Grade 12 Prototype ExaminationPre-calculus 30 Course Code 8426 Barcode Number Month Day Date of Birth 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 Numeric Response Questions This examination is being written in the Record your answer in the numeric subject response section on the answer sheet. A. Chemistry. What is 10% of $2000? (Round to the B. Pre-calculus. nearest dollar.) C. Workplace and Apprenticeship Mathematics. D. Foundations of Mathematics. 1. A B C D i Pre-calculus 30, Prototype Exam November 2016 What is 10% of $248.50? (Round to the What is 10% of 24 125? (Round to the nearest nearest dollar.) 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. ii Pre-calculus 30, Prototype Exam November 2016 Pre-calculus 30 Quadratic Formula b b2 4 ac For ax bx c 0 , x 2 2a Arc Length a r Trigonometry and Trigonometric Identities sin cos tan cot = cos sin 1 1 1 csc = sec cot sin cos tan sin2 cos 2 1 1 tan2 sec2 1 cot 2 csc 2 cos ( A + B) = cos A cos B sin A sin B tan A tan B tan( A B ) cos ( A B) = cos A cos B + sin A sin B 1 tan A tan B sin ( A + B) = sin A cos B + cos A sin B tan A tan B sin ( A B) = sin A cos B cos A sin B tan( A B ) 1 tan A tan B cos2 cos2 sin2 2 tan tan 2 sin 2 2sin cos cos2 2cos2 1 1 tan2 cos2 1 2sin2 - OVER - Ministry of Education 2016-17 Permutations, Combinations, and Binomial Theorem n! n! n Pr Cr C n (n r ) ! n (n r ) ! r ! n r r (a b)n n C0anb0 n C1an1b1 n C2an2b2 ... n Cna0bn n The general term, tk 1 , of the expansion of ( a b)n is: a n kbk OR n Ck ankbk k General Form of Transformed Functions y k af (b( x h)) y a sin b( x c) d y a cos b( x c) d iii Pre-calculus 30, Prototype Exam November 2016 Pre-calculus Insert Ministry of Education iv Pre-calculus 30, Prototype Exam November 2016 Table of Trigonometric Ratios sin cos tan csc sec cot sin cos tan csc sec cot 0 0.0000 1.0000 0.0000 1.0000 45 0.7071 0.7071 1.0000 1.4142 1.4142 1.0000 1 0.0175 0.9998 0.0175 57.298 1.0002 57.290 46 0.7193 0.6947 1.0355 1.3902 1.4396 0.9657 2 0.0349 0.9994 0.0349 28.653 1.0006 28.636 47 0.7314 0.6820 1.0724 1.3673 1.4663 0.9325 3 0.0523 0.9986 0.0524 19.107 1.0014 19.081 48 0.7431 0.6691 1.1106 1.3456 1.4945 0.9004 4 0.0698 0.9976 0.0699 14.335 1.0024 14.300 49 0.7547 0.6561 1.1504 1.3250 1.5243 0.8693 5 0.0872 0.9962 0.0875 11.473 1.0038 11.4301 50 0.7660 0.6428 1.1918 1.3054 1.5557 0.8391 6 0.1045 0.9945 0.1051 9.5668 1.0055 9.5144 51 0.7771 0.6293 1.2349 1.2868 1.5890 0.8098 7 0.1219 0.9925 0.1228 8.2055 1.0075 8.1444 52 0.7880 0.6157 1.2799 1.2690 1.6243 0.7813 8 0.1392 0.9903 0.1405 7.1853 1.0098 7.1154 53 0.7986 0.6018 1.3270 1.2521 1.6616 0.7536 9 0.1564 0.9877 0.1584 6.3925 1.0125 6.3138 54 0.8090 0.5878 1.3764 1.2361 1.7013 0.7265 10 0.1736 0.9848 0.1763 5.7588 1.0154 5.6713 55 0.8192 0.5736 1.4281 1.2208 1.7434 0.7002 11 0.1908 0.9816 0.1944 5.2408 1.0187 5.1446 56 0.8290 0.5592 1.4826 1.2062 1.7883 0.6745 12 0.2079 0.9781 0.2126 4.8097 1.0223 4.7046 57 0.8387 0.5446 1.5399 1.1924 1.8361 0.6494 13 0.2250 0.9744 0.2309 4.4454 1.0263 4.3315 58 0.8480 0.5299 1.6003 1.1792 1.8871 0.6249 14 0.2419 0.9703 0.2493 4.1336 1.0306 4.0108 59 0.8572 