AP Calc BC 2019

20 Please note: Some of the questions in this former practice exam may no longer perfectly align with the AP exam. Even though these questions do not fully represent the 2020 exam, teachers indicate that imperfectly aligned questions still provide instructional value. 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At a Glance Total Time 1 hour and 45 minutes Number of Questions 45 Percent of Total Score 50% Writing Instrument Pencil required Part A Number of Questions 30 Instructions Section I of this exam contains 45 multiple-choice questions and 4 survey questions. For Part A, fill in only the circles for numbers 1 through 30 on the answer sheet. For Part B, fill in only the circles for numbers 76 through 90 on the answer sheet. Because Part A and Part B offer only four answer options for each question, do not mark the (E) answer circle for any question. The survey questions are numbers 91 through 94. Indicate all of your answers to the multiple-choice questions on the answer sheet. No credit will be given for anything written in this exam booklet, but you may use the booklet for notes or scratch work. After you have decided which of the suggested answers is best, completely fill in the corresponding circle on the answer sheet. Give only one answer to each question. If you change an answer, be sure that the previous mark is erased completely. Here is a sample question and answer. Time 1 hour Electronic Device None allowed Part B Number of Questions 15 Time 45 minutes Electronic Device Graphing calculator required Use your time effectively, working as quickly as you can without losing accuracy. Do not spend too much time on any one question. Go on to other questions and come back to the ones you have not answered if you have time. It is not expected that everyone will know the answers to all of the multiple-choice questions. Your total score on the multiple-choice section is based only on the number of questions answered correctly. Points are not deducted for incorrect answers or unanswered questions. Form I Form Code 4PBP4-S 68 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA CALCULUS BC SECTION I, Part A Time—1 hour Number of questions—30 NO CALCULATOR IS ALLOWED FOR THIS PART OF THE EXAM. Directions: Solve each of the following problems, using the available space for scratch work. After examining the form of the choices, decide which is the best of the choices given and fill in the corresponding circle on the answer sheet. No credit will be given for anything written in this exam booklet. Do not spend too much time on any one problem. In this exam: (1) Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f (x ) is a real number. (2) The inverse of a trigonometric function f may be indicated using the inverse function notation f − 1 or with the prefix “arc” (e.g., sin−1 x = arcsin x). -3- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ∫1 (4x 2 1. (A) 27 2 3 ) − x dx = (B) 27 (C) 36 (D) 57 2. Let f be the function defined by f (x ) = x 3 − 3x 2 − 9x + 11. At which of the following values of x does f attain a local minimum? (A) 3 (B) 1 (C) −1 (D) −3 -4- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA d 2(sin x )2 = dx ( 3.  1 (A) 4 cos   2 x )    (B) 4 sin x cos x (C) 2 sin x x (D) ( 2 sin x cos x x ) 4. The position of a particle is given by the parametric equations x(t ) = ln t 2 + 1 and y(t ) = e3−t . What is the velocity vector at time t = 1 ? (A) 1, e2 (B) 1, −e2 (C) 1 2 ,e 2 -5- (D) 1 , −e2 2 GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ∞ en ∑ n is n=1 p 5. (A) p p−e (B) e p−e (C) x f (x ) 2 4 e p ln (D) divergent () p e f ¢(x ) f ≤(x ) g(x ) −3 −2 3 g ¢(x ) g ≤(x ) 5 1 6. The table above gives values of the twice-differentiable functions f and g and their derivatives at x = 2. If h is f ¢(x ) the function defined by h(x ) = , what is the value of h¢(2) ? g(x ) (A) 9 4 (B) 3 5 (C) − 3 2 (D) − 21 4 -6- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ( ) 7. Which of the following is the Maclaurin series for x cos x 2 ? 8. (A) x − x5 x9 x13 + − +L 2! 4! 6! (B) x − x3 x5 x7 + − +L 2! 4! 6! (C) x 3 − x 7 x11 x15 + − +L 3! 5! 7! (D) x 3 − x5 x7 x9 + − +L 3! 5! 