a) 1545 (b) 3(212
- I)
3 dy - 12 24. d2y - 96 36 · dx - x3 - x4 ' dx2 - x 5 - x4
- 0 'f - 11. dy - 0 'f - 0 l,. y - I X - 3 ,- - I X - Or 2 , dx
d2y - 2 = 0 if x = 0 or 21 dx
4. (a) 8·4 em (b) 50·3 cm2
6. 105 850 (c) 4·0 cm2
I. (a) 3x2 + lOx - I (b) 2(x2 + 3x + I)
· (2x + 3)2
(c) .-3 cosec 3x cot 3x
2. (a) x3 - x 2 + h + C
(c) flog. (5x - 3) 3. 13-443 4. (a) 2 (b) -5 5. 26 metres from origin 6. (a) t (b) t
I. 112; 1100 2. 1 square unit
3. (a) x = ± '!} 4. 101·3 km 5. (a) log. 2 - 1 ~ -0·307
(d)~ e2x
(b) 1x1+ fx 1 + C (d) X +loge X
(c) -f (d) 14
(b) a = 2, b =. 3, c = 1
(b) !(e4 - e2
) - 1 ~ 22-605 (c) 2
6. (a) 20 (b) 10·2
I. (a) e2 - 1 ~ 6·389 square units
(b) f(e4 - 1) ~ 84·192 cubic units
2. (a) 6225 (b) 3t 3. 3x + y + 1 = 0
4. r = _!Q_ ~ 1·4 metres 4+n
5. 4·003 square units
222
I. -12 2. (a) (x - 2)(x + 1)
3. 2x- 1 x-2
4. t = 25·64 5. X= 15 6. X= 1!
7. 1- X
x-5 8. (a) (5 - x)(2 + x) 9. 7 ,j2
10. ,j3 + 1 11. (a) 6 = ml-47
12. a = 110, b = -60 13. X= 2, y = 3 14. x = 2 or -It 15. m = 5 16. a= ±5·22 17. -H<x<4! 18. -4 <X<! 19. t 20. 78°28' 21. a = 5,j2 em 22. (a) 2x - 3y + 21 = 0 23. (a) ~ 24. (a) t
-2x- 5 25
' x 2 + 5x + 6 26. -5
5,[5- 3 27. 4
(b) 5(x + 2y)(x - 2y)
(b) (x + 2)(x - a)
(b) 9 = ml-694
(b) 2x + 3y - 1 = 0 (b) rt (b)!
28. x = -1, y = 2 or x = 6, y = 23 29. x = 6 or -6 30. -5! <X< 6! 31. 94·75 m 2
32. 93 33. a= 5, d = 3 34. T50 = 348, S50 = 8825 35. T9
36. X= 2, I'= 3 37. 1830 38. 20475 39. 31·25 40. x = 2or -3! 41. X = 3 ± 2,j2 42. k = -1 43. (a) ~ (b) 6 (c) rt (d) ! 44. x = ±2orx = ±3 45. k = 2-h 46. a> 0, ~ < 0 47. t <X< 2 48. A= 2, B = -9, C = 1 49. 4(x - 1)2 + 2(x - I) + 3 50. k = 12
6 2 3 ) -3 51. (a) 2 - x 3 (b) 8x(x + 3) (c (
2x _ l) 2
52. (a) 6x2 - 16x + 10 (b) ~
1 53. 5 + 2
X
54. -1; 20 55. b = -5 56. 5x + y- 2 = 0 57. X + 2y - 7 = 0 58. a= 1, b = 3, c = -3 59. k = 1 60. llor-12 61. 32 62. u 63. (0, 0) and (- 2, 2) 64. /(2) = 13, /'(2) = 20, /"(2) = 24 65. 108(3x 2) 2
66. x < -3 and x > I 67. (2, -3), radius = 2
x3 68. y = - - tx2 + x + C
3 69. f(x) = x3
- x 2 + x + C I
70. y = 3x - x + C
71. y = 5 + 7x - 2x2
72. 5 73. x 2 = I2y
74. 2t 75. I5t 76. 57t 77. 12~ 78. (t, -5~) 79. X< Jt 80. 29x + y- 13 = 0 81. (0, 1), minimum; ( -~, 1-i\r), maximum
12 82. s = 13
- 2 + 5
x3 83. y = x + 2x2
- ~ - 12
x3 x2 84. y = - - - + 3x - 5
2 2 85. (a) 12 (b) t 86. 36 square units
1 87. (a) 2x3 + 2 + C (b) t(3x + 2)3 + C
X
ss. tv(6x + 1)3 + c 89. 2 square units 90. It square units 91. 21 square units 92. 1·290 square units 93. -if, 94. !lf cubic units ~ 8·378 cubic units 95. (a) 6x - 8e2
x (b) ~ (c) ex(Iog. x + ~) 96. 10·472 ern 97. 372 cm2
98. 49·7 em 100. (a) 3 cos 3x (b) 2 sec2 (2x - 1) (c) 3 sinG- x)
223
101. -sin xe'osx
102. 2(tan x + x sec2 x) 103. -t cos (4x + 2) 104. n- 1 105. 2 square units
106. dy = -sin x = -tan x dx cos x
107. -3 units 108. I = I and I = 5 109. 4 units/second 110. I= 3 Ill. 2 units/second/second 112. 82 metres 113. x= -21+ 12
13 114. s =-- 212
2 115. k = 0·1116; 98000 116. 14 metres/second 117. I 118. (0, 0), maximum; (2, -4), minimum 119. k > !o 120. a= 3, b = -2, c = 9 121. (3t, 4i) 122. k = 3 123. a < 0, d < 0 124. k = 1 125. k = 8; 15 126. x = 4 or I 127. -1 <y < 1;3
128. X=* 129. -234 130. 8 131. a= 4!; r = 2 132. 21, 27, 33, 39, 45 133. X - )' + 2 = 0 134. Perpendicular distance from (0, O) to 3x + 4y = 30 is
equal to the radius. 135. 2 units 136. X - )' + 1 = 0 137. x 2 = 24y 138. (0, -1) 139. x2 = 16y 140. (a) 4e4x-l
141. (a) t log, (3x - 2) + C
142. t square unit 143. t ~ 0·637 square units
(b) 2x(tan 2x + x sec2 2x)
(b) -23 + c e x
144. (a) 40° (b) 70° ,j91.
146. AD= ~em 7 4·8cm 2
147. 144° 148. 15
