g r a d e 11 p r e -c a l c u l u s m at h e m at i c s (30s) · final practice exam answer key 3...
TRANSCRIPT
G r a d e 1 1 P r e - C a l C u l u s M a t h e M a t i C s ( 3 0 s )
Final Practice Exam Answer Key
F i n a l P r a c t i c e E x a m A n s w e r K e y 3 of 34
G r a d e 1 1 P r e - C a l C u l u s M a t h e M a t i C s
Final Practice Exam Answer Key
Name: ___________________________________
Student Number: ___________________________
Attending q Non-Attending q
Phone Number: ____________________________
Address: _________________________________
__________________________________________
__________________________________________
Instructions
The final examination will be weighted as follows: Modules 1–8 100%The format of the examination will be as follows: Module 1: 5 marks Module 5: 19 marks Module 2: 5 marks Module 6: 19 marks Module 3: 5 marks Module 7: 23 marks Module 4: 5 marks Module 8: 19 marks
Time allowed: 2.5 hours
Note: You are allowed to bring the following to the exam: pencils (2 or 3 of each), blank paper, a ruler, a scientific or graphing calculator, and your Final Exam Resource Sheet. Your Final Exam Resource Sheet must be handed in with the exam.
Show all calculations and formulas used. Include units where appropriate. Clearly state your final answer. Diagrams may not be drawn to scale.
For Marker’s Use Only
Date: _______________________________
Final Mark: ________ /100 = ________ %
Comments:
Answer Key
F i n a l P r a c t i c e E x a m A n s w e r K e y 5 of 34
Name:
Answer all questions to the best of your ability. Show all your work.
Module 1: Sequences and Series (5 marks)
1. Write the defining linear function of the following arithmetic sequence. (2 marks) (Lesson 1) 99, 103, 107, . . .
Answer: Points on line: (1, 99), (2, 103), (3, 107) Slope: d = 4 Defining Linear Function:
y – y1 = m(x – x1)y – 99 = 4(x – 1) y = 4x – 4 + 99 y = 4x + 95
2. Use a formula to find the value of
81132
81
=
−
∑k
k
. (3 marks) (Lesson 4)
Answer: Notice k’s initial value is 2. Thus, n is only 7, not 8.
The first few terms are
81 1
381 1
381 1
3
1 2 3
, , , . . . .
Thus, t1 = 27 and
r = 1
3.
St r
r
S
n
n
=−( )−
=
−
−=
1
7
7
11
27 1 13
1 13
27 21862187
=
=2
3
2721862187
23
401327
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s6 of 34
Module 2: Factoring and Rational Expressions (5 marks)
1. Simplify. Identify all non-permissible values. (5 marks) (Lessons 4 and 5)
5 14 192 5 3
3 122 8
2 24 5
2
2 2 2x x
x xx
x xx
x x+ −
+ +÷
−
− −+
−
+ −
Answer:
5 14 192 5 3
3 122 8
2 24 5
5 19
2
2 2 2x x
x xx
x xx
x x
x
+ −
+ +÷
−
− −+
−
+ −
+( ) xxx x
xx x
xx x
−( )+( ) +( )
÷−( )
−( ) +( )+
−( )+( ) −( )
12 3 1
3 44 2
2 15 1
=+( ) −( )+( ) +( )
÷+( )
++( )
=+( ) −
5 19 12 3 1
32
25
5 19