7! lim x→∞ (A) −2 10 − 6x 2 is 5 + 3e x (B) 0 (C) 2 (D) nonexistent -7- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 9. The function f is not differentiable at x = 5. Which of the following statements must be true? (A) f is not continuous at x = 5. (B) lim f (x ) does not exist. x→5 (C) lim x→5 (D) f (x ) − f (5) does not exist. x−5 5 ∫0 f (x) dx does not exist. 10. The second derivative of a function f is given by f ≤(x ) = x(x − 3)5 (x − 10)2. At which of the following values of x does the graph of f have a point of inflection? (A) 3 only (B) 0 and 3 only (C) 3 and 10 only (D) 0, 3, and 10 -8- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA p 11. ⌠ ex − 1 dx =  x ⌡0 e − x (A) e p − p − 1 (B) ln(e p − p) − 1 (D) ln(e p − p) (C) p − ln p x 0 4 8 12 16 f (x ) 8 0 2 10 1 12. The table above gives selected values for the differentiable function f. In which of the following intervals must there be a number c such that f ¢(c) = 2 ? (A) (0, 4) (B) (4, 8) (C) (8, 12) (D) (12, 16) -9- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA dy = x + 2y with initial condition f (0) = 2. What is dx the approximation for f (−0.4) obtained by using Euler’s method with two steps of equal length starting 13. Let y = f (x ) be the solution to the differential equation at x = 0 ? (A) 0.76 (B) 1.20 (C) 1.29 (D) 3.96 14. What is the slope of the line tangent to the curve (A) −3 (B) − 1 3 (C) 1 (D) x + 9 1  y = 2 at the point  ,  ?  4 4  4 3 -10- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ∞ ⌠ 6  dx is  3 2 ⌡1 (x + 3) / 15. (A) 16. If 3 4 (B) 3 (C) 6 (D) divergent dy = 2 − y , and if y = 1 when x = 1, then y = dx (A) 2 − e x−1 (B) 2 − e1−x (C) 2 − e−x -11- (D) 2 + e−x GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 17. Which of the following series converges? (A) ∞ n=1 (B) ∞ (D)  n + 1  2n  ∑ (−1)n  n=1 (C) 1 − n   n  ∑ (−1)n  ∞     ∞ 2 n   n   2 n n (−1)  ∑  3 n n=1 ∑ (−1)n  n=1 -12- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 18. Let R be the region in the first and second quadrants between the graphs of the polar curves f (q ) = 3 + 3 cos q and g(q ) = 4 + 2 cos q , as shaded in the figure above. Which of the following integral expressions gives the area of R ? 6 (A) ∫− 2 ( g(q ) − f (q)) dq (B) ∫0 ( g(q ) − f (q)) dq (C) 1 2 ∫0 ( g(q ) − f (q)) (D) 1 2 ∫0 (( g(q)) p p p 2 2 dq ) − ( f (q ))2 dq -13- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 19. Which of the following statements about the series ∞ (− 1)n n n=1 n2 + 1 ∑ (A) The series can be shown to diverge by comparison with is true? ∞ 1 ∑ n. n=1 (B) The series can be shown to diverge by limit comparison with ∞ 1 ∑ n. n=1 (C) The series can be shown to converge by comparison with ∞ ∑ 1 2 n=1 n . (D) The series can be shown to converge by the alternating series test. -14- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 20. If 4 ∫1 f (x ) dx = 8 and 4 ∫1 g(x ) dx = −2, which of the following cannot be determined from the information given? (A) (B) (C) (D) 1 ∫4 g(x ) dx 4 ∫1 3f (x) dx 4 ∫1 3 f (x) g(x) dx 4 ∫1 (3f (x) + g(x)) dx -15- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA ∞ 21. Which of the following is the interval of convergence for the series (A) [−9, 1) (B) [−5, 5) (C) [1, 9) ∑ (x + 4)n n=1 n ◊ 5n+1 ? (D) (−∞, ∞) 22. What is the slope of the line tangent to the polar curve r = 3q at the point where q = (A) − p 2 (B) − 2 p (C) 0 p ? 2 (D) 3 -16- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 23. The definite integral 4 ∫0 x dx is approximated by a left Riemann sum, a right Riemann sum, and a trapezoidal sum, each with 4 subintervals of equal width. If L is the value of the left Riemann sum, R is the value of the right Riemann sum, and T is the value of the trapezoidal sum, which of the following inequalities is true? (A) L < 4 ∫0 (B) L < T < (C) R < 4 ∫0 (D) R < T < x dx < T < R 4 ∫0 x dx < R x dx < T < L 4 ∫0 x dx < L -17- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 24. The graph of the piecewise linear function f is shown above. What is the value of (A) −8 (B) −6 (C) 0 12 ∫0 f ¢(x ) dx ? (D) 22 -18- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 25. The function f has a continuous derivative. If f (0) = 1, f (2) = 5, and (A) 3 (B) 6 (C) 10 2 ∫0 f (x ) dx = 7, what is 2 ∫0 x ◊ f ¢(x) dx ? (D) 17 