grade 12 prototype examination pre-calculus...

November 2016

Grade 12 Prototype Examination

Pre-calculus 30

Course Code 8426

Barcode Number

Month Day

Date of Birth

i

Pre-calculus 30, Prototype Exam

November 2016

Pre-calculus 30

TIME: Two and One-Half Hours

Calculating devices MUST meet the requirements of the Calculator Use Policy. Before an

examination begins, devices must be removed from their cases and placed on the students’ desks

for inspection by a mathematics or science teacher. Cases must be placed on the floor and left

there for the duration of the examination. Students using a standard scientific or graphing

calculator must clear all information stored in its memory before the examination begins.

Devices such as cell phones, tablets, and iPods may be used as a calculating device if they meet

the requirements of the Calculator Use Policy. The school or writing center must be able to

lock and control the device using a feature such as Guided Access, or a management software that

limits its functionality to permissible graphing and financial applications (apps) with similar

functionality to an approved graphing calculator. It is the student’s responsibility to ensure

their device complies with the Calculator Use Policy in advance of the departmental examination

session.

Do not spend too much time on any one question. Read each question carefully.

The examination consists of 38 multiple-choice questions followed by 7 numeric response questions

of equal value which will be machine scored. Record your answers on the Student Examination

Form which is provided. Each multiple-choice question has four suggested answers, one of which

is better than the others. Select the best answer and record it on the Student Examination Form

as shown in the example below:

Student Examination Form:

Multiple-Choice Questions

This examination is being written in the

subject

A. Chemistry.

B. Pre-calculus.

C. Workplace and Apprenticeship

Mathematics.

D. Foundations of Mathematics.

1. A B C D

Numeric Response Questions

Record your answer in the numeric

response section on the answer sheet.

What is 10% of $2000? (Round to the

nearest dollar.)

ii


November 2016

What is 10% of $248.50? (Round to the

nearest dollar.)

What is 10% of 24 125? (Round to the nearest

whole number.)

Use an ordinary HB pencil to mark your answers on the Student Examination Form. If you

change your mind about an answer, be sure to erase the first mark completely. There should be

only one answer marked for each question. Be sure there are no stray pencil marks on your

answer sheet. If you need space for rough work, use the space in the examination booklet beside

each question.

Do not fold either the Student Examination Form or the examination booklet. Check

that your personal information on the Student Examination Form is correct and

complete. Make any necessary changes, and fill in any missing information. Be sure to

complete the Month and Day of Your Birth section.

- OVER - Ministry of

Education 2016-17

Pre-calculus 30

Quadratic Formula

For 2 0 ax bx c ,

2 4

2

b b acx

a

Arc Length

a r

Trigonometry and Trigonometric Identities

sintan

cos

sin

cos = cot

1csc =

sin

cos

1sec

1cot

tan

1cossin 22 22 sectan1 22 csccot1

cos ( + ) = cos cos sin sin A B A B A B

cos ( ) = cos cos + sin sin A B A B A B

sin ( + ) = sin cos + cos sin A B A B A B

sin ( ) = sin cos cos sin A B A B A B

tan tantan( )

1 tan tan

A BA B

A B

tan tantan( )

1 tan tan

A BA B

A B

sin2 2sin cos

2 2cos2 cos sin 2cos2 2cos 1

2cos2 1 2sin

2

2tantan2

1 tan

iii


November 2016

Permutations, Combinations, and Binomial Theorem

!

( ) !

rn

nP

n r

!

( ) ! !