x xx x x x
x x 112 3 1
3 52 5
2 25 2
( )+( ) +( )
÷+( )
+( ) +( )+
+( )+( ) +( )
=
x xx
x xx
x x
55 19 12 3 1
3 152 5
2 45 2
x xx x
xx x
xx x
+( ) −( )+( ) +( )
÷+
+( ) +( )+
+
+( ) +( )
=+( ) −( )+( ) +( )
÷+
+( ) +( )
5 19 12 3 1
5 192 5
x xx x
xx x
=+( ) −( )+( ) +( )
⋅+( ) +( )
+
=−( )
+
5 19 12 3 1
2 55 19
12 3
x xx x
x xx
xx(( ) +( )
⋅+( ) +( )
=+( ) +( ) −( )
+( ) +( )
≠−
xx x
x x xx x
x
12 5
1
2 5 12 3 1
332
1 4 2 5 1195
, , , , , ,− − −−
F i n a l P r a c t i c e E x a m A n s w e r K e y 7 of 34
Name:
Module 3: Quadratic Functions (5 marks)1. Given the following parabola in vertex form, complete the following questions. (1 mark
each, for a total of 5 marks)
y x= −( ) +
12
3 22
(Lessons 1 to 4)
a) Identify the range. Answer: R: {y|y ³ 2, y Î Â}
b) Identify the direction of the opening. Answer: The parabola opens upward
c) Identify the axis of symmetry. Answer: x = 3
d) Identify the y-intercept. Answer:
y x
x
y
y
y
= −( ) +
=
= −( ) +
= ( )+
=
12
3 2
0
120 3 2
129 2
132
2
2
let
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s8 of 34
e) Sketch the graph of the parabola. Answer:
The vertex is (3, 2), the curve opens up, and its size is wide. The y-values are
12
of the
normal y-values.
x
y
y = (x 3)2 + 212
2
3
132
F i n a l P r a c t i c e E x a m A n s w e r K e y 9 of 34
Name:
Module 4: Solving Rational and Quadratic Equations (5 marks)
1. Solve the following quadratic equation using any method you wish. Explain why you chose the method you used. (2 marks) (Lessons 1 to 4) 2x2 + 25 = 15x
Answer:
2 25 15
2 15 25 0
2 10 5 25 0
2 5 5 5 0
2 5
2
2
2
x x
x x
x x x
x x x
x
+ =
− + =
− − + =
−( )− −( )=
−( )) −( )=
= =
x
x x
5 0
52
5or
You can use any method. As long as you complete the question correctly and explain why you chose the method you did, you will get full marks.
A possible explanation is: I chose to solve this quadratic equation by factoring because it is easily factorable.
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s10 of 34
2. Solve the following rational equation. Identify any non-permissible values. (3 marks)
3 11
xx
xx
+=−
+ (Lesson 6)
Answer: Non-permissible values: x = 0 or x = –1 LCD = x(x + 1)
3 11
11
3 1 1
3 4 1
4
2 2
xx
x xx
xx x
x x x x
x x x
x
+( ) +( )= −
+( ) +( )
+( ) +( )=− ( )
+ + =−
22
2
2
4 1 0
4 2 2 1 0
2 2 1 1 2 1 0
2 1 0
12
+ + =
+ + + =
+( )+ +( )=
+( ) =
=−
x
x x x
x x x
x
x
F i n a l P r a c t i c e E x a m A n s w e r K e y 11 of 34
Name:
Module 5: Radicals (19 marks)
1. Order the following radical expressions from least to greatest. Do not use a calculator. (2 marks) (Lesson 1)
6 3 11 7 9 5 2 15, , ,
Answer:
6 3 36 3 108
11 7 121 7 847
9 5 81 5 405
2 15 4 15 60
60 108 405
= =
= =
= =
= =
< < <Thus, 8847 2 15 6 3 9 5 11 7and < < < .