26. Which of the following series are conditionally convergent? ∞ I. (− 1)n ∑ n=1 n! II. (− 1)n n n=1 III. (− 1)n n=1 n + 2 ∞ ∑ ∞ ∑ (A) I only (B) II only (C) II and III only -19- (D) I, II, and III GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 27. Let R be the region in the first quadrant bounded by the graph of y = x − 1 , the x-axis, and the vertical line x = 10. Which of the following integrals gives the volume of the solid generated by revolving R about the y-axis? 10 (A) p ∫1 (B) p ∫1 10 3 (x − 1) dx (100 − (x − 1)) dx ( 2 )) ⌠ (C) p  10 − y 2 + 1 ⌡0 3 ⌠ (D) p  ⌡0 (  100 − y 2 + 1  ( dy 2 )  dy -20- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 28. If dy d 2y dx = 5 and = sin t 2 , then 2 is dt dt dx () () (A) 2 t cos t 2 (B) () 2 t cos t 2 5 (C) () 2t cos t 2 25 -21- (D) undefined GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 29. Which of the following expressions is equal to lim n →∞ 1 1 1 1 1   + + +L+ n  ? 1 2 3 n  2 +  2 + 2 + 2 +  n  n n n 2 ⌠ 1 dx (A)  ⌡1 x 1 ⌠ 1 dx (B)  ⌡0 2 + x 2 1 ⌠ dx (C)  ⌡0 2 + x 3 1 ⌠ dx (D)  ⌡2 2 + x -22- GO ON TO THE NEXT PAGE. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 1 3 x + L, and the Maclaurin series converges to 3 f (x ) for all real numbers x. If g is the function defined by g(x ) = e f (x), what is the coefficient of x 2 in the 30. A function f has a Maclaurin series given by 2 + 3x + x 2 + Maclaurin series for g ? (A) 1 2 e 2 (B) e2 (C) 5 2 e 2 (D) 11 2 e 2 END OF PART A IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON PART A ONLY. DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO. -23- B B B B B B B B B CALCULUS BC SECTION I, Part B Time—45 minutes Number of questions—15 A GRAPHING CALCULATOR IS REQUIRED FOR SOME QUESTIONS ON THIS PART OF THE EXAM. Directions: Solve each of the following problems, using the available space for scratch work. After examining the form of the choices, decide which is the best of the choices given and fill in the corresponding circle on the answer sheet. No credit will be given for anything written in this exam booklet. Do not spend too much time on any one problem. BE SURE YOU FILL IN THE CIRCLES ON THE ANSWER SHEET THAT CORRESPOND TO QUESTIONS NUMBERED 76–90. YOU MAY NOT RETURN TO QUESTIONS NUMBERED 1–30. In this exam: (1) The exact numerical value of the correct answer does not always appear among the choices given. When this happens, select from among the choices the number that best approximates the exact numerical value. (2) Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f (x ) is a real number. (3) The inverse of a trigonometric function f may be indicated using the inverse function notation f − 1 or with the prefix “arc” (e.g., sin−1 x = arcsin x). -26- GO ON TO THE NEXT PAGE. B B B B B B B B B 76. The function f is continuous on the closed interval [0, 5]. The graph of f ¢, the derivative of f, is shown above. On which of the following intervals is f increasing? (A) [0, 1] and [2, 4] (B) [0, 1] and [3, 5] (C) [0, 1] and [4, 5] only (D) [0, 2] and [4, 5] 77. If 2 dy = 6e−0.08 (t − 5) , by how much does y change as t changes from t = 1 to t = 6 ? dt (A) 3.870 (B) 8.341 (C) 18.017 (D) 22.583 -27- GO ON TO THE NEXT PAGE. B B B B B B B B B 78. The function f has the property that lim− f (x ) = +∞ , lim+ f (x ) = −∞ , lim f (x ) = 2, and lim f (x ) = 2 . x→1 x→1 x→−∞ x→+∞ Of the following, which could be the graph of f ? (A) (B) (C) (D) -28- GO ON TO THE NEXT PAGE. B B B B B B B B B 79. Tara’s heart rate during a workout is modeled by the differentiable function h, where h(t ) is measured in beats per minute and t is measured in minutes from the start of the workout. Which of the following expressions gives Tara’s average heart rate from t = 30 to t = 60 ? 