rn

nC

n r r

rn

nC

r

0 1 1 2 2 0

0 1 2( ) ...n n n n n

n n n n na b C a b C a b C a b C a b

The general term, 1,

kt

of the expansion of ( ) na b is: n k kn

a bk

OR n k k

n kC a b

General Form of Transformed Functions

( ( ))y k af b x h

sin ( )y a b x c d

cos ( )y a b x c d

iv


November 2016

Ministry of Education

Pre-calculus Insert

v

Table of Trigonometric Ratios

sin cos tan csc sec cot

0 0.0000 1.0000 0.0000 1.0000

1 0.0175 0.9998 0.0175 57.298 1.0002 57.290

2 0.0349 0.9994 0.0349 28.653 1.0006 28.636

3 0.0523 0.9986 0.0524 19.107 1.0014 19.081

4 0.0698 0.9976 0.0699 14.335 1.0024 14.300

5 0.0872 0.9962 0.0875 11.473 1.0038 11.4301

6 0.1045 0.9945 0.1051 9.5668 1.0055 9.5144

7 0.1219 0.9925 0.1228 8.2055 1.0075 8.1444

8 0.1392 0.9903 0.1405 7.1853 1.0098 7.1154

9 0.1564 0.9877 0.1584 6.3925 1.0125 6.3138

10 0.1736 0.9848 0.1763 5.7588 1.0154 5.6713

11 0.1908 0.9816 0.1944 5.2408 1.0187 5.1446

12 0.2079 0.9781 0.2126 4.8097 1.0223 4.7046

13 0.2250 0.9744 0.2309 4.4454 1.0263 4.3315

14 0.2419 0.9703 0.2493 4.1336 1.0306 4.0108

15 0.2588 0.9659 0.2679 3.8637 1.0353 3.7321

16 0.2756 0.9613 0.2867 3.6280 1.0403 3.4874

17 0.2924 0.9563 0.3057 3.4203 1.0457 3.2709

18 0.3090 0.9511 0.3249 3.2361 1.0515 3.0777

19 0.3256 0.9455 0.3443 3.0716 1.0576 2.9042

20 0.3420 0.9397 0.3640 2.9238 1.0642 2.7475

21 0.3584 0.9336 0.3839 2.7904 1.0711 2.6051

22 0.3746 0.9272 0.4040 2.6695 1.0785 2.4751

23 0.3907 0.9205 0.4245 2.5593 1.0864 2.3559

24 0.4067 0.9135 0.4452 2.4586 1.0946 2.2460

25 0.4226 0.9063 0.4663 2.3662 1.1034 2.1445

26 0.4384 0.8988 0.4877 2.2812 1.1126 2.0503

27 0.4540 0.8910 0.5095 2.2027 1.1223 1.9626

28 0.4695 0.8829 0.5317 2.1301 1.1326 1.8807

29 0.4848 0.8746 0.5543 2.0627 1.1434 1.8040

30 0.5000 0.8660 0.5773 2.0000 1.1547 1.7321

31 0.5150 0.8572 0.6009 1.9416 1.1666 1.6643

32 0.5299 0.8480 0.6249 1.8871 1.1792 1.6003

33 0.5446 0.8387 0.6494 1.8361 1.1924 1.5399

34 0.5592 0.8290 0.6745 1.7883 1.2062 1.4826

35 0.5736 0.8192 0.7002 1.7434 1.2208 1.4281

36 0.5878 0.8090 0.7265 1.7013 1.2361 1.3764

37 0.6018 0.7986 0.7536 1.6616 1.2521 1.3270

38 0.6157 0.7880 0.7813 1.6243 1.2690 1.2799

39 0.6293 0.7771 0.8098 1.5890 1.2868 1.2349

40 0.6428 0.7660 0.8391 1.5557 1.3054 1.1918

41 0.6561 0.7547 0.8693 1.5243 1.3250 1.1504

42 0.6691 0.7431 0.9004 1.4945 1.3456 1.1106

43 0.6820 0.7314 0.9325 1.4663 1.3673 1.0724

44 0.6947 0.7193 0.9657 1.4396 1.3902 1.0355

sin cos tan csc sec cot

45 0.7071 0.7071 1.0000 1.4142 1.4142 1.0000

46 0.7193 0.6947 1.0355 1.3902 1.4396 0.9657

47 0.7314 0.6820 1.0724 1.3673 1.4663 0.9325

48 0.7431 0.6691 1.1106 1.3456 1.4945 0.9004

49 0.7547 0.6561 1.1504 1.3250 1.5243 0.8693

50 0.7660 0.6428 1.1918 1.3054 1.5557 0.8391

51 0.7771 0.6293 1.2349 1.2868 1.5890 0.8098

52 0.7880 0.6157 1.2799 1.2690 1.6243 0.7813

53 0.7986 0.6018 1.3270 1.2521 1.6616 0.7536

54 0.8090 0.5878 1.3764 1.2361 1.7013 0.7265

55 0.8192 0.5736 1.4281 1.2208 1.7434 0.7002

56 0.8290 0.5592 1.4826 1.2062 1.7883 0.6745

57 0.8387 0.5446 1.5399 1.1924 1.8361 0.6494

58 0.8480 0.5299 1.6003 1.1792 1.8871 0.6249

59 0.8572 0.5150 1.6643 1.1666 1.9416 0.6009

60 0.8660 0.5000 1.7320 1.1547 2.0000 0.5774

61 0.8746 0.4848 1.8040 1.1434 2.0627 0.5543

62 0.8829 0.4695 1.8807 1.1326 2.1301 0.5317

63 0.8910 0.4540 1.9626 1.1223 2.2027 0.5095

64 0.8988 0.4384 2.0503 1.1126 2.2812 0.4877

65 0.9063 0.4226 2.1445 1.1034 2.3662 0.4663

66 0.9135 0.4067 2.2460 1.0946 2.4586 0.4452

67 0.9205 0.3907 2.3558 1.0864 2.5593 0.4245

68 0.9272 0.3746 2.4751 1.0785 2.6695 0.4040

69 0.9336 0.3584 2.6051 1.0711 2.7904 0.3839

70 0.9397 0.3420 2.7475 1.0642 2.9238 0.3640

71 0.9455 0.3256 2.9042 1.0576 3.0715 0.3443

72 0.9511 0.3090 3.0777 1.0515 3.2361 0.3249

73 0.9563 0.2924 3.2708 1.0457 3.4203 0.3057

74 0.9613 0.2756 3.4874 1.0403 3.6279 0.2867

75 0.9659 0.2588 3.7320 1.0353 3.8637 0.2680

76 0.9703 0.2419 4.0108 1.0306 4.1335 0.2493

77 0.9744 0.2250 4.3315 1.0263 4.4454 0.2309

78 0.9781 0.2079 4.7046 1.0223 4.8097 0.2126

79 0.9816 0.1908 5.1445 1.0187 5.2408 0.1944

80 0.9848 0.1736 5.6712 1.0154 5.7587 0.1763

81 0.9877 0.1564 6.3137 1.0125 6.3924 0.1584

82 0.9903 0.1392 7.1153 1.0098 7.1852 0.1405