0.5150 1.6643 1.1666 1.9416 0.6009 15 0.2588 0.9659 0.2679 3.8637 1.0353 3.7321 60 0.8660 0.5000 1.7320 1.1547 2.0000 0.5774 16 0.2756 0.9613 0.2867 3.6280 1.0403 3.4874 61 0.8746 0.4848 1.8040 1.1434 2.0627 0.5543 17 0.2924 0.9563 0.3057 3.4203 1.0457 3.2709 62 0.8829 0.4695 1.8807 1.1326 2.1301 0.5317 18 0.3090 0.9511 0.3249 3.2361 1.0515 3.0777 63 0.8910 0.4540 1.9626 1.1223 2.2027 0.5095 19 0.3256 0.9455 0.3443 3.0716 1.0576 2.9042 64 0.8988 0.4384 2.0503 1.1126 2.2812 0.4877 20 0.3420 0.9397 0.3640 2.9238 1.0642 2.7475 65 0.9063 0.4226 2.1445 1.1034 2.3662 0.4663 21 0.3584 0.9336 0.3839 2.7904 1.0711 2.6051 66 0.9135 0.4067 2.2460 1.0946 2.4586 0.4452 22 0.3746 0.9272 0.4040 2.6695 1.0785 2.4751 67 0.9205 0.3907 2.3558 1.0864 2.5593 0.4245 23 0.3907 0.9205 0.4245 2.5593 1.0864 2.3559 68 0.9272 0.3746 2.4751 1.0785 2.6695 0.4040 24 0.4067 0.9135 0.4452 2.4586 1.0946 2.2460 69 0.9336 0.3584 2.6051 1.0711 2.7904 0.3839 25 0.4226 0.9063 0.4663 2.3662 1.1034 2.1445 70 0.9397 0.3420 2.7475 1.0642 2.9238 0.3640 26 0.4384 0.8988 0.4877 2.2812 1.1126 2.0503 71 0.9455 0.3256 2.9042 1.0576 3.0715 0.3443 27 0.4540 0.8910 0.5095 2.2027 1.1223 1.9626 72 0.9511 0.3090 3.0777 1.0515 3.2361 0.3249 28 0.4695 0.8829 0.5317 2.1301 1.1326 1.8807 73 0.9563 0.2924 3.2708 1.0457 3.4203 0.3057 29 0.4848 0.8746 0.5543 2.0627 1.1434 1.8040 74 0.9613 0.2756 3.4874 1.0403 3.6279 0.2867 30 0.5000 0.8660 0.5773 2.0000 1.1547 1.7321 75 0.9659 0.2588 3.7320 1.0353 3.8637 0.2680 31 0.5150 0.8572 0.6009 1.9416 1.1666 1.6643 76 0.9703 0.2419 4.0108 1.0306 4.1335 0.2493 32 0.5299 0.8480 0.6249 1.8871 1.1792 1.6003 77 0.9744 0.2250 4.3315 1.0263 4.4454 0.2309 33 0.5446 0.8387 0.6494 1.8361 1.1924 1.5399 78 0.9781 0.2079 4.7046 1.0223 4.8097 0.2126 34 0.5592 0.8290 0.6745 1.7883 1.2062 1.4826 79 0.9816 0.1908 5.1445 1.0187 5.2408 0.1944 35 0.5736 0.8192 0.7002 1.7434 1.2208 1.4281 80 0.9848 0.1736 5.6712 1.0154 5.7587 0.1763 36 0.5878 0.8090 0.7265 1.7013 1.2361 1.3764 81 0.9877 0.1564 6.3137 1.0125 6.3924 0.1584 37 0.6018 0.7986 0.7536 1.6616 1.2521 1.3270 82 0.9903 0.1392 7.1153 1.0098 7.1852 0.1405 38 0.6157 0.7880 0.7813 1.6243 1.2690 1.2799 83 0.9925 0.1219 8.1443 1.0075 8.2054 0.1228 39 0.6293 0.7771 0.8098 1.5890 1.2868 1.2349 84 0.9945 0.1045 9.5143 1.0055 9.5667 0.1051 40 0.6428 0.7660 0.8391 1.5557 1.3054 1.1918 85 0.9962 0.0872 11.429 1.0038 11.473 0.0875 41 0.6561 0.7547 0.8693 1.5243 1.3250 1.1504 86 0.9976 0.0698 14.300 1.0024 14.335 0.0699 42 0.6691 0.7431 0.9004 1.4945 1.3456 1.1106 87 0.9986 0.0523 19.080 1.0014 19.106 0.0524 43 0.6820 0.7314 0.9325 1.4663 1.3673 1.0724 88 0.9994 0.0349 28.635 1.0006 28.652 0.0349 44 0.6947 0.7193 0.9657 1.4396 1.3902 1.0355 89 0.9998 0.0175 57.285 1.0002 57.294 0.0175 90 1.0000 0.0000 1.0000 0.0000 v GRADE 12 DEPARTMENTAL EXAMINATION PRE-CALCULUS 30 PROTOTYPE NOVEMBER 2016 VALUE Answer the following 45 questions on the computer sheet entitled (45 2) “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 y 5x 2 ? A. ( x 2) log5 y B. log5 ( x 2) y C. log5 y x 2 D. log y ( x 2) 5 -1- Pre-calculus 30, Prototype Exam 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. g( x ) f ( x 1) 2 B. g( x ) f ( x 1) 2 C. g( x ) f ( x 2) 1 D. g( x ) f ( x 2) 1 4. Functions f ( x ) and g( x ) are defined by f ( x ) 2x 2 3x 2 and g( x ) 4x 2 3. What is the equation of function h( x ) if h( x ) ( g f )x ? A. h( x ) 6x 2 3x 5 B. h( x ) 6x 2 3x 1 C. h( x ) 6x 2 3x 5 D. h( x ) 6x 2 3x 1 -2- Pre-calculus 30, Prototype Exam 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 ). 