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Preface

About the book


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IIT - MATHS

SET - 1


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INDEX

1. BASIC MATHEMATICS...................................................................................

2. ALGEBRA PROGRESSION .............................................................................

3. BASIC TRIGONOMETRY ..............................................................................

4. TRIGONOMETRY EQUATIONS ...................................................................

5. INVERSE TRIGONOMETRY FUNCTION...................................................

6. PROPERTIES OF TRIANGLE .......................................................................

7. CO - ORDINATE GEOMENTRY ...................................................................

2

54

132

164

186

210

232


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1 BASIC MATHEMATICS


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BASIC TRIGONOMETRY

3

Number System

(i) Natural numbers : N = {1, 2, 3, 4, . . . . . }

(ii) Whole numbers : W = {0, 1, 2, 3, 4, . . . . . }

(iii) Integers : Z or I = {. . . . . 3, 2, 1, 0, 1, 2, 3, . . . . .}

Natural numbers are also called positive integers (denoted by Z+or I+)Whole numbers are also called nonnegative integers.

The set of negative integers, Zor I= {. . .. . 3, 2, 1}.

The set of non positive integers is {, 3, 2, 1, 0}.

Zero is neither positive nor negative but it is nonpositive as well as nonnegative.

(iv)Rational numbers:Numbers of the form p/q where p, q Z and q 0 (because division by

zero is not defined). Q represents their set. All integers are rational numbers with q = 1

When q 1 and p, q have no common factor except 1, the rational numbers are calledfractions.

Rational numbers when represented in decimal form are either terminating or non

terminating but repeating.

e.g., 5/4 = 1.25 (terminating)

5/3 = 1.6666 . . . . . (non terminating but repeating)

(v) Irrational numbers: Numbers, which cannot be represented in qp

form.

In decimal representation, they are neither terminating nor repeating all surds fall into this

category

e.g., 2 , 3/115 , p, etc.

Note : p 22/7, 22/7 is only an approximate value of p in terms of rational numbers,

taken for convenience Actually p = 3.14159 . . . . .

(vi)Real numbers: All rational and irrational numbers taken together form the set of real

numbers, represented by R. This is the largest set in the real world of numbers.

Also note that

Integers which give an integer on division by 2 are called even integers otherwise they are

called odd integers.

Zero is considered as even number.

The set of natural numbers can be divided in two ways.

(i) Odd and even natural numbers.

(ii) Prime numbers (which are not divisible by any number except 1 and themselves) andcomposite numbers (which have some other factor apart from 1 and themselves).


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1 is neither prime nor composite

2 is the only even number which is prime

Set Theory

Basic Concept

Set: A set is a welldefined collection of objects or elements. Each element in a set is unique.Usually but not necessarily a set is denoted by a capital letter e.g. A, B, . . . . . U, V etc. and the elements

are enclosed between brackets {}, denoted by small letters a, b, . . . . . x, y etc. For example:

A = Set of all small English alphabets

= {a, b, c, . . . . . x, y, z}

B = Set of all positive integers less than or equal to 10

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

R = Set of real numbers

= {x : < x < }

The elements of a set can be discrete (e.g. set of all English alphabets) or continuous (e.g., set of

real numbers). The set may contain finite or infinite number of elements. A set may contain no elements

and such a set is called Void set or Null set or empty set and is denoted by (phi). The number of

elements of a set A is denoted as n(A) and hence n () = 0 as it contains no element.

Union of sets

Union of two or more sets is the set of all elements that belong to any of these sets. The symbol usedfor union of sets is

i.e., AB = Union of set A and set B = {x : x A or x B(or both)}

e.g. If A = {1, 2, 3, 4} and B = {2, 4, 5, 6} and C {1, 2, 6, 8}, then ABC = {1, 2, 3, 4, 5, 6, 8}

Intersection of sets

It is the set of all the elements, which are common to all the sets. The symbol used for intersection

of sets is i.e., AB = {x : xA and xB} e.g. If A = {1, 2, 3, 4} & B = {2, 4, 5, 6}and C = {1, 2, 6, 8},

then ABC = {2}.

Remember that n(A B) = n(A) + n(B) n (A B)

Difference of two sets

The difference of set A to B denoted as A B is the set of those elements that are in the set

A but not in the set B i.e., A B = {x : x A and xB}

Similarly B A = {x : x B and x A}. In general A B B A

e.g. If A = {a, b, c, d} and B = {b, c, e, f}, then A B = {a, d} and B A = {e. f}


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BASIC TRIGONOMETRY

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LOGARITHM

If a is a positive real number other then 1 and ab= c, then we write logac = b obviously c is positive.

For example log381 = 4 34= 81

Note

The expression logba is meaningful for a > 0 and for either 0 < b < 1. or b > 1

a = b alogb

logab = alog

blog

c

c

logba

1 log

ba

2 1b0ifaa0

1bif0aa

21

21

Brain Teaser 1 : log x2= 2logx, is it true or false?

Formulae

(i) loga|mn| = log

a|m| + log

a|n|

(ii) loganm

= loga|m| loga|n|

(iii) loga|mn| = n log

a|m|

(iv) logab = log

cb log

ac

(v) log ka N = k1

logaN

Modulus Function

Let x R, then the magnitude of x is called its absolute value and in general, denoted by |x| and

defined as |x| =

0x,x

0x,x

Note that x = 0 can be included either with positive values of x or with negative values of x. As we

know all real numbers can be plotted on the real number line, |x| in fact represents the distance of number

x from the origin, measured along the numberline. Thus |x| 0 secondly, any point x lying on the real

number line will have its coordinate as (x, 0). Thus its distance from the origin is 2x

Hence |x| = 2x . Thus we can define |x| as |x| = 2x

e.g if x = 2.5 then |x| = 2.5, if x = 3.8 then |x| = 3.8

Brain Teaser 2 : If x R then 2

1,

2

1,

2

1is

4x

x2

or not defined.

Basic Properties :

||x|| = |x|

|x| > a x > a or x < a if a R+and x Rif a R

|x| < a a < x < a if a R+and no solution if a R{0}

|x + y| |x| + |y|

|xy| = |x||y|


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IIT- MATHS

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|y||x|

yx

, y 0

Intervals

Intervals are basically subsets of R and are of very much importance in calculus as you will get to

know shortly. If there are two numbers a, b R such that a < b, we can define four types of intervals as

follows:

Open interval : (a, b) = {x : a < x < b} i.e., end points are not included.

Closed interval: [a, b] = {x : a x b} i.e., end points are also included. This is possible only when

both a and b are finite.

Openclosed Interval: (a, b] = {x : a < x b}

Closedopen interval:[a, b) = {x : a x < b}

The infinite intervals are defined as follows

(a, ) = {x : x > a}

[a, ) = {x : x a}

(, b] = {x : x b}

intervals are particularly important in solving inequalities or in finding domains etc.

Inequalities

The following are some very useful points to remember

a b either a < b or a = b.

a < b and b < c a < c.

a < b a + c < b + c cR

a < b a > b i.e., inequality sign reverses if both sides are multiplied by a negative number.

a < b and c < d a + c < b + d and a d < b c.

a < b ma < mb if m > 0 and ma > mb if m < 0.

0 < a < b ar< brif r > 0 and ar> brif r < 0.

a

1a 2 a > 0 and equality holds for a = 1.

a

1a 2 a < 0 and equality holds for a = 1.

If a1> b

1, a

2> b

2, a

3> b

3. . . . . , where a

i> 0, b

i> 0, i = 1, 2,

Then a1+ a

2+ a

3+ . . . > b

1+ b

2+ b

3+ . . . and a

1a

2a

3. . . > b

1b

2b

3. . .

If a > b, p and q are some positive integers, then following results are evident.


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BASIC TRIGONOMETRY

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a > b an> bn an< bnwhere n N

a > b al/q> bl/q ap/q> bp/q

Wavy Curve Method

In order to solve inequalities of the form

0

xQxP,0

xQxP , where P(x) and Q(x) are polynomials, we use the following method:

If x1and x

2(x

1< x

2) are two consecutive distinct roots of a polynomial equation, then within this

interval the polynomial itself takes on values having the same sign. Now find all the roots of the polyno-

mial equations P(x) = 0 and Q(x) = 0. Ignore the common roots and write

m321n321

x.....xxx

x.....xxxxf

xQ

xP

,

Where a1, a2, . . . . . an, b1, b2, . . . . . , bmare distinct real numbers. Then f(x) = 0 for x = a1,a

2, . . . . . , a

nand f(x) is not defined for x = b

1, b

2, . . . . . , b

mapart from these (m + n) real numbers

f(x) is either positive or negative. Now arrange a1, a

2, . . . . . , a

n, b

1, b

2, . . . . . , b

min an increasing order say

c1, c

2, c

3, c

4, c

5, . . . . . , c

m+n. Plot them on the real line. And draw a curve starting from right of c

m+nalong

the real line which alternately changes its position at these points. This curve is known as the wavy curve.

The intervals in which the curve is above the real line will be the intervals for which f(x) is positive

and intervals in which the curve is below the real line will be the intervals in which f(x) is negative.

Factor Theorem

Let p(x) be a polynomial of degree greater than or equal to 1 and a be a real number such that p(a)

= 0, then (x a) is a factor of p(x). Conversely, if (xa) is a factor of p(x), then p(a) = 0.

Remainder Theorem

Let p(x) be any polynomial of degree greater than or equal to one and a be any real number. If p(x)is divided by (xa), then the remainder is equal to p(a).

DETERMINANTS

Consider the equations a1x + b

1y = 0 and a

2x + b

2y = 0. These give

2

2

1

1

2

2

1

1

b

a

b

a

b

a

x

y

b

a

a

1b

2 a

2b

1= 0

We express this eliminant as 2211

ba

ba

= 0.


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IIT- MATHS

8

The expression22

11

ba

bais called a determinant of order two, and equals a

1b

2 a

2b

1.

A determinant of order three consisting of 3 rows and 3 columns is written as

333

222

111

cba

cba

cba

and is equal to a1 3322

cb

cb

b1 3322

ca

ca

+ c1 3322

ba

ba

= a1(b

2c

3 c

2b

3) b

1(a

2c

3 c

2a

3) + c

1(a

2b

3 b

2a

3)

The numbers ai, b

i, c

i(i = 1, 2, 3) are called the elements of the determinant.