83 0.9925 0.1219 8.1443 1.0075 8.2054 0.1228

84 0.9945 0.1045 9.5143 1.0055 9.5667 0.1051

85 0.9962 0.0872 11.429 1.0038 11.473 0.0875

86 0.9976 0.0698 14.300 1.0024 14.335 0.0699

87 0.9986 0.0523 19.080 1.0014 19.106 0.0524

88 0.9994 0.0349 28.635 1.0006 28.652 0.0349

89 0.9998 0.0175 57.285 1.0002 57.294 0.0175

90 1.0000 0.0000 1.0000 0.0000

- 1 -


November 2016

GRADE 12 DEPARTMENTAL EXAMINATION

PRE-CALCULUS 30 PROTOTYPE

NOVEMBER 2016

VALUE

(45 2) Answer the following 45 questions on the computer sheet entitled

“Student Examination Form.”

MULTIPLE - CHOICE QUESTIONS

1. Which of the following would be impacted by a change to the parameter

“h ” when graphing ( ( )) ?y a f b x h k

A. vertical stretch

B. horizontal stretch

C. vertical translation

D. horizontal translation

2. Which logarithmic expression is equivalent to 25 ?xy

A. 5

( 2) logx y

B. 5

log ( 2)x y

C. 5

log 2y x

D. log ( 2) 5y

x

- 2 -


November 2016

3. The graph of ( )y f x is transformed to produce the graph of ( )y g x as

shown below:

Which of the following equations best describes ( )?y g x

A. ( ) ( 1) 2g x f x

B. ( ) ( 1) 2g x f x

C. ( ) ( 2) 1g x f x

D. ( ) ( 2) 1g x f x

4. Functions ( )f x and ( )g x are defined by 2( ) 2 3 2f x x x and

2( ) 4 3.g x x What is the equation of function ( )h x if ( ) ( ) ? h x g f x

A. 2( ) 6 3 5h x x x

B. 2( ) 6 3 1h x x x

C. 2( ) 6 3 5h x x x

D. 2( ) 6 3 1h x x x

- 3 -


November 2016

5. The point ( , )x y lies on the graph of ( ).y f x Which of the following

statements about this point is TRUE?

A. It will map to the point ( , )x y on the graph of ( ). y f x

B. It will map to the point ( , )x y on the graph of ( ). y f x

C. It will map to the point ( , )x y on the graph of ( ). y f x

D. It will map to the point ( , )x y on the graph of ( ). y f x

6. What is the equation of the inverse relation of 215 ( 16) ?

2y x

A. 16 2 ( 5)y x

B. 16 2( 5)y x

C. 4 2 ( 5)y x

D. 4 2( 5)y x

- 4 -


November 2016

7. Which of the following fully represents the graph of the rational function

5 13( ) ,

3

xf x

x including all asymptotes?

A.

B.

C.

D.

- 5 -


November 2016

8. What is the single logarithm of 6 4 3 2

2 2 2log log 4 log ,x y xy x y written

in simplest form?

A. 4

2log 4 x

B. 10 3

2log x y

C. 8 4

2log x y

D. 3 2 3 8 4

2log ( )x y xy x y

9. A polynomial function is graphed below:

What are the factors of this polynomial function?

A. 1

( 2)( 1)2

x x x x

B. 1

( 2)( 1)2

x x x x

C. 1

( 2)( 2)( 1)2

x x x x x

D. 1

( 2)( 2)( 1)2

x x x x x

- 6 -


November 2016

10. Four relations are graphed below. Which relation and its inverse are

both functions?

A.

B.

C.

D.

- 7 -


November 2016

11. Functions ( )y f x and ( )y g x are defined by ( ) 2f x x and

( ) 1.g x x Which of the following best represents the graph of

( ( )) ?y f g x

A.

B.

C.

D.

- 8 -


November 2016

12. Which of the following statements is always TRUE about an

exponential function in the form ,xy a where 0 ?a

A. The function has an x-intercept of 1.

B. The range of the function is 0, . y y y

C. The function has a vertical asymptote of 0.x

D. The function increases in value as x increases.

13. What is the solution set of 6 6

log ( 5) log ( ) 2 ?x x

A. 9

B. 4

C. 4, 9

D. no solution

- 9 -


November 2016

14. Which of the following represents the graph of 3( ) 3 2 ?f x x x

A.

B.

C.

D.

- 10 -


November 2016

15. How will the graph of 22 3y x x be transformed if it has been modified

to be 22( 3) 3( 3) 4 ?y x x

A. The new graph will shift 4 units to the right and 3 units up.

B. The new graph will shift 3 units to the right and 4 units up.

C. The new graph will shift 4 units to the left and 3 units down.

D. The new graph will shift 3 units to the left and 4 units down.

16. What is the result when you divide 3 24 9 58 15x x x by 3 ?x

A. 24 3 49x x

B. 24 21 5x x

C. 3 24 21 5x x x

D. 3 24 3 49 132x x x

17. What are the x-intercepts of ( 8)

( ) 4 ?3

f x xx

A. (3,0) and (4,0)

B. ( 5,0) and (4,0)

C. (5,0) and ( 4,0)

D. ( 4,0) and ( 3,0)

18. What is the solution set of 13 1?x x

A. 3

B. 4

C. 4,3

D. 3,4

- 11 -


November 2016

19. A function is defined by

2

2

2 10( ) .