1 6. What is the equation of the inverse relation of y 5 ( x 16)2 ? 2 A. y 16 2 ( x 5) B. y 16 2( x 5) C. y 4 2 ( x 5) D. y 4 2( x 5) -3- Pre-calculus 30, Prototype Exam November 2016 7. Which of the following fully represents the graph of the rational function 5x 13 f (x ) , including all asymptotes? x 3 A. B. C. D. -4- Pre-calculus 30, Prototype Exam November 2016 8. What is the single logarithm of log2 x 6 y4 log2 xy3 4 log2 x 2 y, written in simplest form? A. log 2 4 x 4 B. log2 x10 y3 C. log2 x 8 y4 D. log2 ( x 3 y2 xy3 x 8 y4 ) 9. A polynomial function is graphed below: What are the factors of this polynomial function? 1 A. x ( x 2)( x 1) x 2 1 B. x ( x 2)( x 1) x 2 1 C. x ( x 2)( x 2)( x 1) x 2 1 D. x ( x 2)( x 2)( x 1) x 2 -5- Pre-calculus 30, Prototype Exam November 2016 10. Four relations are graphed below. Which relation and its inverse are both functions? A. B. C. D. -6- Pre-calculus 30, Prototype Exam November 2016 11. Functions y f ( x ) and y g( x ) are defined by f ( x ) 2 x and g( x ) x 1. Which of the following best represents the graph of y f ( g( x )) ? A. B. C. D. -7- Pre-calculus 30, Prototype Exam November 2016 12. Which of the following statements is always TRUE about an exponential function in the form y a x , where a 0 ? A. The function has an x-intercept of 1. B. The range of the function is y y 0, y . C. The function has a vertical asymptote of x 0. D. The function increases in value as x increases. 13. What is the solution set of log6 ( x 5) log6 ( x ) 2 ? A. 9 B. 4 C. 4, 9 D. no solution -8- Pre-calculus 30, Prototype Exam November 2016 14. Which of the following represents the graph of f ( x ) x 3 3x 2 ? A. B. C. D. -9- Pre-calculus 30, Prototype Exam November 2016 15. How will the graph of y 2x 2 3x be transformed if it has been modified to be y 2( x 3)2 3( x 3) 4 ? 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 4x 3 9x 2 58x 15 by x 3 ? A. 4x 2 3x 49 B. 4x 2 21x 5 C. 4x 3 21x 2 5x D. 4x 3 3x 2 49x 132 ( 8) 17. What are the x-intercepts of f ( x ) x 4? x 3 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 x 13 x 1 ? A. 3 B. 4 C. 4, 3 D. 3, 4 - 10 - Pre-calculus 30, Prototype Exam November 2016 2x 2 10x 19. A function is defined by f ( x ) . On the graph of f ( x ), where x 2 3x 10 are the vertical asymptote(s) and the point of discontinuity (hole) located? A. vertical asymptote at x 5; no point of discontinuity 10 B. vertical asymptote at x 2; point of discontinuity at 5, 7 10 C. vertical asymptote at x 2; point of discontinuity at 5, 7 D. vertical asymptotes at x 2 and x 5; 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. V ( x ) x 3 144x B. V ( x ) 4 x 3 576x C. V ( x ) x 3 48x 2 576x D. V ( x ) 4x 3 96x 2 576x - 11 - Pre-calculus 30, Prototype Exam November 2016 21. Which of the following sketches shows an angle of 100° in standard position? A. B. C. D. - 12 - Pre-calculus 30, Prototype Exam November 2016 5 22. The terminal arm of angle is located in quadrant III. If cos , 13 what is the exact value of tan ? 