2. Simplify each of the following. All answers must have rationalized denominators. Assume all variables are non-negative. (Lesson 2)
a) 3 19 2 76 (1 mark)
Answer:
3 19 2 76
3 19 2 4 19
3 19 2 2 19
3 19 4 19
7 19
+
= +
= + ( )
= +
=
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s12 of 34
b) 8 43 2xy x y−( )( )
(2 marks) (Lesson 3)
Answer:
8 4
4 2 4 4
4 2 2
3 2xy x y
x y y y x x x y y
y xy xy x
−( )( )= −
= −
c)
5 8 2 53 2 3
(3 marks) (Lesson 4)
Answer:
5 8 2 53 2 3
5 8 2 53 2 3
3 2 33 2 3
15 8 10 24 6 5 4 159 6 3 6 3 4 9
15 4
−
−
=−
−⋅
+
+
=+ − −
+ − −
=22 10 4 6 6 5 4 15
9 4 3
30 2 20 6 6 5 4 153
30 2 20 6 6 5 4 153
+ − −
− ( )
=+ − −
−
=− − + +
F i n a l P r a c t i c e E x a m A n s w e r K e y 13 of 34
Name:
3. Identify the values for each of the variables for which each radical expression is defined. (1 mark each, for a total of 2 marks)
a) 6 3 x (Lesson 1)
Answer: In order for this radical expression to be defined, the radicand has to be greater than
or equal to zero. 6 – 3x ³ 0 6 ³ 3x 2 ³ x x £ 2 Thus, x £ 2 in order for this radical expression to be defined.
b) 32 2x (Lesson 1)
Answer: Because the index is two, the radicand has to be greater than or equal to zero for this
expression to be defined. As any value squared will always be greater than or equal to zero, x Î Â for this expression to be defined.
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s14 of 34
4. Find the solutions for each of the following equations. Check your solutions for extraneous roots. Determine any restrictions on the variable.
a) 3 2 3− = −x x (4 marks) (Lesson 5)
Answer: Restrictions on the variable:
2 3 0
2 3
32
x
x
x
− ≥
≥
≥
3 2 3
9 6 2 3
8 12 0
6 2 0
6 2
2
2
2
−( ) = −
− + = −
− + =
−( ) −( )=
= =
x x
x x x
x x
x x
x xor
Check x = 2 Check x = 6
LHS RHS LHS RHS3 – x3 – 2
1
2 3
2 2 3
1
1
x−
( )−
3 – x3 – 6–3
2 3
2 6 3
9
3
x−
( )−
LHS = RHS LHS ¹ RHS
Therefore, the only solution is x = 2.
F i n a l P r a c t i c e E x a m A n s w e r K e y 15 of 34
Name:
b) 2 4 3 9x + + = (3 marks) (Lesson 5)
Answer: Restrictions on the variable: 2x + 4 ³ 0 2x ³ –4 x ³ –2
2 4 3 9
2 4 6
2 4 6
2 4 36
2 32
16
2
x
x
x
x
x
x
+ + =
+ =
+ =
+ =
=
=
Check x = 16
LHS RHS
2 4 3
2 16 4 3
6 3
9
x + +
( ) + +
+
9
LHS = RHS \ The solution is x = 16.
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s16 of 34
5. The period, P, measured in seconds, of a pendulum is the time it takes to complete one
full swing. The period can be found using the formula P
L=2
9 8π
., where L measures
the length of the pendulum in metres. How long should a pendulum be to complete one full swing in 4.3 seconds? Round your answer to four decimal places. Check your answer for extraneous solutions. (2 marks) (Lesson 5)
Answer:
P
L=2
9 8π
.
Restrictions: L ³ 0 and P ³ 0 Note: P is positive or zero because it is equal to a positive number, 2p, multiplied by the
principal square root. Also, it would not make sense for either variable to be negative.
PL
L
L
L
L
=
=
=( )
=
=
29 8
4 3 29 8
4 3 29 8
18 494
9 818 49
2 2
2
π
π
π
π
.
..
..
... ××
=
9 84
4 5899
2.
.π
L
Check: L ³ 0
LHS RHSP
4.3 29 8
24 5899
9 8
4 3
π
π
L.
..
.
LHS = RHS
The pendulum should be 4.5899 m long.