Function

In the study of natural phenomena and the solution of technical and mathematical problems, it is

necessary to consider the variation of one quantity as dependent on the variation of another. For example,

in studies of motion, the path traversed is regarded as a variable, which varies with time. Here we say that

the distance traversed is a function of time. The area of a circle, in terms of its radius R, is pR2. If R takes

on various numerical values, the area assumes different numerical values. So the variation of one variable

brings about a variation in the other. Hence area of the circle is a function of the radius R.

If to each value of variable x (within a certain range) there corresponds a unique value of another

variable y, then we say that y is a function of x, or, in functional notation y = f(x). The variable x is called

the independent variable or argument. And the variable y is called the dependent variable. The relation

between the variable x and y is called a functional relation. The letter f in the functional notation y = f(x)

indicates that some kind of operation must be performed on the value of x in order to obtain the values of y.

f(x)

y2

y1

x1

x2

x3 x

y = f(x)

f(x) L

C

B

A

x0 x

y = f(x)

y3

y2

y1

Fig (a) Fig (b)

These figures show the graph of two arbitrary curves. In the figure any line drawn parallel to y-

axis would meet the curve at only one point. That means each element of X would have one and only one

image. Thus the figure (a) would represent the graph of a function.

In the figure (b) certain line (e.g. line L) would meet the curve in more than one points (A, B and

C). Thus element x0of X would have three distinct images. Thus this curve will not represent a function.

The set of all possible values which the independent variable (here x) is permitted to take for a

given functional dependence to be defined is called the domain of definition or simply the domain of the


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BASIC TRIGONOMETRY

9

function.

e.g. The function y = sin x is defined for all values of x. Therefore its domain of definition is the

infinite interval < x < .

The function y =1x

1

is defined for all x > 1 its domain is (1, ).

Elementary Functions:

(i) Constant function: y = c where c is a constant, defined for all real x.

(ii) Power function: xy

(a) a is positive integer. The function is defined in the infinite interval < x < .

(b) a is negative integer. The function is defined for all values of x except for x = 0.

(iii) General exponential function:y = ax, where a is positive not equal to unity. This function

is defined for all values of x.(iv)Logarithmic function:y = log

ax, a > 0 but a 1. This function is defined for all x > 0.

(v) Trigonometric function: y = sinx, y = cosx defined for all real x

y = tanx, y = secx, defined for R (2n + 1)2

.

y = cotx, y = cosecx, defined for R n , where nl

It must be noted that in all these function the variable x is expressed in radians. All these

function have a very important property that is Periodicity.

` Is sec2 tan2 = 1 valid for all R (real) ?

(vi) Algebraic function:

(a) Polynomial function: y = a0xn+ a

1xn1+ + a

n, where a

0, a

1 a

nare real constants

(a00) and n is a positive integer, called the polynomial of degree n.

e.g. y = ax + b, a 0 (a linear function)

y = ax2+ bx + c, a 0 (a quadratic function)

A polynomial function is defined for all real values of x.

(b) Rational Function y =m

1m1

m0

n1n

1n

0

b...xbxb

a...xaxa

e.g:y = a/x (inverse variation)

The rational function is defined for all values of x except for those where the denominatorbecomes zero.

(c) Irrational function e.g. y = 2

2

x51

xx2


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IIT- MATHS

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Differential Calculus

Let y = f(x) be a function. Putting the values of x in this relation, we obtain the corresponding

values of y. Suppose we start putting some values of x in increasing order. The respective values of y

that we obtain may turn out to be in increasing order, or in decreasing order, or they may remain constant,

or they may even have a mixed trend, depending upon the type of function.

Let us take two values of x: x1and x2(x1< x2). So, y1= f(x1) and y2= f(x2)

Then, the quantity

12

12

xx

yywill tell us the average rate of change of y w.r.t. x in the interval [x

1, x

2] .

Let y2> y

1

12

12

xx

yy

is positive Function is increasing on an average.

if y2< y

1

12

12

xx

yy

is negative Function is decreasing on an average.

If y2= y

1

12

12

xx

yy

is zero Function is constant on an average.

As you can see, if x1and x

2are sufficiently far apart, the quantity

12

12

xx

yycan not give the exact

idea of the variation of y w.r.t. x in the interval [x1, x

2]. it just provides an overall information. For

example if y2= y

1it does not necessarily mean that y is same for all x in the interval [x

1, x

2]. Thus, to

obtain a sufficiently accurate information, we have to choose x1and x

2 sufficiently close to each other.

This sufficiently close is the key word here. To know the rate of change of y w.r.t. x at x = x1, we take x2

very near to x1(as much as possible), i.e., 2x tends to x1and then calculate

12

12

xx

yy

. In the limiting case,

we say that x2nearly coincides with x

1and represent it as x

2 x

1. We use the notation

1xxdx

dy

for

12

12

xx

yy

as x2 x

1. dx means small change in x (near x = x

1) and dy means the corresponding change in y. We call

dx

dythe derivative or the differential coefficient of y w.r.t. x.

y2 y1

x2 x1

x1 x2

y

x

(You can understand it physically by taking x as time and y as displacement of a body,


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BASIC TRIGONOMETRY

11

Thendx

dydenotes the magnitude of velocity).

dx

dyis also represented as f (x) or

dx

)x(df

Graphically, 1xxdx

dy

(i.e., dxdy

computed at x = x1) denotes the slope of the tangent to the curve y =

f(x) at x = x1

We will not here derive the formulae for (x) of various functions, but we give the results of the

derivations here,

Basic Differentiation Formulae

y = constant 0dx

dy y = tan1x 2

x1

1

dx

dy

y = xn 1nnxdx

dy y = cot1x 2x11

dx

dy

y = sinx xcosdx

dy y = cosec1x

1x|x|

1

dx

dy2

y = cosx xsindx

dy y = sec1x

1x|x|

1

dx

dy2

y = tanx xsecdxdy 2 y = ax alnadx

dy x

y = cotx xeccosdx

dy 2 y = ex xedx

dy

y = sin1x 2x1

1

dx

dy

y = log

ax alnx

1

dx

dy

y = cos1x 2x1

1

dx

dy

y = ln x

x

1

dx

dy

Some Important Theorems

The following are very important theorems, which can be applied directly.

Theorem 1:

If a function is of the form y = k f(x), where k is a constant, thendx

)x(dfk

dx

dy

Theorem 2:

The derivative of the sum or difference of a finite number of differentiable functions is equal

to the sum or difference of the derivatives of these functions.


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IIT- MATHS

12

i.e., if y = u (x) + v (x) + w(x) then y= u(x) + v(x) + w(x).

Theorem 3

The derivative of the product of two differentiable functions is equal to the product of the

derivative of the first function with the second function plus the product of the first function

with the derivative of the second function: i.e., if y = uv, then y= uv + uv.

This formula can be extended for the derivatives of the product of any (finite) number of

functions.

Theorem 4

If y = )x(v)x(u

, then y = 2vuvvu

dx

dy

Theorem 5

If y = uv, where u and v are functions of x, then y= vuv-1u+ uvln u.v

Derivative of a Composite FunctionGiven a composite function y = f(x), i.e., a function represented by

y = F(u), u = f (x) or y = F[f(x)],

then y=dx

du

du

dF

dx

dy

This is called the chain rule. The rule can be extended to any number of composite function;

e.g. if

y = f(u(v)), then y=dx

dv

dv

du

du

df

dx

dy .

Parametric Representation of a Function and its Derivatives

We find the trajectory of a load dropped from an aeroplane moving horizontally with uniform

velocity v0at an altitude y

0. We take the co-ordinate system as shown and assume that the load is dropped

at the instant the aeroplane cuts the y-axis. Since the horizontal translation is uniform, the position of the

load at any time t, is given x = v0t, y = y

0

2

gt 2

X

(x, y)

v0

Y

Those two equations are called the parametric equations of the trajectory because the two vari-

ables x and y have been expressed in terms of the third variable t (parameter) i.e. two equations x = (t),

y = (t)


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BASIC TRIGONOMETRY

13

where t assumes values that lie in a given interval (t1, t

2)

Then )t()t(

)dt/d(

)dt/d(

dt/dx

dt/dy

dx

dy

Second Derivative of a Function

The second derivative of y w.r.t. x is the function obtained by differentiatingdx

dyw.r.t. x.

It is represented as 22

dx

ydor y or f (x). e.g. If y = x5then

dx

dy= 5x4

So,

dx

dy

dx

d

dx

yd2

2

= )x5(dx

d 4 = 5.4 x3= 20x3

The acceleration a of a particle is the second derivative of the distance s (given as a function oftime).

i.e. if s = f(t) then v =dt

ds= f (t) and a = 2

2

dt

sd=

dt

dv= f (t)

THE BEHAVIOUR OF FUNCTIONS

The following behaviours of a function are important to study

Increasing and Decreasing Functions

(i) Increasing Functions

If y = f(x) and x2> x

1implies y

2> y

1for any x belonging to the interval [a, b], then y is said

to be an increasing function of x. (x) increases in [a, b] f (x) > 0 x in (a, b).

(ii) Decreasing Function

If x2> x

1y

2< y

1for any x belonging to [a, b], then y is said to be a decreasing function

of x.

f(x) decreases in [a, b] f (x) < 0 x in (a, b)

For a constant function, f (x) = 0For a non-decreasing function f (x) 0

For a non increasing function, f (x) 0

Maxima and Minima of Functions

A function f is said to have a maxima at x = x0if f(x) < f(x

0), x in the immediate neighbourhood

of x0.

Similarly, a function f is said to have a minima at x = x0 if f(x) > f(x

0), x in the immediate

neighbourhood of x0

We have used the word immediate here because a given function may have any number of high


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IIT- MATHS

14

and low points. It is just like moving on an uneven surface (which has many bumps and depressions).

Mathematically, these bumps are called the points of local maxima and the depressions are called the

points of local minima. The highest of all the bumps is the global maxima and the lowest of all the

depressions is the global minima. We state here the preliminary methods only to find the maxima and

minima of functions.

(a) Finddxdy or f (x)

(b) Find the points at which it becomes zero. These points are called critical points.

To find the points of maxima and minima we resort to either of the following tests.

(a) First Derivative Test

Suppose x = x0is a critical point i.e., f (x

0) = 0.

If f (x) changes sign from positive to negative in the neighbourhood of x = x0

Maxima at x0

If f (x) changes sign from negative to positive in the neighbourhood of x = x0

Minima at x0

(b) Second Derivative Test

(i) Find 22

dx

ydor f(x)

(ii) Compute the value of f (x) at the critical pointsIf it is positive Minima at those values of x.