60 (A) ∫30 h(t ) dt (B) 1 30 ∫30 h(t) dt (C) 1 30 ∫30 h¢(t) dt (D) h¢(30) + h¢(60) 2 60 60 -29- GO ON TO THE NEXT PAGE. B B B B B B B B B 80. The function f has derivatives of all orders for all real numbers with f (2) = −1, f ¢(2) = 4 , f ≤(2) = 6, and f ¢¢¢(2) = 12 . Using the third-degree Taylor polynomial for f about x = 2, what is the approximation of f (2.1) ? (A) −0.570 (B) −0.568 (C) −0.566 81. Let f be a function with derivative given by f ¢(x ) = (D) −0.528 x 3 + 1 . What is the length of the graph of y = f (x) from x = 0 to x = 1.5 ? (A) 4.266 (B) 2.497 (C) 2.278 (D) 1.976 -30- GO ON TO THE NEXT PAGE. B B B B B B B B B 82. Let h be a continuous function of x. Which of the following could be a slope field for a differential equation of dy = h(x ) ? the form dx (A) (B) (C) (D) -31- GO ON TO THE NEXT PAGE. B B B B B B B B B   k 3 + x for x < 3   f (x ) =   16  for x ≥ 3  2  k − x  83. Let f be the function defined above, where k is a positive constant. For what value of k, if any, is f continuous? (A) 2.081 (B) 2.646 (C) 8.550 (D) There is no such value of k. 84. Let f be a differentiable function. The figure above shows the graph of the line tangent to the graph of f at x = 0 . Of the following, which must be true? (A) f ¢(0) = − f (0) (B) f ¢(0) < f (0) (C) f ¢(0) = f (0) (D) f ¢(0) > f (0) -32- GO ON TO THE NEXT PAGE. B B B B B B B B B 85. A referee moves along a straight path on the side of an athletic field. The velocity of the referee is given by v(t ) = 4(t − 6) cos(2t + 5), where t is measured in minutes and v(t) is measured in meters per minute. What is the total distance traveled by the referee, in meters, from time t = 2 to time t = 6 ? (A) 3.933 (B) 14.578 (C) 21.667 (D) 29.156 -33- GO ON TO THE NEXT PAGE. B B B B B B B B B 86. The graph of the function f shown above consists of three line segments. Let h be the function defined by h(x ) = x ∫0 f (t) dt . At what value of x does h attain its absolute maximum on the interval [−4, 3] ? (A) −4 (B) −2 (C) 0 (D) 3 -34- GO ON TO THE NEXT PAGE. B B B B B B B B B 87. The position of a particle moving in the xy-plane is given by the parametric equations x(t ) = e− t and y(t ) = sin(4t) for time t ≥ 0. What is the speed of the particle at time t = 1.2 ? (A) 1.162 (B) 1.041 (C) 0.462 (D) 0.221 1 4 2 3 1 2 1 x − x + x − x . For how many values of x in the open 4 3 2 2 interval (0, 1.565) is the instantaneous rate of change of f equal to the average rate of change of f on the closed 88. Let f be the function defined by f (x ) = interval [0, 1.565] ? (A) Zero (B) One (C) Three (D) Four -35- GO ON TO THE NEXT PAGE. B B B B B B B B B 89. The population P of rabbits on a small island grows at a rate that is jointly proportional to the size of the rabbit population and the difference between the rabbit population and the carrying capacity of the population. If the carrying capacity of the population is 2400 rabbits, which of the following differential equations best models the growth rate of the rabbit population with respect to time t, where k is a constant? (A) dP = 2400 − kP dt (B) dP = k(2400 − P) dt (C) dP 1 = k (2400 − P) dt P (D) dP = kP(2400 − P) dt 90. A region is bounded by two concentric circles, as shown by the shaded region in the figure above. The radius of the outer circle, R, is increasing at a constant rate of 2 inches per second. The radius of the inner circle, r, is decreasing at a constant rate of 1 inch per second. What is the rate of change, in square inches per second, of the area of the region at the instant when R is 4 inches and r is 3 inches? (A) 3p (B) 6p (C) 10p (D) 22p -36- GO ON TO THE NEXT PAGE. B B B B B B B B END OF SECTION I IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON PART B ONLY. DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO. _______________________________________________________ MAKE SURE YOU HAVE DONE THE FOLLOWING. • PLACED YOUR AP NUMBER LABEL ON YOUR ANSWER SHEET • WRITTEN AND GRIDDED YOUR AP NUMBER CORRECTLY ON YOUR ANSWER SHEET • TAKEN THE AP EXAM LABEL FROM THE FRONT OF THIS BOOKLET AND PLACED IT ON YOUR ANSWER SHEET -37- B AP Calculus BC Exam ® 2019 SECTION II: Free Response DO NOT OPEN THIS BOOKLET OR BREAK THE SEALS ON PART B UNTIL YOU ARE TOLD TO DO SO. At a Glance Total Time 1 hour and 30 minutes Number of Questions 6 Percent of Total Score 50% Writing Instrument Either pencil or pen with black or dark blue ink Weight The questions are weighted equally, but the parts of a question are not necessarily given equal weight. Part A Instructions Number of Questions The questions for Section II are printed in this booklet. Do not break the seals on Part B until you are told to do so. Write your solution to each part of each question in the space provided. Write clearly and legibly. Cross out any errors you make; erased or crossed-out work will not be scored. 