F i n a l P r a c t i c e E x a m A n s w e r K e y 17 of 34
Name:
Module 6: Systems of Equations and Inequalities (19 marks)
1. Solve the following system of equations algebraically and graphically. (6 marks)
y x
y x
= −( )
=− +( ) +
3
3 1 12
2
2
(Lesson 2)
Answer: Algebraic Solution
y x y x
y x x y x x
y x x
= −( ) =− +( ) +
= − + =− + +( )+
=− − +
3 3 1 12
6 9 3 2 1 12
3 6 9
2 2
2 2
2
Substitute the expression for y in Equation (1) into the y-variable of Equation (2).
x x x x
x
x
2 2
2
6 9 3 6 9
4 0
0
− + =− − +
=
=
Solve for y:
y x
y
y
= −( )= −( )=
3
0 3
9
2
2
Thus, the solution to this system of equations is (0, 9).
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s18 of 34
Graphing Solution The parabola, y = (x – 3)2, has its vertex at (3, 0), it opens up, and has normal shape. The parabola, y = –3(x + 1)2 + 12, has its vertex at (–1, 12), opens down, and its shape is
narrower. The y-coordinates are multiplied by 3.
x
y
1 3
y = (x 3)2
y = 3(x 1)2 12
2
12
9
16
3
1
The solution to this system of equations is the point at which the two functions intersect, or (0, 9).
F i n a l P r a c t i c e E x a m A n s w e r K e y 19 of 34
Name:
2. The sum of two numbers is 22. Their product is 117.a) Write a system of equations to represent this problem. (1 mark) (Lesson 1) Answer: Let the first number be represented by m and the second number be represented by
n. The system of equations is:
m n
mn
+ =
=
22
117
b) Solve the system of equations to find the two numbers. (3 marks) Answer: Substitute m = 22 – n into nm = 117. (22 – n)(n) = 117 22n – n2 – 117 = 0 n2 – 22n + 117 = 0 (n – 13)(n – 9) = 0 n = 13 or n = 9
Substituting into the equation m = 22 – n, you get: n = 13 n = 9 m = 22 – n m = 22 – n m = 22 – 13 m = 22 – 9 m = 9 m = 13
Therefore, the two numbers are 9 and 13.
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s20 of 34
3. Solve the following inequalities by graphing. (2 marks each for a total of 6 marks) a) y < 3x – 8 (Lesson 3) Answer: The slope of the line is 3. The y-intercept is –8. The boundary line is dotted and the
shading is below the line.
x
y
8
y < 3x 8
2
b) 0 £ x2 – 4 (Lesson 4) Answer: You are interested in the domain where the expression is greater than or equal to 0. Write x2 – 4 ³ 0 and write the corresponding function as y = x2 – 4. The parabola has
its vertex at (0, –4), opens up, and has normal shape.
x
y
4
22
y = x2 4
The parabola crosses the x-axis at –2 and 2. The parabola lies on or above the x-axis when x £ –2 and when x ³ 2. Therefore, the solution set is {x|x £ -2 or x ³ 2, x Î Â}.
F i n a l P r a c t i c e E x a m A n s w e r K e y 21 of 34
Name:
c) y > (x + 1)2 – 3 (Lesson 5) Answer: The parabola has its vertex at (–1, –3), opens up, and has normal shape. The
boundary line is dotted and the region above the parabola is shaded.
x
y
2
4
y > (x + 1)2 3
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s22 of 34
4. Use the sign analysis method to solve the following quadratic inequality. (3 marks) 0 £ 2x2 – 5x – 3 (Lesson 4)
Answer: Write as 2x2 – 5x – 3 ³ 0. You want to consider all values of the expression that are greater than or equal to zero. Factoring the corresponding quadratic equation: 2x2 – 5x – 3 = 0 2x2 – 6x + x – 3 = 0 2x(x – 3) + 1(x – 3) = 0 (2x + 1)(x – 3) = 0 2x + 1 = 0, x – 3 = 0
x x= − =1
23or
The critical points are
12
and 3.