3 10

x xf x

x x On the graph of ( ),f x where

are the vertical asymptote(s) and the point of discontinuity (hole) located?

A. vertical asymptote at 5;x no point of discontinuity

B. vertical asymptote at 2;x point of discontinuity at 10

5,7

C. vertical asymptote at 2;x point of discontinuity at 10

5,7

D. vertical asymptotes at 2x and 5;x no point of discontinuity

20. A square piece of cardboard measures 24 inches per side. Julie cuts

4 smaller squares from each corner of the cardboard as shown by the

shading in the diagram below. She then bends up the sides of the

remaining cardboard to form an open box with no top.

If the squares she cut out had sides of x inches, what is the polynomial

expression, V(x), that represents the volume of the box?

A. 3( ) 144V x x x

B. 3( ) 4 576V x x x

C. 3 2( ) 48 576V x x x x

D. 3 2( ) 4 96 576V x x x x

- 12 -


November 2016

21. Which of the following sketches shows an angle of 100° in standard

position?

A.

B.

C.

D.

- 13 -


November 2016

22. The terminal arm of angle is located in quadrant III. If 5

cos ,13

what is the exact value of tan ?

A. 12

5

B. 12

13

C. 12

13

D. 12

5

23. What is the exact value of cos15

A. 6 2

4

B. 6 2

4

C. 6 2

2

D. 6 2

2

24. What are all of the non-permissible values for in tan csc sec ?

A. 0, n

B.

3

0, ,2 2

C. ,n n I

D. ,2

n n I

- 14 -


November 2016

25. What are the possible values of if 2 3

csc ,3

where 0 < 2 ?

A. 11

,6 6

B. 2

,3 3

C. 5 7

,6 6

D. 4 5

,3 3

26. Which of the following statements is TRUE for all permissible values of ?

A. 2 2tan sec sin

B. 2(1 tan ) cos 1

C. 2 2(1 cos ) csc 1

D. 2cos tan csc sin

27. What is the phase shift (horizontal translation) of 2 cos 3 26

y

with respect to cos ?y

A. 2 units to the left

B. 6

units to the left

C. 2 units to the right

D. 6

units to the right

- 15 -


November 2016

28. Which of the following shows all of the angles that are coterminal

with

,6

where 4 < 4 ?

A. 11

6

B. 13 11

,6 6

C. 13 11 23

, ,6 6 6

D. 25 13 11 23

, , ,6 6 6 6

29. The graph of cosy has been transformed so that the resultant graph

has the following characteristics:

amplitude is 4

phase shift (horizontal translation) is 2

to the left

vertical displacement is 2 units down

Which of the following equations would produce a graph that has all of

the above characteristics?

A. 2 cos 3 42

y

B. 2 cos 3 42

y

C. 4 cos 3 22

y

D. 4 cos 3 22

y

- 16 -


November 2016

30. To the nearest degree, what is the general solution for 5 cot 6 0 ?

A. 130 180 ,n n I

B. 130 360 ,n n I

C. 140 180 ,n n I

D. 140 360 ,n n I

31. What is 1 (cos )(cot )(csc ) x x x simplified?

A. 2csc x

B. 2cos x

C. 2sec x

D. 2sin x

32. What is the solution set of 22 sin sin 0, where 0 2 ?

A. 5

0, , ,6 6

B. 11

0, , ,6 6

C. 4

0, , ,3 3

D. 5

0, , ,3 3

- 17 -


November 2016

33. Mr. Wallace challenged his math club to derive the unit circle equation

using a method of their choice. Submissions from 4 of the club members

are shown below:

Andrew

2 2 2

2 2 2

2 2 2

2 2

( )

1

1

a b c

x y r

x y

x y

Becky

2 2

2 1 2 1

2 2

2 2

2 2

2 2 2

2 2

( ) ( )

1 ( 0) (0 ( ))

1 ( ) (0 )

1

1

1

d x x y y

x y

x y

x y

x y

x y

Colin

2 2 2

2 2 2

2 2 2

2 2

( )

1

1

a b c

x y r

x y

x y

Darla

2 2

2 1 2 1

2 2

2 2

2 2 2

2 2

( ) ( )

1 ( 0) ( 0)

1 ( )

1

1

d x x y y

x y

x y

x y

x y

- 18 -


November 2016

Which students submitted a fully correct derivation of the unit circle

equation?

A. Andrew and Darla

B. Andrew and Becky

C. Colin and Becky

D. Colin and Darla

34. The point (6, )b is on the terminal arm of an angle, , in standard

position. If 4

tan3

and is a first-quadrant angle, what is the exact

value of ?b

A. 3

B. 4

C. 6

D. 8

35. You can purchase a Roughrider license plate made up of 1 letter, either

R or S, and 4 digits. The letter can go in any of the 5 spots and

repetition of digits is allowed. How many different Roughrider license

plates of this form can be made?

A. 20 000

B. 65 610

C. 100 000

D. 248 832

- 19 -


November 2016

36. Which of the following represents the 2nd term in the expansion

of 7(3 4) ?x

A. 6 1

7 13 4C x

B. 5 2

7 23 4C x

C. 2 5

7 23 4C x

D. 1 6

7 13 4C x

37. Which of the following shows one way to calculate for the number of

different arrangements that can be made using all of the letters of the

word TRAMPOLINE?