12 A. 5 12 B. 13 12 C. 13 12 D. 5 23. What is the exact value of cos15 6 2 A. 4 6 2 B. 4 6 2 C. 2 6 2 D. 2 24. What are all of the non-permissible values for in tan csc sec ? A. 0, n 3 B. 0, , 2 2 C. n, n I D. n, n I 2 - 13 - Pre-calculus 30, Prototype Exam November 2016 2 3 25. What are the possible values of if csc , where 0 < 2 ? 3 11 A. , 6 6 2 B. , 3 3 5 7 C. , 6 6 4 5 D. , 3 3 26. Which of the following statements is TRUE for all permissible values of ? A. tan2 sec sin2 B. (1 tan2 ) cos 1 C. (1 cos2 ) csc2 1 D. cos tan csc sin2 27. What is the phase shift (horizontal translation) of y 2 cos 3 2 6 with respect to y cos ? A. 2 units to the left B. units to the left 6 C. 2 units to the right D. units to the right 6 - 14 - Pre-calculus 30, Prototype Exam November 2016 28. Which of the following shows all of the angles that are coterminal with , where 4 < 4 ? 6 11 A. 6 13 11 B. , 6 6 13 11 23 C. , , 6 6 6 25 13 11 23 D. , , , 6 6 6 6 29. The graph of y cos has been transformed so that the resultant graph has the following characteristics: amplitude is 4 phase shift (horizontal translation) is to the left 2 vertical displacement is 2 units down Which of the following equations would produce a graph that has all of the above characteristics? A. y 2 cos 3 4 2 B. y 2 cos 3 4 2 C. y 4 cos 3 2 2 D. y 4 cos 3 2 2 - 15 - Pre-calculus 30, Prototype Exam November 2016 30. To the nearest degree, what is the general solution for 5 cot 6 0 ? A. 130 180n, n I B. 130 360n, n I C. 140 180n, n I D. 140 360n, n I 31. What is 1 (cos x )(cot x )(csc x ) simplified? A. csc2 x B. cos2 x C. sec2 x D. sin2 x 32. What is the solution set of 2 sin2 sin 0, where 0 2 ? 5 A. 0, , , 6 6 11 B. 0, , , 6 6 4 C. 0, , , 3 3 5 D. 0, , , 3 3 - 16 - Pre-calculus 30, Prototype Exam 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: a 2 b2 c 2 ( x )2 y 2 r 2 Andrew x 2 y2 12 x 2 y2 1 d ( x 2 x1 )2 ( y2 y1 )2 1 ( x 0)2 (0 ( y ))2 Becky 1 ( x )2 (0 y)2 1 x 2 y2 12 x 2 y2 1 x 2 y2 a 2 b2 c 2 ( x )2 y 2 r 2 Colin x 2 y2 12 x 2 y2 1 d ( x 2 x1 )2 ( y2 y1 )2 1 ( x 0)2 ( y 0)2 Darla 1 ( x )2 y 2 12 x 2 y2 1 x 2 y2 - 17 - Pre-calculus 30, Prototype Exam 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 4 position. If tan and is a first-quadrant angle, what is the exact 3 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 - 18 - Pre-calculus 30, Prototype Exam November 2016 36. Which of the following represents the 2nd term in the expansion of (3x 4)7 ? 