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...

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

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 s4 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


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

, , , , , ,− − −−


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


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


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.


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


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

+

= +

= + ( )

= +

=


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

+ − −

− ( )

=+ − −

−

=− − + +


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.


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.


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.


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.


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).


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).


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.


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}.


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


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


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


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

°=

°=

°=


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


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°.


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


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


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+ +


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


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


Draw the curve:

x

y

1

x = 3

14

y =1

2x 6


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).


The parabola has its vertex at (1, –9) and opens up.

x

y

y = fx

9

12 4

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...

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