A. 10 10

C

B. 10 10

P

C. 10 1

C

D. 10 1

P

38. What is the solution set for r given 9

84 ?r

C

A. 3

B. 4

C. 3, 6

D. 4, 5

- 20 -


November 2016

NUMERIC RESPONSE QUESTIONS

Record your answer in the Numeric Response section of the

“Student Examination Form.”

39. Function f(x) and g(x) are graphed below. What is the value of ( ( 3)) ?g f

y g(x)

y f (x)

= ( )y g x

( )y f x

40. The graphs of 10 different polynomial functions are sketched below:

How many of these polynomials have a negative leading coefficient and

an odd degree?

- 21 -


November 2016

41. Kim drew a sketch of the shot put field she was setting up. What is the

measure of the central angle, x, in her sketch of her field? (Round to the

nearest degree.)

42. Heidi marks 2 points on an inner tube, one on the inside and the other

on the outside, as shown in the left diagram below. When the tire is

rolled along the floor, Heidi measures and records the heights of the 2

points. Her graph of the data is the sinusoidal curves shown below:

What is the difference in the radii of the 2 circles formed by the inner

tube? (Round to the nearest inch.)

- 22 -


November 2016

43. Scott bought a new car for $35 000. Each month the car depreciates by

1.2%. Scott plans to sell the car once it reaches a value of $20 000. The

equation to determine the time it takes for the car to depreciate to this

amount is 20 000 35 000 0.988 ,t

where t represents the number of

months.

How many months will it take for Scott’s car to depreciate to $20 000?

(Round to the nearest month.)

44. What is the approximate solution to 4 cos 1 3 cos 4, where

270 < 360 ? (Round to the nearest degree.)

45. A test has 2 parts. Students must answer 18 out of 20 multiple-choice

questions in Part A and 2 out of 4 long-answer questions in Part B.

How many different combinations of questions are there for this test?

- i -

Pre-calculus 30, Prototype Exam Answer Key

November 2016

GRADE 12 DEPARTMENTAL EXAMINATION

Pre-calculus 30

PROTOTYPE EXAM − Answer Key

(See Explanation of Answers)

1. D 11. C 21. C 31. A 41. 38 − 39

2. C 12. B 22. D 32. A 42. 15

3. A 13. A 23. B 33. A 43. 46 – 47

4. D 14. D 24. D 34. D 44. 295 − 296

5. A 15. D 25. D 35. C 45. 1140

6. B 16. B 26. C 36. A

7. A 17. B 27. B 37. B

8. B 18. A 28. C 38. C

9. C 19. C 29. D 39. 8

10. B 20. D 30. C 40. 1

Explanation of Answers

1. D.

The parameters , , , anda b h k in the function ( ( ))y a f b x h k correspond

to the following transformations:

a corresponds to a vertical stretch about the x-axis.

b corresponds to a horizontal stretch about the y-axis.

h corresponds to a horizontal translation.

k corresponds to a vertical translation.

Thus, when parameter h is changed, the graph translates to the left or right,

thus undergoing a horizontal translation.

2. C.

In general:

log y

bx y b x

Therefore, 2

5

5

log ( 2)

xy

y x

- ii -


November 2016

3. A.

To translate a function 1 unit to the right, each individual x must have

a value of 1 subtracted from it. This means that ( )y f x will become

( 1).y f x

To translate a function 2 units down, the entire equation must have a

value of 2 subtracted from it. This means that ( 1)y f x will become

( 1) 2y f x or ( ) ( 1) 2g x f x

4. D.

2 2

2 2

2

( ) ( )

( ) ( )

(4 3) ( 2 3 2)

4 3 2 3 2

6 3 1

h x g f x

g x f x

x x x

x x x

x x

5. A.

Given the function ( ) :y f x

( )y f x is a reflection in the x-axis and each point ( , )x y will map

to a point ( , )x y ; and,

( )y f x is a reflection in the y-axis and each point ( , )x y will map

to the point ( , ).x y

The only correct statement is A.

- iii -


November 2016

6. B.

To find the inverse equation of 215 ( 16) ,

2y x exchange x and y. The

equation of the inverse relation can then be found by solving for y.

2

2

15 ( 16)

2

2( 5) ( 16)

2( 5) 16

16 2( 5)

16 2( 5)

x y

x y

x y

x y

y x

7. A.

The graph of

5 13( )

3

xf x

x crosses the x-axis at the point when 0 :y

5 13 130 ; 5 13 0; .

3 5

xx x

x The x-intercept is at

13, 0 .

5

The function crosses the y-axis at the point when 0 :x

5 0 13 13 13.

3 30 3y

The y-intercept is at

130, .

3

The function has a vertical asymptote at the point where the denominator is

equal to 0, which occurs at 3.x

Since the degree of the polynomials in the numerator and the denominator

are equal, the horizontal asymptote can be found from the ratio of the leading

coefficients 5 .y

The graph that has all of these characteristics is A.