4 6 1 A. 7 C1 3x 3x 4 5 2 B. 7 C2 3x 4 2 5 C. 7 C2 3x 4 1 6 D. 7 C1 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 C10 B. 10 P10 C. 10 C1 D. 10 P1 38. What is the solution set for r given 9 Cr 84 ? A. 3 B. 4 C. 3, 6 D. 4, 5 - 19 - Pre-calculus 30, Prototype Exam 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 g( f ( 3)) ? g(x) y= g( x ) y ff(x) (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? - 20 - Pre-calculus 30, Prototype Exam 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.) - 21 - Pre-calculus 30, Prototype Exam 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 t amount is 20 000 35 000 0.988 , 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? - 22 - Pre-calculus 30, Prototype Exam 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 a, b, h, and 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: logb x y by x Therefore, y 5x 2 log5 y ( x 2) -i- Pre-calculus 30, Prototype Exam Answer Key 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 y f ( x 1). To translate a function 2 units down, the entire equation must have a value of 2 subtracted from it. This means that y f ( x 1) will become y f ( x 1) 2 or g( x ) f ( x 1) 2 4. D. h( x ) ( g f )x g( x ) f ( x ) (4 x 2 3) ( 2x 2 3x 2) 4 x 2 3 2x 2 3x 2 6 x 2 3x 1 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. - ii - Pre-calculus 30, Prototype Exam Answer Key November 2016 6. B. 1 To find the inverse equation of y 5 ( x 16)2 , exchange x and y. The 2 equation of the inverse relation can then be found by solving for y. 1 x 5 ( y 16)2 2 2( x 5) ( y 16)2 2( x 5) y 16 16 2( x 5) y y 16 2( x 5) 7. A. 5x 13 The graph of f ( x ) crosses the x-axis at the point when y 0 : x 3 5x 13 13 13 0 ; 5x 13 0; x . The x-intercept is at , 0 . x 3 5 5 The function crosses the y-axis at the point when x 0 : y 5 0 13 13 13 13 . The y-intercept is at 0, . 03 3 3 3 The function has a vertical asymptote at the point where the denominator is equal to 0, which occurs at x 3. 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 y 5 . The graph that has all of these characteristics is A. - iii - Pre-calculus 30, Prototype Exam Answer Key November 2016 8. B. log 2 x 6 y4 log 2 xy3 4 log 2 x 2 y 1 log 2 ( x y ) log 2 xy3 log 2 ( x 2 y )4 6 4 2 log 2 x 3 y2 log 2 xy3 log 2 x 8 y4 ( x 3 y2 )( x 8 y4 ) log 2 xy3 x11 y6 log 2 xy3 log 2 x 10 y3 9. C. The given polynomial function has: 3 zeros with a multiplicity of 1 (at 1 x , 0, and 1 ) because the graph crosses the 2 x-axis. 1 zero with a multiplicity of 2 (at x 2 ) because the graph is tangent to the x-axis. The factors that give zeroes with a multiplicity of 1, and will appear once in 1 the factored form of the polynomial function, are x , x , and ( x 1); the 2 factor ( x 2) will appear twice since the zero at 2 has a multiplicity of 2. Therefore the factored form of this polynomial function is 1 x ( x 2)( x 2)( x 1) x . 2 - iv - Pre-calculus 30, Prototype Exam Answer Key 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 x 1 for x in f ( x ), f ( g( x )) 2 x 1 . The domain of this function is x|x 1, x R and the range of the function is y|y 0, 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. -v- Pre-calculus 30, Prototype Exam Answer Key November 2016 12. A. The graph of the exponential function y a x passes from quadrant II into quadrant I and is: decreasing for 0 a 1 increasing for a 1 y ax , 0 x 1 y ax , a 1 Each graph has a y-intercept of 1, a horizontal asymptote of y 