If it is negative maxima at those values of x.

If the function is defined in an interval [a, b], then to find the maxima and minimum

values i.e., global maxima and global minimum of the function in that interval we com

pare the values of the function (i.e., y) at all the critical points and also the end points (i.e.,

y = f(a) and f(b)). Then the largest among them gives the global maximum values and

smallest gives the global minimum values.

Integral Calculus: The Antiderivative of Function

A function F(x) is called the antiderivative of the function f(x) on the interval [a, b] if, at all points

of the interval f(x) = F(x).

For example, the antiderivative of the function f(x) = x2 is3x3

, as'

3

x3

= x2. The function

13

xand2

3

x 33 are also antiderivatives of f(x) = x2. Infact, C

3x3

, where C is an arbitrary constant, is

the antiderivative of x

2

. So if a function f(x) possesses an anti-derivative F(x), then it possesses infinitelymany antiderivatives, all of them being contained in the expression F(x) + C, where C is a constant.


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BASIC TRIGONOMETRY

15

If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is called the indefinite

integral of the function f(x) and is denoted by the symbol f(x) dx. Thus, by definition f(x) dx = F(x) +

C, if F(x) = f(x). If a function f(x) is continuous on an interval [a, b], then this function has an antideriva-

tive. The process of finding the antiderivative of a function f(x) is called integration. Two different inte-

grals of a function differ by a constant.

Standard Elementary IntegralsIn the following integrals, C stands for an arbitrary constant.

)1n(,c1n

xdxx

1nn sec x dx = ln |sec x + tan x| + c

1n(C1n

))x(f(dx)x(f))x(f(

1nn ) 22 xa

dx=

a

1tan1 c

a

x

c|x|lndxx1

cxtan

x1

dx 12

cedxe xx c

a

xsin

xa

dx 122 orcos1 a

x+ C

sin x dx = cosx + c 2x1dx

= sin1 x + c

cos x dx = sinx + c 1xx

dx2 = sec

1x + c or cosec1x + c

sec2x dx = tanx + c

cosec2x dx = cotx + c

tan x dx = ln|cosx| + c = ln |secx| + c

cot x dx = ln |sin x| + c = ln |cosecx| + c

e.g. (i)

12

1x

dxx

12

1

2

1

+ c =3

2. x3/2+ c (ii)

c

x1

c12

xdxxdx

x

1 1222

The following points are to be noted:

dxx1

lnx + c if x is positive = ln (x) + c if x is negative becausedxd

(ln (x)) =x

1

(

1) =x

1 C|x|lndxx

1

2x1

1dx = sin1x or cos1x. It does not mean that sin1x = cos1x.

The only legitimate conclusion is that they differ by some constant. In fact sin1x (cos1x)

= sin1x + cos1x = /2.

If a is a constant, then a f(x) dx = a )x(f dx


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IIT- MATHS

16

[f(x) g(x)] dx, = f(x) dx g(x) dx

Methods of Integration

(i) Integration by Substitution

This method consists of expressing the integral f(x) dx, where x is the independent variable, in

terms of another integral where some other, say t, is the independent variable; x and t being connected

by the relation x =f (t). i.e., f(x)dx = f[j (t)] j (t) dt. This method is useful only when a relation x = (t)

can be so selected that the new integrand f(x)dt

dxis of a form whose integral is known

(ii) Integration by parts

dx)x()x(f)ii()i(

= f (x)

dxdx)x(dx

dfdx)x(

Integral of the product of two function = first function integral of secondintegral of (derivative

of first integral of second).Definite Integral

The difference in the values of an integral of a function f(x) for two assigned values of the inde-

pendent variable x, say a, b, is called the definite integral of f(x) over the interval (a, b) and is denoted by

b

a

.dx)x(f Thus b

a

),a(F)b(Fdx)x(f where F(x) is the antiderivative of f(x). Or, we write

).a(F)b(F|dx)x(F|dx)x(f ba

b

a

a is called the lower limit and b the upper limit of integration.

Note:

b

a

a

b

dx)x(fdx)x(f

b

a

c

a

b

c

dx)x(fdx)x(fdx)x(f

where c is any point inside or outside the interval (a, b). Geometrically definite integral represents area under curve.


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BASIC TRIGONOMETRY

17

Ex amp l e 1 :

In a DABC, the medians AD, BE and CF pass through G.

(a) If BG = 6, what is BE ?

(b) If FG = 4, what is GC ?

So l u t i o n :

(a) We have, BG =3

2BE

6 =3

2BE BE = 9

(b) We have,

GC = 2 FGGC = 2 4 = 8

Ex amp l e 2 :

Triangles ABC and DBC are on the same base BC with A, D on opposite sides of lne BC, suchthat ar. (DABC) = ar. (DDBC). Show that BC bisects AD.

So l u t i o n :

Since Ds ABC and DBC are equal in area and have a common side BC.

Therefore the altitudes corresponding to BC are equal

i.e.,AE = DF

Now, in Ds AEO and DFO,

We have

1 = 2 (vertically opposite angles)

AEO = DFO (90 each)

and AE = DF

DAEO @ DDFO (by A.A.S)

A

C

D

O F

2

1

BE

AO = OD

SOLVED SUBJECTIVE EXAMPLES


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IIT- MATHS

18

BC bisects AD

Ex amp l e 3 :

Solve |x2 3x 4| = x23x 4

So l u t i o n :

We know |x| = x when x 0

So, x2 3x 4 0

(x 4) (x + 1) 0 x 4 or x 1.

Ex amp l e 4 :

Solve 184x3= (54 2 )3x4

So l u t i o n :

Given equation is 184x3= 254( )3x4

Taking log on both the sides, we get(4x 3)log 18 = (3x 4) log(18.3 2 ) (since 3 2 = 18 )

or, (4x 3) log 18 = (3x 4) log (18)3/2or, 4x 3 = (3x 4)2

3

or, 8x 6 = 9x 12, or x = 6

Ex amp l e 5 :

Solve for x if log3x + log

9(x2) + log

27(x3) = 3

So l u t i o n :

log3x + log

9(x2) + log

27(x3) = 3

33log3xlog3

3log2

xlog2

3log

xlog 33log

xlog3 logx = log3

x = 3

Ex amp l e 6 :

If r be the ratio of the roots of the equation ax2

+ bx + c = 0, show that ac

b

r

)1r( 22

So l u t i o n :

Let a and ra be the roots of the equation ax2+ bx + c = 0

So, a + ra =a

b a(1 + r) =

a

ba = )1r(a

b

(1)

Also, a ra =a

cra2=

a

c

a

c

)1r(a

b.r

22

2

[Using (1)]


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BASIC TRIGONOMETRY

19

ac

b

r

)1r( 22

Ex amp l e 7 :

If x is real, prove that 3x2 5x + 4 is always positive

So l u t i o n :

3x2 5x + 4 = 3

3

4x

3

5x2

= 036

23

6

5x3

36

2548

6

5x3

22

since square of real number is always non-negative.

Ex amp l e 8 :

Find the area of the largest circle that can be inscribed in a square of side 14 c.m.

So l u t i o n :

BC = 14 c.m.

radius of circle = 72

14 c.m.

A

D C

B7

7

Now the area of circle = 2r = 2)7( = 49 c.m2

Ex amp l e 9 :

If f(x) = 22

x1

x

, x R, find the range of f(x)

So l u t i o n :

f(x) = 22

x1

x

= 2

2

x1

11x

= 2

x1

11

Clearly f(x) [0, 1)

Ex amp l e 10 :

If x = 2 ln cot t and y = tan t + cot t, finddx

dy

So l u t i o n :

Sincedt/dx

dt/dy

dx

dy


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IIT- MATHS

20

Now t2sin4

tsintcos

2

tcot

teccos2

dt

dx 2

Also 4t2sin

t2cos

tcostsin

tcostsinteccostsec

dt

dy222

2222

Hence t2sin t2cos4 t2sint2sin t2cos4dxdy 2 = cot 2t

Ex amp l e 11 :

xcosxsin

xcosxsin22

33

dx

So l u t i o n :

xcosxsin

xcosxsin

22

33

dx = dx

xcosxsin

xcosdx

xcosxsin

xsin

22

3

22

3

= tanx secx dx + cotx cosecx dx = secx cosec x + c

Ex amp l e 12 :

Evaluate: 2/1

02/32

1

dx)x1(

xsin

So l u t i o n :

Let x = sinq dx = cos q dq and q = sin1

x; when x = 0, sinq = 0When x = 2/1 , sinq = p/4

2

1

0

4

0

4

0

232/32

1

dseccos

dcosdx

x1

xsin

=

4/

0

4/0 dtan.1tan (Integrating by parts) = 2ln

2

1

4

Ex amp l e 13 :

Solve : log| x |

| x | = 0

So l u t i o n :

We have

log| x |

| x | = 0

|x| = 1

but | x | 1 (being in base of logarithm)

x

Ex amp l e 14 :


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BASIC TRIGONOMETRY

21

The maximum and minimum value of f (x) = 2x3 24x + 107 in the interval [1, 3].

So l u t i o n :

We have f(x) = 2x3 24x + 107

So, f (x) = 6x2 24

Now, f (x) = 0 6x2

24 = 0 x = 2Butx = 2 [1, 3]

So x = 2 is the only stationary point.

Now, (1) = 2 14 + 107 = 85, f(2) = 2(2)3 24 (2) + 107 = 75

And, (3) = 2 (3)3 24 3 + 107 = 89.

Hence, the maximum value of f (x) is 89 which attains at x = 3 and the minimum value is 75which is attained at x = 2.

Ex amp l e 15 :

Find area bounded by y = cosx, xcos , y-axis and x =2

.

So l u t i o n :

The represented area =/ 2

0

cosxdx

/ 2

0sin x 1 0

= 1 sp. unit.