- iv -


November 2016

8. B.

6 4 3 2

2 2 2

13 42

2 2 2

3 2 3 8 4

2 2

6 4

2

3 2 8 4

2 3

11 6

2 3

10 3

2

2

log log 4 log

log log log

log log log

( )( )log

log

lo

) ( )

g

(

x y xy x y

xy

x y xy x y

x y x y

xy

x y

x

x y x y

y

x y

9. C.

The given polynomial function has:

3 zeros with a multiplicity of 1 (at

1

,0, and 12

x ) because the graph crosses the

x-axis.

1 zero with a multiplicity of 2 (at 2x )

because the graph is tangent to the x-axis.

The factors that give zeroes with a multiplicity of 1, and will appear once in

the factored form of the polynomial function, are 1

, , and ( 1);2

x x x

the

factor ( 2)x will appear twice since the zero at 2 has a multiplicity of 2.

Therefore the factored form of this polynomial function is

1( 2)( 2)( 1) .

2x x x x x

- v -


November 2016

10. B.

A relation is a function only if there is only one value of y in the range for

each value of x in the domain. A graph of a relation will be a function if it

passes the vertical line test. Options A and B both pass the vertical line test

and these relations are, therefore, functions.

A relation will have an inverse that is a function if there is only one value of x

in the domain for each value of y in the range. The graph of the original

function will pass the horizontal line test if the inverse is also a function.

Options B and D both pass the horizontal line test and the inverses of these

relations are, therefore, functions.

The only option that passes both the vertical and the horizontal line tests is

B.

11. C.

To find ( ( ))f g x substitute 1x for x in ( (( )) 2), 1 .f g xx xf

The domain of this function is | 1, x x x R and the range of the function

is | 0, .y y y R

The graph is a translation of the basic function y x with a vertical stretch

of 2 and a horizontal shift of 1 to the left. This corresponds to option C.

- vi -


November 2016

12. A.

The graph of the exponential function xy a passes from quadrant II into

quadrant I and is:

decreasing for 0 1a increasing for 1a

, 0 1xy a x

, 1xy a a

Each graph has a y-intercept of 1, a horizontal asymptote of 0,y and

depending on the value of ,a the function may be increasing, decreasing, or

constant (if 1).a Thus the only statement that is always true is B.

- vii -


November 2016

13. A.

6 6

6 6

2

2

log 5 log 2

log 5 log 36

5 36

5 36

5 36 0

9 4 0

9 and 4

x x

x x

x x

x x

x x

x x

x x

Check for extraneous roots by substituting each value into the original

equation:

6 6

6 6

6

6

check: 9

log (9 5) log (9) 2

log (4) log (9) 2

log (4 9) 2

log (36) 2

2 2

x

9x is a solution to this logarithmic equation.

Check: 4x

6 6

6 6

log ( 4 5) log ( 4) 2

log ( 9) log ( 4) 2

4x is extraneous because you cannot take the logarithm of a negative

number.

- viii -


November 2016

14. D.

The function 3( ) 3 2f x x x is a cubic function (an odd degree) with a

positive leading coefficient, and therefore will extend down into quadrant III

and up into quadrant I.

Factoring the function completely results in 2( ) ( 1) ( 2).f x x x The roots of

this equation are 1,x which has a multiplicity of 2, and 2,x which has

a multiplicity of 1. Therefore, the graph will touch, but not cross, the x-axis

at 1,x and will cross the x-axis at 2.x

15. D.

There is a value (3) being added to each individual x so the graph will shift 3

units to the left. The entire equation has 4 being subtracted from it, so the

graph will shift 4 units down.

16. B.

Using synthetic division, we have:

34x 29x 58x 15

↓ ↓ ↓ ↓

3 4 9 −58 −15

12 63 15

4 21 5 0

The quotient is: 24 21 5.x x

- ix -


November 2016

17. B.

Method One – With Technology

Graphing ( 8)

43

y xx

gives the graph shown below. The graph crosses

the x-axis twice (once at 5 and once at 4). Therefore, the

x-intercepts are ( 5,0) and (4,0).

Method Two – Algebraic Manipulation

Finding the x-intercepts can be done by setting the equation equal to zero

and solving for x.

2

2

( 8)4 0

3

( 8)( 3) ( 3) 4( 3) 0( 3)

( 3)

3 ( 8) 4 12 0

20 0

( 5)( 4) 0

5 and 4

xx

x x x x xx

x x x

x x

x x

x x

Checking for extraneous roots by substituting each value into the original

equation shows that both 5x and 4x are permissible. Therefore, the

x-intercepts are ( 5,0) and (4,0).

- x -


November 2016

18. A.

2

2

2

13 1

13 ( 1)

13 2 1

12 0

( 4)( 3) 0

4 3

x x

x x

x x x

x x

x x

x x

Therefore, 4 x is extraneous and the solution set is 3 .

Verification:

?

?

4 13 4 1

9 3

3 3

?

?

3 13 3 1

16 4

4 4

- xi -


November 2016

19. C.

Method One – With Technology

Graphing 2

2

2 10( )

3 10

x xf x

x x

gives the graph shown below.

The graph has a vertical asymptote at 2.x Checking the table of

values associated with this graph shows an error when 5.x Since this

does not appear as an asymptote, it must be a point of discontinuity (hole).

Simplifying the equation gives 2 ( 5)

( )( 2)( 5)

x xf x

x x

or

2( ) .