0, and depending on the value of a, the function may be increasing, decreasing, or constant (if a 1). Thus the only statement that is always true is B. - vi - Pre-calculus 30, Prototype Exam Answer Key November 2016 13. A. log 6 x 5 log 6 x 2 log 6 x 5 x log 6 36 x 5 x 36 x 2 5x 36 x 2 5x 36 0 x 9x 4 0 x 9 and x 4 Check for extraneous roots by substituting each value into the original equation: check: x 9 log 6 (9 5) log 6 (9) 2 log 6 (4) log 6 (9) 2 log 6 (4 9) 2 log 6 (36) 2 22 x 9 is a solution to this logarithmic equation. Check: x 4 log 6 ( 4 5) log 6 ( 4) 2 log 6 ( 9) log 6 ( 4) 2 x 4 is extraneous because you cannot take the logarithm of a negative number. - vii - Pre-calculus 30, Prototype Exam Answer Key November 2016 14. D. The function f ( x ) x 3 3x 2 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 f ( x ) ( x 1)2 ( x 2). The roots of this equation are x 1, which has a multiplicity of 2, and x 2, which has a multiplicity of 1. Therefore, the graph will touch, but not cross, the x-axis at x 1, and will cross the x-axis at x 2. 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: 4x 3 9x 2 58x 15 ↓ ↓ ↓ ↓ 3 4 9 −58 −15 12 63 15 4 21 5 0 The quotient is: 4x 2 21x 5. - viii - Pre-calculus 30, Prototype Exam Answer Key November 2016 17. B. Method One – With Technology ( 8) Graphing y x 4 gives the graph shown below. The graph crosses x 3 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. ( 8) x 4 0 x 3 ( 8) x ( x 3) ( x 3) 4( x 3) 0( x 3) ( x 3) x 2 3x ( 8) 4 x 12 0 x 2 x 20 0 ( x 5)( x 4) 0 x 5 and x 4 Checking for extraneous roots by substituting each value into the original equation shows that both x 5 and x 4 are permissible. Therefore, the x-intercepts are ( 5, 0) and (4, 0). - ix - Pre-calculus 30, Prototype Exam Answer Key November 2016 18. A. x 13 x 1 Verification: x 13 ( x 1)2 ? ? x 13 x 2 2x 1 4 13 4 1 3 13 3 1 ? ? x 2 x 12 0 9 3 16 4 ( x 4)( x 3) 0 3 3 44 x 4 x 3 Therefore, x 4 is extraneous and the solution set is 3 . -x- Pre-calculus 30, Prototype Exam Answer Key November 2016 19. C. Method One – With Technology 2x 2 10x Graphing f ( x ) 2 gives the graph shown below. x 3x 10 The graph has a vertical asymptote at x 2. Checking the table of values associated with this graph shows an error when x 5. Since this does not appear as an asymptote, it must be a point of discontinuity (hole). 2x ( x 5) 2x Simplifying the equation gives f ( x ) or f ( x ) . By ( x 2)( x 5) x 2 substituting x 5 into this reduced equation, the coordinates of the point 10 of discontinuity (hole) can be found to be 5, . 7 Method Two – Algebraic Manipulation 2x 2 10x 2x ( x 5) The equation f ( x ) 2 can be rewritten as f ( x ) or x 3x 10 ( x 2)( x 5) 2x f (x ) . x 2 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 x 2 because ( x 2) is a factor of the denominator but not the numerator - xi - Pre-calculus 30, Prototype Exam Answer Key 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 2x 2 10x f (x ) 2 there will be a hole when x 5 because the numerator x 3x 10 and denominator have a common factor of ( x 5). By substituting x 5 into the reduced equation, the coordinates of the point of discontinuity 10 (hole) can be found to be 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: width 24 2x height x length 24 2x V ( x ) length width height V ( x ) (24 2x ) (24 2x ) x V ( x ) (576 96x 4 x 2 ) x V ( x ) 576x 96x 2 4 x 3 V ( x ) 4 x 3 96x 2 576x - xii - Pre-calculus 30, Prototype Exam Answer Key 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. 