26/315

IIT- MATHS

22

1. Find the solution to the inequality )4x()5x()1x()3x()3x( 52

> 0

2. Solve for x

(a) |x 4| > 7 (b) |x| > x

3. (a) Solve for x: log2x > 3

(b) Which is greater: log23 or log1/25

4. If the roots of (1 + m) x2 2 (1 + 3m) x + (1 + 8m) = 0 are equal, then find the value of m.

5. For every x R, prove that 2x2 6x + 9 is always positive.

6. Differentiate the following with respect to x.

(i) sinx + cosx

(ii) xlogx(iii) Differentiate sin2x + cos2x with respect to x

7. If y = acos t + sin t , then prove that2

22

d yy 0

dt

8. (a) Find domain of the definition of the following:

(i) x (ii) elog x(b) Draw the graph of the following:

(i) f(x) = x3 (ii) f(x) = logex

9. Find the intervals of increase and decrease of (x 3) (x + 1).

10. Evaluate : (i) 2(x 2x 1) dx . (ii)2x(2xe ) dx

(ii) dx)xcosx(sin2/

0

LEVEL - I

REVIEW YOUR CONCEPTS

SECTION - I

SUBJECTIVE


27/315

BASIC TRIGONOMETRY

23

1. Find the solution common to both the inequalities 0)4x4x(

|4x|)2x3x()1x(72

523

& 1 < |x 3| < 5

2. Solve for x

(i) |x 3| + |x + 2| = 3 (ii) (a) Solve for x, 57x2

2x

3. If a2+ 4b2= 12 ab, then prove that log (a + 2b) = )2log4bloga(log21

.

4. Find Domain of definition (i)1x3x

sin

(ii) loge

)4x()2x()3x()1x(

5. Differentiate cos (4x3

3x) with respect to x.

6. If for the function h, given by h(x) = kx2+ 7x 4, h(5) = 97, find k.

7. Find the intervals of increase and decrease of the function y = cosx,

x2

8. Integrate the following

2

0

2 dxxcos.xsin .

9. In the figure given below, find the value of angle P

36 54

A

B D

R P

C

24

10. In the figure given below, find the value of x

P

B

T

Ax4

5

LEVEL - II

BRUSH YOUR CONCEPTS


28/315

IIT- MATHS

24

1. Solve: log0.3

(x 1) < log0.09

(x 1).

2. Find the set of all solution of the equation 2|y|= 12 1y

3. Evaluate: 3log7log7log5log 5353 7537

4. Evaluate:7

1log 5

3

10

15

log 0.1

.

5. Find domain of definition (i) f(x) =2x3x

)3x(log2

2

(ii) f(x) =

|x|2

1|x|

(iii)

4

xx5logy

2

10 (iv)x1x1

2x2x

y

6. Evaluate: 2q

5cos x 3sin x dxcos x

.

7. Find the intervals of decrease and increase of (x + 2) ex.

8. In the figure given below, ABCD is square and triangles BCX and DYC are equilateral triangle.Find the value of y.

A

B

D C

x

y

9. (i) In the figure given below. Find QSR

S

R

Q

50O T P

LEVEL - III

CHECK YOUR SKILLS


29/315

BASIC TRIGONOMETRY

25

(ii) In the given figure, O is the centre of the circle. If OCA = 26, then find ODB

O

A

C B

D

10. ABC is a triangle in which BC is produced to D. CA is produced to E, DCA = 108 and EAB= 124. Then find ABC.


30/315

IIT- MATHS

26

1. If log16x + log4x + log2x = 14, then x =(a) 16 (b) 32

(c) 64 (d) none

2. If | 4 3x | 2

1then x is equal to

(a)

2

3,

6

7(b)

2

3,

6

7

(c)

23,

67 (d) none of these

3. The product of all the roots of the equation x2- |x| 6 = 0 is

(a) 9 (b ) 6

(c) 9 (d) 36

4. The value of p for which (x 1) is a factor of x3+ (p + 1)2 x2 10 is given by

(a) 4, 2 (b) 2, 4(c) 2, 4 (d) none of these

5. The domain of definition of the function f(x) =1

x x is

(a) R (b) (0,)

(c) (, 0) (d) none of these

6. The differential coefficient of f(x) = log (logx) with respect to x is

(a) xlogx

(b)x

xlog

(c) (xlogx)1 (d) xlogx

7. The function f(x) = tanx x,

(a) always increases (b) always decreases

(c) never decreases (d) some times increases and some times decreases

8. The function f(x) = x3 3x is

(a) increasing on (, 1] [1, ) and decreasing on (1, 1)

SECTION - II

OBJECTIVE

LEVEL - I


31/315

BASIC TRIGONOMETRY

27

(b) decreasing on (, 1] [1, ) and increasing on (1, 1)

(c) increasing on (0, ) and decreasing on (, 0)

(d) decreasing on (0, ) and increasing on ( , 0)

9. The minimum value of 2 22( 3) 27x is

(a) 227 (b) 2(c) 1 (d) none of these

10. The maximum value of x3 3x in the interval [0, 2] is

(a) 2 (b) 0

(c) 2 (d) 1

11. Evaluate

/ 4

0

sin

cos 3 3cos

xx

dxx x

(a)2

1(b)

6

1

(c)8

1(d)

12

1

12. If 10log 3 = 0.477, the number of digits in 340is

(a) 18 (b) 19

(c) 20 (d) 21

13. The interior and its adjacent exterior angle of a triangle are in the ratio 1 : 2. What is the sum of

the other two angles of the triangle?

(a) 112 (b) 110

(c) 120 (d) 90

14. In the figure PAQ is a tangent of the circle with centre O at a point A if OBA = 32. The value

of x and y is

A

P

Q

C

xB

y

O


32/315

IIT- MATHS

28

(a) 30, 50 (b) both 40

(c) both 58 (d) 30, 60

15. The difference between the interior and exterior angles of a regular polygon is 132. The numberof sides of the polygon is

(a) 12 (b) 8

(c) 15 (d) 20


33/315

BASIC TRIGONOMETRY

29

1. If A = log2log

2log

4256 + 22log 2, then A equals to

(a) 2 (b) 3(c) 5 (d) 7

2. If logkA . log

5k = 3, then A =

(a) 53 (b) k3

(c) 12 (d) 243

3. If the product of the roots of the equation 22 log5 4 0x x is 8, then l is

(a) 22 (b) 2 2(c) 3 (d) none of these

4. The number of real roots of the equation (x 1)2

+ (x 2)2

+ (x 3)2

= 0(a) 3 (b) 2(c) 1 (d) 0

5. If a and b are roots of the equation x2+ x + 1 = 0 then a2+ b2=(a) 1 (b) 2(c) 1 (d) 3

6. The domain of definition of the function f (x) =2 9

log

x

x

(a) R (b) (1, )(c) (0, 1) (1, (d) [1, )

7. If x = a cos3, y = a sin3,2

dy1+ =

dx

(a) tan2 (b) sec2(c) sec (d) |sec|

8. The function y = x3

3x2

+ 6x 17(a) increases everywhere(b) decreases everywhere(c) increases for positive x and decreases for negative x(d) increases for negative x and decreases for positive x

9. The largest value of 2x3 3x2 12x + 5 for 2 x 4 occurs at x =(a) 2 (b) 1(c) 2 (d) 4

10. The greatest value of f(x) = cos (xex+ 7x2 3x), x[ 1, ) is(a) 1 (b) 1(c) 0 (d) none of these

LEVEL - II


34/315

IIT- MATHS

30

11. 21

x

x

edx

e

(a) cot1ex+c (b) tan1ex+ c(c) tanex+ c (d) sin1ex+ c

12. Eva luate/ 4

2

0

tanx

x dx

(a)2

3 (b)

62

(c) 14

(d) 5

2

13. If the area of parallelogram ABCD is 32 sq. cm. Area of DAMN is equal to

D M

N

A B

C

(a) 8 cm2 (b) 2 cm2

(c) 4 cm2

(d) none of these

14. In the given figure if AX = 5 cm, XD = 7 cm, CX = 10 cm find BX

A B

C O

DP

T

x

(a) 3 cm (b) 3.5 cm(c) 4 cm (d) 4.5 cm

15. If A, B, C are three consecutive points on the arc of a semicircle such that the angles subtended by

the chords AB and AC at the centre O is 90 and 100 respectively. Then the value of angle BAC

is equal to

(a) 5 (b) 10(c) 20 (d) 30


35/315

BASIC TRIGONOMETRY

31

1. Find the solution to the inequality2 5

( 3) ( 3) ( 1) 0( 5)( 4)

x x xx x

Solution:

2 5(x 3) (x 3) (x 1)0

(x 5) (x 4)

By wavy curve method,

+ ++

5 3 1 3 4

x ( , 5) ( 3 1) (4 )

2. Solve for x

(a) |x 4| > 7 (b) |x| > x

Solution:

(a) |x 4| > 7

x 4 > 7 or x 4 < 7

x > 11 or x < 3

3 11

x ( , 3) (11, )

(b) |x| > x

Case I : x > 0,

x > x which is not possible.

Case II: x < 0

x < x 2x < 0 x < 0

x( , 0) or R

3. (a) Solve for x: log2x > 3

(b) Which is greater: log23 or log

1/25

Solution:

(a) log2x > 3 x > 23 [ base is greater than 1]

SUBJECTIVE SOLUTIONS

CHECK YOUR SKILLS

LEVEL - I (CBSE Level)


36/315

IIT- MATHS

32

x > 8 x(8, )

(b) log23 or log

1/25

log1/25 = log25 < 0. [ log1/ax = logax]

log23 is greater than log1/25.

4. If the roots of (1 + m) x2 2 (1 + 3m) x + (1 + 8m) = 0 are equal then find the value of m.

Solution:

Given equation is

(1 + m) x2 2(1 + 3m) x + (1 + 8m) = 0

roots are equal , then D = 0

4 (1 + 3m)2 4 (1 + m) (1 + 8m) = 0

1 + 9m2 + 6m 1 9m 8m2= 0

m2 3m = 0

m (m 3) = 0

m = 0, 3.

5. For every x R, prove that 2x2 6x + 9 is always positive.

Solution:

Let f(x) = 2x2 6x + 9 = 2(x2 3x + 9/2)

= 2( x 3/2)2 + 9/4 > 0 [ square of real number is always nonnegative]

Hence f(x) is always positive.