x +
x +
x++
+
2x + 1
(2x + 1)(x 3)
Factor Sign Diagram
3x 3
12
312
The solution set contains the points where 2x2 – 5x – 3 ³ 0 and is
x x x x| , .≤− ≥ ∈ ℜ
12
3or
F i n a l P r a c t i c e E x a m A n s w e r K e y 23 of 34
Name:
Module 7: Trigonometry (23 marks)
1. P(2, –3) is a point on the terminal side of angle q in standard position. Determine sin q, cos q, and tan q for the following point. Also, determine the distance from the origin to the point P(2, –3). Rationalize any denominators. (5 marks) (Lesson 1)
Answer:
r
yr
xr
yx
= + −( ) =
= =−
=−
= = =
= =−
2 3 13
313
3 1313
213
2 1313
2 2
sin
cos
tan
θ
θ
θ33
232
=−
x
y
2
r
P(2, 3)
3
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s24 of 34
2. Given the following angles in standard position, determine the exact values of the sine, cosine, and tangent ratios. Show how you determined the exact values. (2 marks each, for a total of 4 marks) (Lesson 3)a) 240° Answer: qr = 240° – 180° = 60°
2
60°
3
1x
y
sin
cos
tan
2403
2
2401
2
2403
13
°=−
°=−
°=−−
=
b) 45° Answer: qr = 45°
145°
21
x
y
sin
cos
tan
4512
4512
45 1
°=
°=
°=
F i n a l P r a c t i c e E x a m A n s w e r K e y 25 of 34
Name:
3. Sketch the angle 342° in standard position and find its reference angle. Determine the other angles that have the same reference angle as the given angle for q in the interval [0°, 360°]. (3 marks) (Lesson 2)
Answer:
x
y
= 342°
The reference angle for 342° is 360° – 342° = 18°. The other angles that have this same reference angle are:
18° in Quadrant I 180° – 18° = 162° in Quadrant II 180° + 18° = 198° in Quadrant III
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s26 of 34
4. Determine the solution set for each of the following trigonometric equations over the interval [0°, 360°]. Round to the nearest degree where necessary. (2 marks each, for a total of 4 marks) (Lesson 2)
a) cos θ =15
Answer:
θ
θ
r
r
=
= °
−cos 1 15
78
As cos q > 0, q must be in Quadrants I or IV. In Quadrant I, q = 78°. In Quadrant IV, q = 360° – 78° = 282°.
b) sin q – 1 = 0 Answer: sin q = 1 This is a quadrantal angle through the point (0, 1) and only occurs when q is 90°.
F i n a l P r a c t i c e E x a m A n s w e r K e y 27 of 34
Name:
5. In DABC, ÐC = 41°, c = 6, and a = 8. Find all possible values for b and ÐB. Draw a diagram and round off answers to two decimal places. (5 marks) (Lessons 4 and 6)
Answer: Since AB < BC, find h: h = 8 sin 41° = 5.25 Since h < AB, there are two solutions. Draw two triangles and solve.
41°
8 6
B1
C1 A1b1 41°
86
B2
C2 A2b2
641
8
6 8 41
8 4168 41
61
sin sin
sin sin
sinsin
sinsin
°=
= °
=°
∠ =°
−
A
A
A
A
∠ = ° ∠A with A in Quadrants I and II.related 61 01. ,
∠ = °
∠ = °− ° + °( )
= °
°=
°
A
B
1
1
1
61 01
180 41 61 01
77 99
641 77 99
.
.
.
sin sin .b
b11
1
6 77 9941
8 95
=°
°
=
sin .sin
.b units
∠ = °− °
= °
∠ = °− ° + °( )= °
A
B
2
2
180 61 01118 99
180 41 118 9920 01
641
..