5 For cos , x 5, r 13, then: 13 x 2 y2 r 2 ( 5)2 y2 132 y 169 25 y 12 For any angle in quadrant III, both the x- and y-coordinates will be negative. y 12 12 Therefore, y 12 and tan . x 5 5 - xiii - Pre-calculus 30, Prototype Exam Answer Key November 2016 23. B. cos( A B ) cos A cos B sin A sin B cos(15) cos(45 30) cos 45 cos 30 sin 45 sin 30 2 3 2 1 2 2 2 2 6 2 4 4 6 2 4 24. D. sin tan cos cos 0 3 This occurs at , , ... 2 2 1 csc sin sin 0 This occurs at 0, 2 , ... 1 sec cos cos 0 3 This occurs at , , ... 2 2 Therefore the non-permissible values are n, n I 2 - xiv - Pre-calculus 30, Prototype Exam Answer Key November 2016 25. D. The cosecant function is the reciprocal of sine. Thus: 2 3 3 3 3 3 csc sin 3 2 3 2(3) 2 The sine function is negative in quadrants III and IV. The reference angle for 3 sin is . Therefore, the angles within the given domain that are 2 3 2 3 solutions to the equation csc are: 3 and 2 3 3 4 5 and 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. ? ? (1 cos2 ) csc2 1 cos tan csc sin 2 ? ? (1 cos2 15) csc2 15 1 cos15 tan 15 csc 15 sin 2 15 ? ? (0.066987298)(14.928203) 1 (0.9659258)(0.2679491)(3.863703) (0.0669872) 1 1 1 0.0669872 ? ? tan 2 sec sin 2 (1 tan 2 ) cos 1 ? ? tan 2 15 sec 15 sin 2 15 (1 tan2 15) cos15 1 ? ? (0.0717968)(1.0352762) (0.066987298) 1 (1.07179667)(0.9659258263) 0.07432949 0.066987298 1.03527618 1 Verifying this numerically using 15 , the correct response is (1 cos2 ) csc2 1 . - xv - Pre-calculus 30, Prototype Exam Answer Key November 2016 27. B. The phase shift (horizontal translation) is represented by the value of c in y a cos b( x c) d. If c 0, the phase shift will be to the right. If c 0, the phase shift will be to the left. For y 2 cos 3 2, the value of c is . Therefore, the phase shift 6 6 (horizontal translation) is to the left. 6 28. C. The angles that are coterminal with can be represented by 6 2 n, n N . Substitute values for n to determine the angles in the given 6 domain. n 1 2 3 13 25 37 2 n 6 6 6 6 11 23 35 2 n 6 6 6 6 The angles in the domain 4 < 4 that are coterminal with are 6 13 11 23 , , and . 6 6 6 - xvi - Pre-calculus 30, Prototype Exam Answer Key November 2016 29. D. The equation in question must have the following characteristics: An amplitude of 4 means the possible values of a in y a cos 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 y cos to the left by . This means that the value of c in y a cos b( c) d 2 is . B and D meet this criteria. 