6. Differentiate with respect to x(i) sinx + cosx

(ii) xlogx

Solution:

(i) Let y = sinx + cosx

dy

cos x sin xdx

d

dx

(sinx + cosx) = cosx sinx

(ii) Let y = x logx

dy d dx (log x) log x (x)

dx dx dx =

1x log x1

x

d(xlogx)

dx= 1 + logx

8. (a) Find domain

(i) x (ii) log ex

(b) Draw graph

(i) f(x) = x3 (ii) f(x) = logex


37/315

BASIC TRIGONOMETRY

33

Solution:

(a) (i) Let f(x) = x

f(x) is real for all x 0

Df = [0, ]

(ii) Let f(x) = elog x

f(x) is real for all x 1

Df= [1, )

(b) Let f(x) = x3

It is odd function so graph will be symmetric about the origin

x

y

(ii) DfR

(iii) R fR

(iv) f (x) = 3x2

f (x) = 0 x = 0

(v) f(x) = 6x

(vi)ve, x 0

f (x) 6xve, x 0

(ii) Let f(x) = logx

(i) f (x) =1

x, f(x) = 2

1

x

(ii) Df R {0}

y

x

y = logx

(1, 0)

(iii) Rf R

9. Differentiate with respect to x

(i) sin2x + cos2x

(ii) x = a (cost + log tan2t

), y = a sint.

Solution:

(i) Let y = sin2x + cos2x

2dy d d(sin x)

dx dx dx (sinx) +

2d d(cos x) (cos x)dx dx

= 2sinx cosx + 2cosx (sinx) = 2sinx cosx 2sinx cosx = 0

2 2d (sin x cos x) 0

dx

d

(1) 0

dx

[ d

dx

(constant) = 0]

(ii) x = a (cost + log tant

2), y = asint


38/315

IIT- MATHS

34

Diff. w.r.to t

dx

adt

21 t 1sin t sec

t 2 2tan2

,dy

acostdt

= a[ sint + cosect]

dy cos tdx cos ect sin t

2

cos t.sin t1 sin t

= 2cost.sint

cos t= tant

10. Find the intervals of increase and decrease of (x 3) (x + 1).

Solution:

Let f(x) = (x 3) (x + 1) = x2 2x 3

f (x) 2x 2 = 2(x 1)

for increasing, f (x) 0

2(x 1) 0 x 1

x [1, )

for decreasing, f (x) 0

x ( ,1]

11. Find the point of local maxima and minima of the following(i) x2+ 3x (ii) logex + x

So l u t i o n :

Let f(x) = x2+ 3x = x(x + 3)

f (x) = 2x + 3, for local maxima or local minima we must have, f (x) = 0

3

x2

f (x) = 2 > 0 3/2

f(x) is local minima at 3x2

(ii) f(x) = logx + x

1f (x) 1

x

for local maxima or local minima, we must have f (x) 0

1

1 0x

x = 1

21f (x) 0x

f(x) is local maxima at x = 1.


39/315

BASIC TRIGONOMETRY

35


(i) x2+ 2x + 1 (ii) 22 xxe

(iii)/2

0

(sin cos )x

x x dx

(iv)2

1

x dx

Solution:

(i) Let I = 2(x 2x 1) dx 2x dx 2 x dx 1.dx

3 2x x2 x C

3 2

=3

2x x x C,3

where C is integral constant.

(ii) Let I = 2x2x e dxPut x2= t2xdx = dt

tI e dt = et+ C =

2xe C, where C is integral constant.

(iii)

/ 2

0

I (sin x cos x) dx

=

/ 2 / 2

0 0sin xdx cos x dx

= / 2 / 20 0[sinx]cosx

= cos cos 0 sin sin 02 2

= [0 1] + [1 0] = 2

(iv) Let2

1

I | x |dx

Let f(x) = |x| = x, x 0x, x 0

0 2

1 0

I f (x)dx f (x) dx

0 2

1 0

xdx xdx

0 22 2

1 0

x x

2 2

1 1[0 1] (4 0)

2 2

5

2

13. In the figure given below. Find QSR


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IIT- MATHS

36

S

R

Q

50O T P

Solution:

QSR + PRQ + QPR + QOR = 360

90 + 90 + 50 + QOR = 360 QOR = 130

QSR =1

2 QOR = 65

14. In the given figure, O is the centre of the circle. If OCA = 26, then find ODB

O

A

C B

D

Solution:

OCA = 26 = DBA

OB = OD

OBD = ODB = 26 O

A

C B

26 26

D

15. ABC is a triangle in which BC is produced to D. CA is produced to E, DCA = 108 and EAB= 124. Then find ABC.

Solution:

DCA = 108 ACB = 72

EAB = 124

E

B C

A

56D

124

108 BAC = 56 ABC = 180 (72 + 56) = 62


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BASIC TRIGONOMETRY

37

1. Find the solution common to both the inequalities

0)4x4x(

|4x|)2x3x()1x(72

523

& 1 < |x 3| < 5

Solution:

Case I : x < 43 5 5

14

(x 1) (x 2) (x 1) (x 4)0

(x 2)

3 5 5

14

(x 1) (x 2) (x 1) (x 4)0

(x 2)

4

+ + +

2 1 1

x ( , 4) Case II : x > 4

4

+ + +

2 1 1

Similarly, x ( 4 2) ( 1,1) x ( , 4) ( 4, 2) ( 1,1) ... (i)

And 1 < | x 3| < 5Case I : x < 3

1 < x + 3 < 5 2 < x < 2 2 < x < 2 x (2, 2)Case II: x > 31 < x 3 < 5 4 < x < 8

x (4,8)

x ( 2, 2) (4,8) ... (ii)from (i) & (ii)

4 2 1 2 4 8

x ( 1,1)

2. Solve for x

(i) |x 3| + |x + 2| = 3 (ii) 57x2

2x

Solution:

(i) |x 3| + |x + 2| = 3Case I:

BRUSH UP YOUR CONCEPTS

LEVEL - I


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38

x < 2x + 3 x 2 = 32x = 2 x = 1, which is impossibleCase II:

2 x < 3x + 3 + x + 2 = 3, which is not possible

Case III:x 3x 3 + x + 2 = 32x = 4x = 2, which is impossible

x

(ii)x 2

52x 7

x 2 5(2x 7) x 2 10x 350 0

2x 7 2x 7

9x 37

02x 7

(9x + 37) (2x + 7) > 0

37/9

+ +

7/2

37 7

x ( , ) ( , )9 2

3. If a2+ 4b2= 12 ab, then prove that log (a + 2b) = 1 (log log 4log 2)2

a b

Solution:

a2+ 4b2= 12aba2+ 2a2b + 4b2 = 16 ab(a + 2b)2= 16abtaking log on both the sides,2 log (a + 2b) = log16 + log a + logb

log (a + 2b) =1

2

(loga + logb + 4log2)

4. Find Domain (i)3

sin1

x

x

(ii) loge

( 1)( 3)( 2)( 4)x x

x x

Solution:

(i) Let f(x) =x 3

sinx 1

f(x) is defined for

x 3

0x 1

(x 1) (x + 3) 0


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BASIC TRIGONOMETRY

39

+3

x

1

x ( ,3] (1, )

(ii) Let f(x) =e

(x 1) (x 3)log

(x 2)(x 4)

f(x) is defined for

(x 1) (x 3)

0(x 2)(x 4)

+ + +

31

42

x ( , 3) ( 2,1) (4, )

5. Differentiate with respect to x

(i) 2sin

(1 )

x x

x

+ log 2(1 )x

(ii) (sinx)x

(iii) cos (4x33x)Solution:

(i) Let2

2

xsinxy log 1 x

1 x

2

2 2 2 2

dy (1 x ) (sin x x cos x) (2x)x sin x 1 (x 2x)

dx (1 x ) 1 x 2 1 x

2 2 2

2 2(1 x ) x cos x (1 x )sin x 2x sin x 1(1 x )2 1 x

2 2

2 2 2

(1 x ) x cos x (1 x )sin x 1

(1 x ) 1 x

(ii) Let y = (sinx)x. Taking log on both the sides, we getlogy = x log sinx

1 dy x cos x. log sin x

y dx sin x = x cotx + log sinx

dydx

= (sinx)x [x cotx + log sinx]

x xd {(sin x) } (sin x ) [x cot x log sin x]dx

(iii) Let y = cos(4x3 3x)

dy

dx= sin (4x3 3x) [12x2 3]

d

dx (cos 4x3 3x) = 3 (1 4x2) sin (4x3 3x)

6. If for the1 function h, given by h(x) = kx2+ 7x 4, h(5) = 97, find k.Solution:


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40

h(x) = kx2+ 7x 4

h (x) 2kx + 7

h (5) = 10k + 7

97 = 10k + 7 k = 9 [ h (5) = 97]

7. (i) Find the intervals of increase and decrease of the function y = cosx,

2

x

(ii) Find the point of local maxima & minima of the function, f (x) = x1

x .

Solution:

(i) Let f(x) = cosx, x2

f (x) = sinxf(x) increases, if f (x) 0 sinx 0 sinx 0

O/2x ,02

f(x) decreases, if f (x) 0

sinx 0 sinx 0

x [0, ]

O /2

(ii) 1

f (x) xx

21

f (x) 1x

3

2f (x)

x

For local maximum or local minimum we must have f (x) 0

1 21

0x

x = 1

at x = 1, f (x) 1 0

Hence local minima at x = 1

at x = 1, f (x) 1 0 Hence local maxima at x = 1.

8. Integrate the following (i)

2

0

2 dxxcos.xsin (ii) 8

13/2

x

dxx

e3

.

Solution:

(i) Let I =

/ 22

0

sin x.cos x dx

put cosx = t =0 2

1

t dt sinx dx = dt when x = 0, t = 1, when x = , t 02


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BASIC TRIGONOMETRY

41

=

11 32

00

1tt dt33

(ii) Let I =

38 x

2 / 31

edx

x put x1/3 = t 2 /3

1 1. dx dt

3 x when x = 8, t = 2 ,

x = 1, t = 1 =2

t t 2

1

3 e dt 3(e ) , = 3(e2 e) = 3e (e 1)

9. In the figure given below, find the value of angle P

36 54

A

B D

R P

C

24

So l u t i o n :

ACD = ABC + BAC = 36 + 24 = 60 P = RCD + RDC = 60 + 52 = 104.

10. In the figure given below, find the value of x

P

B

T

Ax4

5

Solution:

PT2 = PA.PB

PA =5 5 25

4 4

AB = PA PB

x =25 9

44 4

units

Hence x =9

4units.