..
sin °°=
°
=°
°
=
b
b
b
2
2
2
20 016 20 01
413 13 3 14
sin .sin .sin
. .units or units
41°
8 6h
B
C A
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s28 of 34
6. Find the measure of ÐA in DABC (to the nearest degree). (2 marks) (Lesson 5)
A B
C
4 6
7
Answer:
a b c bc2 2 2
2 2 2
2 2 2
2
6 4 7 2 4 7
6 4 7 2 4 7
= + −
= + − ( )( )
− − =− ( )( )
−
cos
cos
cos
A
A
A
229 56
0 517857
0 517857
59
1
=−
=
= ( )
∠ = °
−
cos
. cos
cos .
A
A
A
A
F i n a l P r a c t i c e E x a m A n s w e r K e y 29 of 34
Name:
Module 8: Absolute Value and Reciprocal Functions (19 marks)
1. Evaluate. (1 mark each, for a total of 2 marks) (Lesson 1)a) 4| 3(–7) + 4 | – 3 Answer: 4| 3(–7) + 4 | – 3 = 4| –17 | – 3 = 4(17) – 3 = 65
b) –2| –2(3) + 1 | + 5 Answer: –2| –2(3) + 1 | + 5 = –2| –5 | + 5 = –2(5) + 5 = –5
2. Solve the following absolute value equation algebraically. (4 marks) (Lesson 3) | x2 – x | = 6
Answer: Case 1: x2 – x = 6 and x2 – x ³ 0 x2 – x – 6 = 0 x(x – 1) = 0 (x – 3)(x + 2) = 0 x = 0, 1 x = 3, – 2
x Î (–¥, 0] È [1, ¥)
Since x = 3 is in the interval [1, ¥) and x = –2 is in the interval (–¥, 0], both are solutions.
Case 2: –(x2 – x) = 6 and x2 – x < 0 –x2 + x = 6 –x2 + x – 6 = 0 x2 – x + 6 = 0 b2 – 4ac = (–1)2 – 4(1)(6) = 1 – 24 = –23 There is no solution. Therefore, the solutions to this absolute value equation are x = 3 and x = –2
0 1 x+ +
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s30 of 34
3. Solve the following absolute value equation graphically. (4 marks) | x + 4 | = 5 (Lesson 3)
Answer: From the left side of the equation, graph y = |x + 4|. From the right side, graph y = 5.
The two graphs intersect at the points (–9, 5) and (1, 5). The solutions for the equation are x = –9 and x = 1.
x
y
4
y = 5
y = |x + 4|
9
5
4
F i n a l P r a c t i c e E x a m A n s w e r K e y 31 of 34
Name:
4. Sketch the graph of the reciprocal of the given function. State the equation(s) of the horizontal and vertical asymptote(s). (4 marks) (Lesson 4) y = 2x – 6
Answer:
The reciprocal function is:
f x
x( )=
−1
2 6
y-intercept: Find f(0):
f 0
12 0 6
16
16
( )=( )−
=−
=−
Vertical asymptote: 2x – 6 = 0 2x = 6 x = 3
Horizontal asymptote: The horizontal asymptote is the line y = 0.
Invariant points:
Let 2 6 1
2 7
72
x
x
x
− =
=
=
or
2 6 1
2 5
52
x
x
x
− =−
=
=
The invariant points are
72
152
1, , .
−
and
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s32 of 34
Draw the curve:
x
y
1
x = 3
14
y =1
2x 6
F i n a l P r a c t i c e E x a m A n s w e r K e y 33 of 34
Name:
5. Given the graph of y
f x=
( )1
, sketch the graph of y = f(x). (5 marks) (Lesson 5)
x
y
x = 4
x = 2
1, 19
3
Answer: The vertical asymptotes of this reciprocal graph are at x = –2 and x = 4, and are thus the
x-intercepts in the graph of f(x). In other words, f(x) is of the form f(x) = a(x + 2)(x – 4).
The key point is
1
19
, .−
The corresponding point in the graph of the function, f(x), is
(1, –9).
The invariant points can be read from the graph and are approximately (–2.2, 1), (4.2, 1), (–1.8, –1), and (3.8, –1).