2 A vertical displacement of 2 units down. This means that the value of d in y a cos 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 6 cot 5 5 tan 6 39.8 5 Since the period for the tangent function is 180, tan will also be true 6 for 140.2 ( 39.8 180). Therefore, the general solution to this equation is 140 180n, n I . - xvii - Pre-calculus 30, Prototype Exam Answer Key November 2016 31. A. cos x 1 1 (cos x )(cot x )(csc x ) 1 (cos x ) sin x sin x cos2 x 1 sin 2 x 2 1 cot x csc2 x 32. A. 2 sin 2 sin 0 sin (2 sin 1) 0 sin 0 and 2 sin 1 0 1 0, and sin 2 5 , 6 6 5 Therefore, the solution set is: 0, , , . 6 6 - xviii - Pre-calculus 30, Prototype Exam Answer Key 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: d ( x 2 x1 )2 ( y2 y1 )2 1 ( x 0)2 ( y 0)2 1 ( x )2 ( y ) 2 1 x 2 y2 12 x 2 y2 1 x 2 y2 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. - xix - Pre-calculus 30, Prototype Exam Answer Key November 2016 34. D. 4 For a first-quadrant angle where tan , y 4 and x 3. 3 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. y 2(4) 8. 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 - xx - Pre-calculus 30, Prototype Exam Answer Key November 2016 36. A. n n ( r 1) r 1 The rth term of the expansion of ( a b)n is: a b r 1 For the binomial (3x 4)7 , a 3x , b 4, and n 7. The 2nd term is thus: 7 7(21) 3x ( 4)21 2 1 7 3x 7(1) ( 4)1 1 7C1 (3x )6 ( 4)1 37. B All 10 letters are being used taken 10 at a time, and there are no repetitions. n Pn 10 P10 3 628 800 38. C. 9 C0 1 C1 9 9 9C2 36 9 C3 84 9C4 126 9 C5 126 9C6 84 9 C7 36 C8 9 9 9 C9 1 Therefore, r 3 and r 6 are both solutions to this equation. - xxi - Pre-calculus 30, Prototype Exam Answer Key November 2016 39. Numeric Response: 8 Using the graphs for f ( x ) and g( x ) as shown: f ( 3) 2 and then g(2) 8. g( f ( 3)) 8. g(x) y= g( x ) y ff(x) (x ) 40. Numeric Response: 1 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. - xxii - Pre-calculus 30, Prototype Exam Answer Key November 2016 41. Numeric Response: 38 – 39 a r a r 0.75 m 2.25 m 2 0.6 radians 180 0.6 radians radians 38.197 42. Numeric Response: 15 The radius of each inner tube will be equal to the amplitude of each curve. maximum value minimum value amplitude . 2 (50 0) The path of the point on the large inner tube has an amplitude of 25, 2 which means the radius of the large inner tube is 25 inches. (35 15) The path of the point on the small inner tube has an amplitude of 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. - xxiii - Pre-calculus 30, Prototype Exam Answer Key November 2016 43. Numeric Response: 46 – 47 20 000 35 000 (0.988)t 20 000 (0.988)t 35 000 4 (0.988)t 7 4 log log(0.988)t 7 4 log (t ) log(0.988) 7 4 log t 7 log(0.988) t 46.3542781 months Therefore, it will take between 46 and 47 months for a $35 000 car to depreciate to $20 000. 44. Numeric Response: 295 – 296 4 cos 1 3cos 4 7 cos θ 3 3 cos 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 45. Numeric Response: 1140 20C18 4C2 (190)(6) 1140 - xxiv - Pre-calculus 30, Prototype Exam Answer Key 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 Multiple- Numeric Content Area Outcomes choice Response Questions Questions Algebra 6 − 11 1 – 20 39 −40, 43 Trigonometry 1−5 21 – 34 41 − 42, 44 Permutations, Combinations, 12 − 13 35 − 38 45 and the Binomial Theorem - xxv - Pre-calculus 30, Prototype Exam Answer Key November 2016