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42

1. Solve, log0.3

(x 1) < log0.09

(x 1)Solution:

log

0.3 (x 1) < log

0.09(x 1)

log0.3 (x 1) 1 x > 2

x (2, )

2. Find the set of all solution of the equation 12 2 1y y Solution:

2|y| = 2y1+ 1Case I: y < 0 2y= 2y1+ 1

1 =2y

y2 22

(2y)2 + 2.2y 2 = 0

a2 + 2a 2 = 0 , where a = 2y

2 4 8 2 2 3a

2 2

a = 1 3 , 1 + 3

a = 1 3 , which is not possible

a = 3 1

2y= 3 1

log22y = log

2( 3 1 )

y = log2( 3 1 )

Case II: y < 02y= 2y1 + 1

2.2y = 2y + 2 2y= 2 y = 1

set of solution 1, log2 ( 3 1)

3. Evaluate: 3 5 3 55 7 7 37log 3log 5log 7 log

Solution:

Let y =5 7 7 3

3 5 3 5log log log log7 3 5 7

= 3 5 3 5log 5 log 7 log 5 log 77 3 7 3 [ b blog c log aa c ]

LEVEL - III

CHECK YOUR SKILLS


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BASIC TRIGONOMETRY

43

= 0Hence the result.

4. Evaluate: 5 0.17 101 1

log log3 2

Solution:

Let 7 101 1

3 log 5 log 0.1y 5

5

1/ 3log 7

10

15

log 10

= (7 + 1)1/3 = (8)1/3 = 2

Hence the result.

5. Find domain (i) f(x) = (ii) f(x)1

2

x

x

Solution:

(i) f(x) = 22log (x 3)

x 3x 2

Let g(x) = log2(x + 3) and h(x) = x2+ 3x + 2

g(x) is defined forx + 3 > 0 x > 3 Dg= (3, )And h(x) is defined for x2+ 3x + 2 0 (x + 2) (x + 1) 0 Dh = R {2, 1} Df = Dg Dh = (3, ) R {2, 1} = (3, ) {2, 1}

(ii) f(x) =| x | 1

2 | x |

f(x) is defined for

2

| x | 1 (| x | 1) (2 | x |)0 0

2 | x | (2 | x |)

(2 |x|) (|x| 1) 0 2 |x| > 0 or |x| 1 0 |x| < 2 or |x| 1 2 < x < 2 or x 1 or x 1

Df = (2, 1] [1, 2)

6. Find the domain

(i)

4xx5

logy2

10 (ii) 3/1)x(sinecxcosy

(iii)x1x1

2x2x

y

Solution:

(i)2

105x xy log f (x)

4


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44

f(x) is defined for

log10

25x x 0

4

2

25x x 1 5x x 4 04

x2

5x + 4 0 (x 1) (x 4) 0either x 1 or x 4

Df= [1, 4]

(ii) Let f(x) = cosecx + (sinx)1/3

Let g(x) = cosecx and h(x) = (sinx)1/3

g(x) is defined forcosecx > 0D

g= (2n , (2n + 1) ).

And h(x) is defined for R

Df = Dg Dh = (2n , (2n + 1) )

(iii) Letx 2 1 x

f(x)x 2 1 x

Let g(x) =x 2

x 2

and h(x) =1 x

1 x

g(x) is defined for

x 20

x 2

either x 2 0

or x + 2 < 0x 2 or x + 2 < 0

Dg = ( , 2) [2, )

h(x) is defined for

1 x x 10 0

1 x x 1

either 1 x 0 or 1 + x > 0

x 1 or x > 1

Dn = (1, 1] Df = Dg Dh =


(a) 5 cos x 3 sinx +xcos

q2 (b)

4

02

dxxcos

xsin32

Solution:

(a) Let I = 2a

(5cos x 3sin x )dxcos x 25 cos x dx 3 sin xdx 9 sec x dx


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BASIC TRIGONOMETRY

45

= 5 sinx + 3cosx + 9 tanx + C,where C is integral constant.

(b) Let/ 4

20

2 3sin xI dx

cos x

=

/ 42

02 sec xdx 3 tan xsec x dx

= / 4 / 40 02[tan x] 3[sec x]

= 2[1 0] + 3 [ 2 1] = 3 2 1

8. Differentiate 2 1xe w.r.t. x.

Solution:

Let y = 2x 1e

2x 1dy e

dx

21

2 x 1 (2x 0)

2x 1

2

x.e

x 1

9. (a) Find the intervals of decrease and increase of (x + 2) ex.(b) Find the greatest and least values of the following functions on the intervals

y = 3x4+ 6x2 1 (2 x 2)

y = 1x3x23x 2

3

(1 x 5)

Solution:

(a) Let f(x) = (x + 2) ex

f (x) = ex

(x + 2) + ex

= ex

(x + 1)for decreasing f (x) 0

ex (x + 1) 0 ex (x + 1) 0 ( ex 0] x + 1 > 0

x > 1 x [ 1, )

for increasing , f (x) 0

ex(x + 1) 0ex (x + 1) 0

x + 1

0 [ ex

0]

x 1

x ( , 1] (b) (i) y = 3x4 + 6x2 1, 2 x 2

3dy 12 x 12x

dx

22 2

2

d y36x 12 12(x 1) 0

dx

dy

0dx 12x (x2 1) = 0 x = 0, 1 2least value y = 2, at x = 0


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46

greatest value y = 25 at x = 2

(ii)3x

y3

2x2+ 3x + 1 , 1 x 5

2dy x 4x 3dx

2

2

d y2x 4 2(x 2)

dx

dy 0dx

x2 4x + 3 = 0

(x 1) (x 3) = 0 x = 1, 3, 5

at x = 1,2

2

d y2 0

dx

maximum at x = 1

maximum value of y =7

3

at x = 3,2

2d y 2 0dx

minimum at x = 5

minimum value of23

y3

10. In the figure given below, ABCD is square and triangles BCX and DYC are equilateraltriangle. Find the value of y.

AB

D C

x

y

So l u t i o n :

BCX and DYC are equilateral

AB = BC = CD = DA = CX = DY = CY = BX CBX = 60 ABX = 150 BAX = 15

DAY = 75 = DYAA [ AD DY]


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BASIC TRIGONOMETRY

47

Illustration 1:

If A = {a, b, c} and B = {b, c, d} then evaluate AB, AB, A B and B ASolution:

AB = {x : x A or x B} = {a, b, c, d}AB = {x : x A and x B} = {b, c}A B = {x : x A and x B} = {a}

B A = {x : x B and x A} = {d}

Illustration 2:

Find the logarithms of 0.0625 to the base 2.Solution:

Suppose 2x= 0.0625 = 421

16

1

or 2x= 24 or x = 4

Illustration 3:

Solve the equation a2xb3x= c5, where a, b, c R+

Solution:

Equation is a2x.b3x= c5

Taking log on both sides, we have2x log a + 3x logb = 5 logc or x (2loga + 3 logb) = 5logc

x = blog3alog2

clog5

Illustration 4:

Solve for x, log1/2

(x 2) > 2.

Solution:

log1/2

(x 2) > log1/2

4

1 0 < x 2 <

4

12 < x <

4

9

So x 49,2

Illustration 5:

Solve |2x 1| < 3.Solution:

|2x 1| < 3 3 < 2x 1 < 3 2 < 2x < 4 1 < x < 2.

Illustration 6:


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48

If f(x) =

7x1x5x2x3x

then find x such that

(i) f(x) > 0. (ii) f(x) < 0.Solution:

Given f(x) =

7x1x5x2x3x

Illustration 7:

If xx5x

= 24,, then find the value of x.

Solution:

Here, xx5x

= 24 x2 5x = 24

x2 8x + 3x 24 = 0 (x 8) (x + 3) = 0 x = 3, 8

Illustration 8:

Find the value of x and y when xy32

= 4 and 27

12

yx

Solution:

Since, 4y3x24xy32

(1)

and 27y2x

27

12yx (2)

Solving (1) and (2), we get x =2

5, y = 3.

Illustration 9:

Find the derivative of y =x

1)x(x3 3/1

Solution:Here we have y = u + v + w,

where u = 3/1xv,x3 and w =x

1

Hence we can use theorem 2

23/211

13

11

2

1

x

1

)x(

131

x3

21

x.1x31

x21

3dxdy

Illustration 10:

Find the derivative of y = (a + x) exw.r.t. x.Solution:


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BASIC TRIGONOMETRY

49

Using theorem 3

)xa(dx

dee

dx

d)xa(

dx

dy xx

= )1xa(e1.ee)xa( xxx

Illustration 11:

Find the derivative of y =xa

xa

w.r.t. x

Solution:

Illustration 12:

Differentiate the following w.r.t x.(i) y = xx

(ii) y =2x)x(sin

Solution:(i) y = xx

)xln1(x1.xlnx1.x.xdx

dy xx1x

(ii)2x)x(siny

dx

dxxsinln)x(sin)x(sin

dx

d)x(sinx

dx

dy 2x1x2 22

= x2 .x2.xsinln)x(sinxcos)x(sin22 x1x

Illustration 13:

Find the derivative of the following functions w.r.t. x(i) y = sinx2

(ii) y = (lnx)3

(iii) y = sin (lnx)3

(iv) y = cos1(lnx)Solution:

(i) y = sin (x)2. Let u = x2y = sin u

x2.ucos)x(dxd)u(sinduddxdy2 2x cosx2

(ii) y = (ln x)3. Let u = ln x y = u3dx

dy= 22

3

)x(lnx

3

x

1u3

dx

du

du

)u(d

(iii) y = sin (lnx)3 Let u = ln x, v = u3 y = sin v

)x(lndx

d

du

)u(d

dv

)v(sind

dx

du

du

dv

dv

dy

dx

dy 3 = cos v. ])x[(lncos

x

)x(ln3

x

1u3 3

22

(iv) Let u = ln x = cos1u

222 )x(ln1x

1

x

1

u1

1

dx

du

u1

1

dx

dy


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IIT- MATHS

50

Illustration 14:

The function y of x is given by, x = a cos t, y = a sin t. Find the derivative of y w.r.t. x.Solution:

tcottsin

tcos

)tcosa(

)tsina(

dx

dy

If we want to compute dx

dy

at a particular t, say t = 4

, then 14cotdx

dy

4/x

Illustration 15:

Find 22

dx

yd, where y = sin2x

Solution:

dx

dy= 2sinxcosx = sin2x

Illustration 16:

Find the interval of increase and decrease of the function y = x4.Solution:

y = x4y= 4x3

For x > 0, y> 0 the function increases in (0, ).For x < 0, y< 0 the function decreases in (, 0)

Illustration 17:

Separate the intervals in which f(x) = 2x3 15x2+ 36x + 1 is increasing or decreasing

Solution:

We have f (x) = 6x2 30x + 36 = 6 (x 2) (x 3)Thus for x < 2, f (x) > 0over 2 < x < 3, f(x) < 0 and for x > 3, f(x) > 0Hence the given function is increasing in (, 2) and (3, ), and decreasing in (2, 3)Is tanx always increasing Rx ?