2

xf x

x

By

substituting 5x into this reduced equation, the coordinates of the point

of discontinuity (hole) can be found to be 10

5, .7

Method Two – Algebraic Manipulation

The equation 2

2

2 10( )

3 10

x xf x

x x

can be rewritten as

2 ( 5)( )

( 2)( 5)

x xf x

x x

or

2( ) .

2

xf x

x

Functions will have vertical asymptotes along lines where they are

undefined, which happens when there is a non-zero value being divided by

zero. For this equation, this occurs at 2x because ( 2)x is a factor of

the denominator but not the numerator

- xii -


November 2016

Holes or points of discontinuity occur when a point is indeterminate, which

occurs when there is a value of zero being divided by zero. For 2

2

2 10( )

3 10

x xf x

x x

there will be a hole when 5x because the numerator

and denominator have a common factor of ( 5).x By substituting 5x

into the reduced equation, the coordinates of the point of discontinuity

(hole) can be found to be 10

5, .7

20. D.

The volume of a box can be found by multiplying its length, width, and

height. The dimensions of the open box made by Julie are shown in the

diagram below:

2

2 3

3 2

( ) length width height

( ) (24 2 ) (24 2 )

( ) (576 96 4 )

( ) 576 96 4

( ) 4 96 576

V x

V x x x x

V x x x x

V x x x x

V x x x x

height x

length 24 2x

width 24 2x

- xiii -


November 2016

21. C.

An angle in standard position has its centre at the origin and its initial arm

along the positive x-axis. The terminal arm for a 100 angle is located in

Quadrant II.

22. D.

For 3

os1

5c , 5,x 13,r then:

2 2 2

2 2 2( 5) 13

169 25

12

x y r

y

y

y

For any angle in quadrant III, both the x- and y-coordinates will be negative.

Therefore, 12y and 12 12

tan .5 5

y

x

- xiv -


November 2016

23. B.

cos( ) cos cos sin sin

cos(15 ) cos(45 30 )

cos 45 cos 30 sin 45 sin 30

32 2 1

2 2 2 2

6 2

4 4

6 2

4

A B A B A B

24. D.

sintan

cos

cos 0

3This occurs at , , ...

2 2

1csc

sin

sin 0

This occurs at 0, 2 , ...

1sec

cos

cos 0

3This occurs at , , ...

2 2

Therefore the non-permissible values are ,2

n n I

- xv -


November 2016

25. D.

The cosecant function is the reciprocal of sine. Thus:

2 3 3 3 33csc sin

3 2(3) 22 3

The sine function is negative in quadrants III and IV. The reference angle for

3sin

2 is .

3

Therefore, the angles within the given domain that are

solutions to the equation 2 3

csc3

are:

and 23 3

4 5and

3 3

26. C.

Using verification, students can choose an angle to substitute into each

possible identity to determine which ones actually are identities. The table

below shows each option being verified using 15 .

?

2 2

?2 2

?

(1 cos ) csc 1

(1 cos 15 ) csc 1

(0.066987298)(14.9

5

28203) 1

1 1

1

?2

?2

?

(0.9659258)(0.2679491)(3.863703) (0.0669872)

0.0669872

cos tan csc sin

cos15 tan15 csc15 sin 15

1

?

2 2

?2 2

?

(0.0717968)(1.0352762) (0.066987298)

tan sec sin

tan 15 sec15 sin 15

0.07432949 0.066987298

?

2

?2

?

(1.07179667)(0.9659258263) 1

(1 tan ) cos 1

(1 tan 15 ) cos15 1

1.03527618 1

Verifying this numerically using 15 , the correct response is 2 2(1 cos ) csc 1 .

- xvi -


November 2016

27. B.

The phase shift (horizontal translation) is represented by the value of c in cos ( ) .y a b x c d

If 0,c the phase shift will be to the right.

If 0,c the phase shift will be to the left.

For 2 cos 3 2,6

y

the value of c is .6

Therefore, the phase shift

(horizontal translation) is 6

to the left.

28. C.

The angles that are coterminal with 6

can be represented by

, .6

2 n n N

Substitute values for n to determine the angles in the given

domain.

n 1 2 3

62 n

13

6

25

6

37

6

62 n

11

6

23

6

35

6

The angles in the domain 4 < 4 that are coterminal with 6

are

13,

6

11,

6

and

23.

6

- xvii -


November 2016

29. D.

The equation in question must have the following characteristics:

An amplitude of 4 means the possible values of a in

cos ( )y a b x c d are 4 and 4. Options C and D meet this criteria.

A phase shift (horizontal translation) that moves the graph of cosy

to the left by .2

This means that the value of c in cos ( )y a b c d

is

.2

B and D meet this criteria.

A vertical displacement of 2 units down. This means that the value of

d in cos ( )y a b c d is 2. Options C and D meet this criteria.

The only equation that meets all the criteria is D.

30. C.

5 cot 6 0

6cot

5

5tan

6

39.8

Since the period for the tangent function is 180 ,5

tan6

will also be true

for 140.2 ( 39.8 180 ). Therefore, the general solution to this equation is

140 180 , .n n I

- xviii -


November 2016

31. A.

2

2

2

2

cos 11 (cos )(cot )(csc ) 1 (cos )

sin sin

cos1

sin

1 cot

csc

xx x x x

x x

x

x

x

x

32. A.

22 sin sin 0

sin (2 sin 1) 0

sin 0 and 2 sin 1 0

10, and sin

2

5,

6 6

Therefore, the solution set is: 5

0, , ,6 6

.