Illustration 18:

Test y = 1x4for maximum and minimum

Solution:Here y= 4x3= 0 for x = 0

x = 0 is the critical point.

Now y= 12 x2= 0 at x = 0

It is thus impossible to determine the character of the critical point by means of the sign of thesecond derivative. Thus we investigate the character of the given function in an interval containing pointx = 0.

For x < 0, y> 0 the function is increasing for x < 0

For x 0, y< 0 the function is decreasing for x > 0.

Consequently, at x = 0, the function has a maximum i.e., yx = 0= 1


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BASIC TRIGONOMETRY

51

Illustration 19:

Determine the maximum and minimum of the function y = x3 3x + 3 on the interval [3,2

3].

Solution:

For the given function, y= 3x2 3

For the critical points, 3x2 3 = 0 x = 1Then y= 6x > 0 at x = 1, y < 0 at x = 1

Hence there is maximum at x = 1 at which y = 1 + 3 + 3 = 5

Also there is minimum at x = 1 at which y = 13 + 3 = 1

Now at x = 3, y = 27 + 9 + 3 = 15

and at x = 3/2, y = 15/8

Hence the minimum value of the given function is 15 at x = 3 and the maximum value is 5 at

x = 1. It should be noted that the values are actually the largest and smallest values of the functionin the given interval.

Illustration 20:

Evaluate:

(i) (a0+ a

1x + a

2x2) dx (ii)

dxex

2xcos x

(iii) dxx1x

2

2

(iv) dx1x

x2

4

Solution:

(i) (a0+ a1x + a2x2

) dx = a0dx + a1x dx + a2x2

dx= a

0x + a

1 c

3x

a2

x 32

2

(ii)

dxedx

x

12dxxcosdxe

x

2xcos xx = sin x + 2 log |x| ex+ c

(iii)

222

2

2

2

x1

dxdxdx

x1

11dx

x1

11xdx

x1

x= x tan1x + c

(iv)

dx

x1

11xdx

x1

11xdx

1x

x2

22

4

2

4

= cxtanx

3

x

x1

dxdxdxx 1

3

22

Illustration 21:

Integrate the following w.r.t. x.(i) sin2x cosx

(ii) 83

x1

x

Solution:

(i) Let sin x = t cos x dx = dt


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52

c3xsin

c3t

dttdxxcosxsin33

22

(ii) Let x4= t = 4 x3dx = dt x3dx = dt

dt)t1(41

dxx1

x28

3

= cxtan4

1cttan

4

1

t1

dt

4

1 4112

Illustration 22:

Evaluate

dxxa

122

Solution:

Let x = a sin dx = a cosda2 x2= a2cos2

cax

sind.1cosa

dcosa

xa

dx 122

Note:

a cbxsinb1)cbx(adx 122

Illustration 23:

Evaluate

(i) dxxex (ii) dxxsin 1

Solution:

(i) let f(x) = x, (x) = ex cexedxe.1e.xdxxe xxxxx

(ii) Let f(x) = sin1x. (x) = 1

dxxx1

1x.xsindx1.xsindxxsin

2

111

= x sin1x cx1xsinxdx)x1(x 212/12

Note: dx)x(fedx)x(fedx))x(f)x(f(e xxx

= f(x) ex ce)x(fdx)x(fedxe)x(f xxx

Illustration 24:

Evaluate :

2/

02

dxxsin1

xcos

Solution:

Let sin x = t cos x dx = dt

When x = 0, t = 0, x =

/2, t = 1

2/

0

1

0

10

122

)t(tant1

dt

xsin1

xdxcos= tan11 tan10 =

4

0 = /4


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2 ARITHMETIC PROGESSION


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ARTHMETIC PROGESSION

55

SQUENCE AND SERIES

A succession of numbers a1, a

2, a

3..., a

n, ... formed, according to some definite rule, is called a

sequence.

ARITHMETIC PROGESSION (A.P.)

A sequence of numbers {an} is called an arithmetic progression, if there is a number d, such that d= a

n-a

n1for all n. d is called the common difference (C.D.) of the A.P.

(i) Useful Formulae

If a = first term, d = common difference and n is the number of terms, then

(a) nth term is denoted by tnand is given by

tn= a + (n 1) d.

(b) Sum of first n terms is denoted by Snand is given by

n nS [2a (n 1)d]2

or ( )2

n

nS a l , where l= last term in the series i.e., l= t

n = a + (n 1) d.

(c) Arithmetic mean A of any two numbers a and ba b

A2

.

(d) Sum of first n natural numebrs ( n )n(n 1)

n2

, where n N .

(e) Sum of squares of first n natural numbers ( 2n )2 ( 1) (2 1)

6

n n nn

(f) Sum of cubes of first n natural numbers 3( n )

23 n(n 1)n

2

(g)Middle term: If the number of terms is n, and

n is odd, thenth

n 1

2

term is the middle terms

n is even, then 2

thn

andth

n1

2

terms are middle terms.

(h) If terms are given in A.P., and their sum is known, then the terms must be picked

up in following way

For three terms in A.P., we choose them as (a d), a, (a + d)

For four terms in A.P. , we choose them as (a 3d), (a d), (a + d), (a + 3d)


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56

For five terms in A.P., we choose them as (a 2d), (a d), a, (a + d), (a + 2d) etc.

(ii) Useful Properties

(a) If tn= an + b, then the series so formed is an A.P.

(b) If Sn= an2+ bn + c, then series so formed is an A.P.

(c) If every term of an A.P. is increased or decreased by some quantity, the resulting terms` will also be in A.P.

(d) If every term of an A.P. is multiplied or divided by some non-zero quantity, the result

ing terms will also be in A.P.

(e) In an A.P. the sum of terms equidistant from the beginning and end is constant and

equal to sum of first and last terms.

(f) Sum and difference of corresponding terms of two A.P.s will form an A.P.

(g) If terms a1, a2, ..., an, an+1, ..., a2n+1are in A.P., then sum of these terms will be equal to(2n + 1)a

n+1.

(h) If terms a1, a

2, ..., a

2n1, a

2n are in A.P. The sum of these terms will be equal to

(2n)n n 1a a

2

.

GEOMETRIC PROGRESSION (G.P.)

A sequence of the numbers {an}, in which 1 0a , is called a geometric progression, if there is a

number 0r such thatn

n 1

a ra

for all n.r is called the common ratio (C.R.) of the G.P..

(i) Useful Formulae

If a = first term, r = common ratio and n is the number of terms, then

(a) nth term, denoted by tn, is given by t

n= arn1

(b) Sum of first n terms denoted by Snis given by

n

n

a(1 r )S

1 r

orna(r 1)

r 1

ora r

1 r

l

,

where lis the last term (the nth term) in the series, r 1 In case r = 1, Sn= na.

(c) Sum of infinite terms (S )a

S (for | r | 1)1 r

Note: When |r| 1, the series is divergent and so its sum is not possible.

(d) Geometric mean (G.M.)

G ab is the geometric mean of two positive numbers a and b.

(e) If terms are given in G.P. and their product is known, then the terms must be picked up

in the following way.


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For three terms in G.P., we choose them asa

, a , a r r

For four terms in G.P., we choose them as3

3

a a, , ar, ar

r r

For five terms in G.P., we choose them as2

2

a a, , a, ar, ar

r retc.

(ii) Useful Properties

(a) The product of the terms equidistant from the beginning and end is constant, and it isequal to the product of the first and the last term.

(b) If every term of a G.P. is multiplied or divided by the some non-zero quantity, the

resulting progression is a G.P.

(c) If a1, a

2, a

3... and b

1, b

2, b

3, ... be two G.P.s of common ratio r

1and r

2respectively, then

a1b

1, a

2b

2... and 31 2

1 2 3

aa a, ,

b b b... will also form a G.P. Common ratio will be r

1r

2and 1

2

r

r

respectively.

(d) If a1, a

2, a

3, ... be a G.P. of positive terms, then loga

1, loga

2, loga

3, ... will be an A.P. and

conversely.

HARMONIC PROGRESSION (H.P.)

(a) A sequence is said to be in harmonic progression, if and only if the reciprocal of its

terms form an arithmetic progression.

For example

1 1 1, ,2 4 6

... form an H.P., because 2, 4, 6, ... are in A.P..

(b) If a, b are first two terms of an H.P. then

n

1t

1 1 1(n 1)

a b a

(c) Harmonic mean H of any two numbers a and b

2 2abH1 1 a ba b

, where a, b are two non-zero numbers.

(d) If terms are given in H.P. then the terms could be picked up in the following way

(i) For three terms in H.P, we choose them as

1 1 1, ,

a d a a d

(ii) For four terms in H.P, we choose them as


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1 1 1 1, , ,

a 3d a d a d a 3d

(iii) For five terms in H.P, we choose them as1 1 1 1 1

, , , ,a 2d a d a a d a 2d

Insertion of Means Between two numbers

Let a and b be two given numbers.

(i) Arithmetic Means

If a, A1, A

2, ... A

n, b are in A.P., then A

1, A

2, ... A

nare called n A.M.s between a and b. If d is

the common difference, then

b = a + (n + 2 1) d d =b a

n 1

Ai= a + id = a + i

b a a(n 1 i) ib,

n 1 n 1

i = 1, 2, 3, ..., n

Note: The sum of n-A. M s, i.e., AA1+ A2+ ... + An=n

(a b)2

(ii) Geometric means

If a, G1, G

2... G

n, b are in G.P., then G

1, G

2... G

nare called n G.M.s between a and b. If r is the

common ratio, then

b = a.rn+1 r =1

(n 1)b

a

Gi= ari=

in 1 i i

n 1n 1 n 1

ba a .b ,

a

i = 1, 2, ..., n

Note: The product of n-G. M s i.e., G1

G2

... Gn

=

n

ab

(iii)Harmonic Means

If a, H1, H

2... H

n, b are in H.P., then H

1, H

2... H

nare called n H.M.s between a and b. If d is the

common difference of the corresponding A.P., then

1 1 a b(n 2 1) d d

b a ab(n 1)

i

1 1 1 a bid i

H a a ab(n 1)

ab(n 1), i 1, 2, 3, ..., n

b(n i 1) ia


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MEANS OF NUMBERS

Let a1, a

2, ... a

nbe n given numbers

(i) Arithmetic Means (A.M.) = 1 2 na a ... a

n

(ii) Geometric Means (G.M.) = (a1a2... an)1/n

(iii) Harmonic Mean (H.M.) =1 2 n

n

1/ a 1/ a ... 1/ a

If weights of a1, a

2, ... , a

nare w

1, w

2, ..., w

nrespectively, then their weighted arithmetic mean,

weighted geometric mean and weighted harmonic mean are respectively defined by

1 2 n 1 2 n

1w w w w w ...w1 1 2 2 n n

1 2 n1 2 n

a w a w ... a w(a .a ...a )

w w ... w

1 2 n

1 2 n

1 2 n

w w ... wand

w w w...

a a a

.