- xix -


November 2016

33. A.

Andrew has a correct derivation using the Pythagorean Theorem.

Becky’s derivation is not correct as she attempted to use the distance

formula but made incorrect substitutions for the second point. To be

fully correct, the derivation is:

2 2

2 1 2 1

2 2

2 2

2 2

2 2 2

2 2

( ) ( )

1 ( 0) ( 0)

1 ( ) ( )

1

1

1

d x x y y

x y

x y

x y

x y

x y

Colin’s derivation is not fully correct as his circle has a radius of 2 and

is not a unit circle. Thus, his substitution for r is incorrect.

Darla has a correct derivation using the distance formula.

Andrew and Darla are the 2 students with the correct derivations.

- xx -


November 2016

34. D.

For a first-quadrant angle where 4

tan3

, 4 and 3.y x

For the point (6, ),b the x-value has been doubled. Since the ratio between x

and y must be constant, the y-value will also be doubled.

2(4) 8.y

35. C.

Decisions must be made about which of the 2 letters (R or S) to include, which

of the 5 possible spots to place the letter, and which digits (0 9 ) should fill

the other four spots.

Since repetition is allowed, the number of different Roughrider license plates is:

(2 10 10 10 10) (10 2 10 10 10) (10 10 2 10 10) (10 10 10 2 10)

(10 10 10 10 2)

20 000 20 000 20 000 20 000 20 000

100 000

OR

(2 5 10 10 10 10)

100 000

- xxi -


November 2016

36. A.

The thr term of the expansion of ( )na b is: ( 1) 1

1

n r rna b

r

For the binomial 7(3 4) ,x 3 , 4, and 7.a x b n The 2nd term is thus:

7 (2 1) 2 1

7 (1) 1

6 1

7 1

73 ( 4)

2 1

73 ( 4)

1

(3 ) ( 4)C

x

x

x

37. B

All 10 letters are being used taken 10 at a time, and there are no repetitions.

10 103 628 800

n nP P

38. C.

9 0 9 1 9 2 9 3 9 4

9 5 9 6 9 7 9 8 9 9

1 9 36 84 126

126 84 36 9 1

C C C C C

C C C C C

Therefore, 3r and 6r are both solutions to this equation.

- xxii -


November 2016

39. Numeric Response: 8

Using the graphs for ( )f x and ( )g x as shown:

( 3) 2f and then (2) 8.g

( ( 3)) 8.g f

y g(x)

y f (x)

= ( )y g x

( )y f x


Polynomial functions with an odd degree will extend up into quadrant II and

down into quadrant IV when the leading coefficient is negative, such as:

There is only 1 graph that meets this criteria.

- xxiii -


November 2016

41. Numeric Response: 38 – 39

0.75 m

2.25 m

2

0.6 radians

1800.6 radians

radians

38.197

a r

a

r


The radius of each inner tube will be equal to the amplitude of each curve.

maximum value minimum valueamplitude .

2

The path of the point on the large inner tube has an amplitude of (50 0)

25,2

which means the radius of the large inner tube is 25 inches.

The path of the point on the small inner tube has an amplitude of (35 15)

10,2

which means the radius of the small inner tube is 10 inches.

Therefore, the difference between the two radii is: 25 in 10 in 15 in.

- xxiv -


November 2016


20 000 35 000 (0.988)

20 000(0.988)

35 000

4(0.988)

7

4log log(0.988)

7

4log ( ) log(0.988)

7

4log

7

log(0.988)

46.3542781 months

t

t

t

t

t

t

t

Therefore, it will take between 46 and 47 months for a $35 000 car to

depreciate to $20 000.


4cos 1 3cos 4

7 cos θ 3

3cos

7

64.6

This is the reference angle in quadrant I. We need to find the corresponding

angle in quadrant IV to meet the restrictions given in domain 270 360 :

360 64.6 295.4


20 18 4 2

(190)(6)

1140

C C

- xxv -


November 2016

Question by Outcome

Question Outcome Question Outcome Question Outcome

1. PC30.7 16. PC30.10 31. PC30.5

2. PC30.9 17. PC30.11 32. PC30.4

3. PC30.7 18. PC30.11 33. PC30.2

4. PC30.6 19. PC30.11 34. PC30.2

5. PC30.8 20. PC30.10 35. PC30.12

6. PC30.8 21. PC30.1 36. PC30.13

7. PC30.11 22. PC30.2 37. PC30.12

8. PC30.9 23. PC30.5 38. PC30.13

9. PC30.10 24. PC30.5 39. PC30.6

10. PC30.8 25. PC30.2 40. PC30.10

11. PC30.6 26. PC30.5 41. PC30.1

12. PC30.9 27. PC30.3 42. PC30.3

13. PC30.9 28. PC30.1 43. PC30.9

14. PC30.10 29. PC30.3 44. PC30.4

15. PC30.7 30. PC30.4 45. PC30.13

Content Area Outcomes

Multiple-

choice

Questions

Numeric

Response

Questions

Algebra 6 − 11 1 – 20 39 −40, 43

Trigonometry 1 − 5 21 – 34 41 − 42, 44

Permutations, Combinations,

and the Binomial Theorem 12 − 13 35 − 38 45

grade 12 prototype examination pre-calculus...

Documents