RELATION BETWEEN A, G AND H

If A, G and H are A.M., G.M. and H.M. of positive numbers a1, a

2... a

n( 1 2 na a ... a ) then

1 na H G A a ...(1)

Note:

(i) The equality at any place in (1) holds if and only if the numbers a1, a

2, ..., a

nare all equal

(ii) (1) is true for weighted means also

(iii) G2= AH, if n = 2.

ARITHMETIC MEAN OF mthPOWER

Let a1, a

2 . . . , a

n be n positive real numbers and let m be a real number, then

mm m m1 2 n 1 2 na a ... a a a ... a , if m R [0,1].

n n

However if m (0, 1), thenmm m m

1 2 n 1 2 na a ... a a a ... a

n n

Obviously ifmm m m

1 2 n 1 2 na a ... a a a ... am {0,1}, thenn n

ARITHMETIC-GEOMETRIC SERIES

A series whose each term is formed, by multiplying corresponding terms of an A.P. and a G.P., iscalled an Arithmetic-geometric series.


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e.g. 1 + 2x + 4x2 + 6x3+ ..... ; a + (a + d) r + (a + 2d)r2+ .....

(i) Summation of n terms of an Arithmetic-Geometric Series

Let Sn= a + (a + d) r + (a + 2d)r2+ ... + [a + (n 1)d] rn1, d 0 , r 1

Multiply by r and rewrite the series in the following way

rSn = ar + (a + d)r2+ (a + 2d)r3+ ... + [a + (n 2)d]rn1+ [a + (n 1)d ]rnon subtraction,S

n(1 r) = a + d(r + r2+ ... + rn1) [a + (n 1)d]rn

or,n 1

nn

dr(1 r )S (1 r) a [a (n 1)d].r

1 r

or,n 1

nn 2

a dr(1 r ) [a (n 1)d]S .r

1 r (1 r) 1 r

(ii) Summation of Infinite Series

If |r| < 1, then (n 1)rn, rn1 0, as n .

Thus S = S = 2a dr

1 r (1 r)

SUM OF MISCLENIOUS SERIES

(i) Defference Method

Let T1, T

2, T

3 ... T

nbe the trms of a sequence and let (T

2 T

1) =

1T , (T3 TT2) = 2T ... ,

(Tn Tn1) = n 1T .

Case I: If 1 2 n 1T , T ,....T are in A.P. then TTnis a quadratic in n. If 1 2 2 3T T , T T , ... are in

A.P., then Tnis a cubic in n.

Case II: If 1 2 n 1T , T ,....T are not in A.P., but if 1 2 n 1T , T ,..., T are in G.P., then TTn= arn+ b,

where r is the C.R. of the G.P.1 2 3T , T , T .....and a, b R.

Again if1 2 n 1T ,T ,...T are not in G.P. but 2 1 3 2 n 1 n 2T T ,T T ,...T T are in G.P., then TTn is of the

form arn+ bn + c; r is the C.R. of the G.P. 2 1 3 2 4 3T T ,T T T T ... and a, b, c R.

(ii) Vn Vn1Method

Let T1, T

2, T

3 , ... be the terms of a sequence. If there exists a sequence V

1, V

2, V

3... satisfying

Tk= V

k V

k1, k 1,

thenn n

n k k k 1 n 0k 1 k 1

S T (V V ) V V

.


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Multiple Choice Questions with Single Answer:

1. If a1, a

2, a

3... are in AP, then a

p, a

q, a

rare in AP, if p, q, r are in

(a) AP (b) GP

(c) HP (d) none of these

2. Let trdenote the rth term of an AP. If t

m=

1

mand n

1t =

n, then t

mn equal to

(a)

1

mn (b)

1 1

m n

1

(c) 1 (d) 0

3. If p, q, r, s N and they are four consecutive terms of an AP, then the pth, qth, rth, sth terms of

a GP are in

(a) AP (b) GP


4. Let a1, a

2, a

3, ... be in AP and a

p, a

q, a

rbe in GP. Then a

q: a

pis equal to

(a)r p

q p

(b)

q p

r q

(c)r q

q p

(d) none of these

5. If a, b, c are in G.P., then a + b, 2b, b + c are in

(a) A.P. (b) G.P.

(c) H.P. (d) none of these

6. If a, b, c, d are nonzero real numbers such that (a2+ b2+ c2) (b2+ c2+ d2)

(ab + bc + cd)2, then

a, b, c, d are in

(a) AP (b) GP(c) HP (d) none of these

OBJECTIVE

SECTION - I

LEVEL - I


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7. If 4a2+ 9b2+ 16c2= 2(3ab + 6bc + 4ca), where a, b, c are nonzero numbers, then a, b, c are in

(a) AP (b) GP


8 If a, b, c are in H.P., then c, c a, c b are in :

(a) A.P. (b) G.P.


9. Let S be the sum, P be the product and R be the sum of the reciprocals of n terms of a GP. Then

P2Rn: Snis equal to

(a) 1 : 1 (b) (common ratio)n : 1

(c) (first term)2: (common ratio)n (d) none of these

10. If a1, a

2,a

3are in AP, a

2, a

3, a

4are in GP and a

3, a

4, a

5are in HP, then a

1, a

3, a

5are in

(a) AP (b) GP


11. If x > 1, y > 1, z > 1 are three numbers in GP then1 1 1

, ,1+ ln x 1+ ln y 1+ ln z are in

(a) AP (b) HP(c) GP (d) none of these

12. If a, a1, a

2, a

3, ... a

2n1, b are in AP, a, b

1, b

2, b

3..., b

2n1, b are in GP and a, c

1, c

2, c

3, ... , c

2n1, b are

in HP, where a, b are positive, then the equation anx2 b

nx + c

n= 0 has its roots

(a) real and unequal (b) real and equal

(c) imaginary (d) none of these

13. The product of n positive numbers is unity. Then their sum is

(a) a positive integer (b) divisible by n

(c) equal to n +1

n(d) never less than n

14. If x > 0 and a is known positive number, then the least values of ax +a

xis

(a) a2 (b) a

(c) 2a (d) none of these


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15. If p, q, r be three positive real numbers, then the value of (p + q) (q + r) (r + p) is

(a) > 8 pqr (b) < 8 pqr

(c) 8 pqr (d) none of these

16. If a1, a

2, a

3.. a

nare in H.P., then

1 2 n

2 3 n 2 3 n 2 3 n 1

a a a, ,....

a a ....a a a ....a a a ....a

are in

(a) A.P. (b) G.P.


17. Sn = 1 + n

1 1 1 1..... ,

2 3 4 2 1

then

(a) S100

< 100 (b) S200

< 200

(c) S200

> 100 (d) S50

> 25

18. Given Sn

n

r rr 0 r 0

1 1,s .

2 2

If S Sn

1

1000 , then least value of n is

(a) 8 (b) 9

(c) 10 (d) 11

19. If x15 x13+ x11 x9+ x7 x5+ x3 x = 7,(x > 0), then

(a) x16is equal to 15 (b) x16is less than 15

(c) x16greater than 15 (d) none of these

20.n 1

1(n 1) (n 2) (n 3)......(n k)

is equal to

(a)1

(k 1) k 1 (b)1

k k

(c)1

(k 1) k (d)1

k

21. Maximum value of n for whichn n

14 1

11 n

2

is

(a) 4 (b) 5

(c) 6 (d) 7


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22. The nth term of the series1 7 1 20

2 1 1 .....2 13 9 23

is

(a)20

(5n 3) (b)20

(5n 3)

(b) 20 (5n + 3) (d) 220(5n 3)

23. If1 1 1 1

+ + +a a - b c c - b

= 0 and a c b then a, b, c are in

(a) A.P. (b) G.P.

(c) H.P. (d) None of these

24. If 2p + 3q + 4r = 15, the maximum value of p3q5r7will be

(a) 2180 (b)4 5

15

5 .3

2

(c)5 7

17

5 .7

2 .9(d) 2285

25. The minimum value of x4 21

x is

(a) 31/ 3

1

4

(b)1

2

(c)1/ 3

12

3

(d) 2


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Multiple Choice Questions with one or more than one correct Answers:

1. Total number of positive real values of x such that x, [x], {x} are in H.P, where [.] denotes the

greatest integer function and {.} denotes fraction part, is equal to

(a) zero (b) 1(c) 2 (d) none of these

2. In the sequence 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, ..... , where n consecutive terms have the

value n, the 1025th term is

(a) 210 (b) 211

(c) 29 (d) 28

3. Sr denotes the sum of the first r terms of an AP. Then S3n: (S2n Sn) is(a) n (b) 3n

(c) 3 (d) independent of n

4. a, b, c are distinct real numbers such that a, b, c are in A.P. and a2, b2, c2are in H.P. then

(a) 2b2= ac (b) 4b2= ac

(c) 2b2= ac (d) 4b2= ac

5. If x1, x2, ... xnare in H.P. then

n-1

r r+1r=1 x x is equal to

(a) (n 1)x1x

n(b) nx

1x

n

(c) (n + 1) x1x

n(d) none of these

6. If ax = by = czand x, y, z are in GP then logc b is equal to

(a) logba (b) log

ab

(c) z/y (d) none of these

7. The value ofn

r=1

1

a + rx + a + (r -1)x is

(a)n

a a nx (b)a nx a

x

(c)n( a nx a)

x

(d) none of these

8. Sides a, b, c of a triangle are in G.P. If5c 3b a

ln , ln and lna 5c 3b

are in A.P., then triangle

must be

LEVEL - II


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(a) Isosceles (b) Equilateral(c)

iit 1 unlocked maths

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