2 Time 30 minutes Electronic Device Graphing calculator required Percent of Section II Score 33.33% Part B Number of Questions 4 Time 1 hour Electronic Device None allowed Percent of Section II Score 66.67% Manage your time carefully. During Part A, work only on the questions in Part A. You are permitted to use your calculator to solve an equation, find the derivative of a function at a point, or calculate the value of a definite integral. However, you must clearly indicate the setup of your question, namely the equation, function, or integral you are using. If you use other built-in features or programs, you must show the mathematical steps necessary to produce your results. During Part B, you may continue to work on the questions in Part A without the use of a calculator. As you begin each part, you may wish to look over the questions before starting to work on them. It is not expected that everyone will be able to complete all parts of all questions. • Show all of your work, even though a question may not explicitly remind you to do so. Clearly label any functions, graphs, tables, or other objects that you use. Justifications require that you give mathematical reasons, and that you verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit. • Your work must be expressed in standard mathematical notation rather than calculator syntax. For example, 5 1 x2 dx may not be written as fnInt(X2, X, 1, 5). • Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If you use decimal approximations in calculations, your work will be scored on accuracy. Unless otherwise specified, your final answers should be accurate to three places after the decimal point. • Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f x is a real number. Form I Form Code 4PBP4-S 68 CALCULUS BC SECTION II, Part A Time—30 minutes Number of questions—2 A GRAPHING CALCULATOR IS REQUIRED FOR THESE QUESTIONS. -3- GO ON TO THE NEXT PAGE. 1 1 1 1 1 1 1 1 1 1 1. The rate at which cars enter a parking lot is modeled by E (t ) = 30 + 5(t − 2)(t − 5)e− 0.2t . The rate at which cars leave the parking lot is modeled by the differentiable function L. Selected values of L(t ) are given in the table above. Both E (t ) and L(t ) are measured in cars per hour, and time t is measured in hours after 5 A.M. (t = 0). Both functions are defined for 0 £ t £ 12. (a) What is the rate of change of E (t ) at time t = 7 ? Indicate units of measure. Do not write beyond this border. Do not write beyond this border. (b) How many cars enter the parking lot from time t = 0 to time t = 12 ? Give your answer to the nearest whole number. -4- Continue question 1 on page 5. 1 1 1 1 1 1 1 1 1 1 (c) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approximate 12 L(t ) dt . Using correct units, explain the meaning of 12 ∫2 L(t ) dt in the context of this problem. (d) For 0 £ t < 6 , 5 dollars are collected from each car entering the parking lot. For 6 £ t £ 12, 8 dollars are collected from each car entering the parking lot. How many dollars are collected from the cars entering the parking lot from time t = 0 to time t = 12 ? Give your answer to the nearest whole dollar. -5- GO ON TO THE NEXT PAGE. Do not write beyond this border. Do not write beyond this border. ∫2 2 2 2 2 2 2 2 2 2 2 2. A laser is a device that produces a beam of light. A design, shown above, is etched onto a flat piece of metal Both x and y are measured in centimeters, and t is measured in seconds. The laser starts at position (0, 0) at time t = 0, and the design takes 3.1 seconds to complete. For 0 £ t £ 3.1, dy = 4 cos(2.5t). dt dx = 3 cos t 2 and dt ( ) (a) Find the speed of the laser at time t = 3 seconds. (b) Find the total distance traveled by the laser from time t = 1 to time t = 3 seconds. -6- Continue question 2 on page 7. Do not write beyond this border. Do not write beyond this border. using a moving laser. The position of the laser at time t seconds is represented by (x(t), y(t )) in the xy-plane. 2 2 2 2 2 2 2 2 2 2 (d) What is the difference between the y-coordinates of the laser’s highest position and lowest position for 0 £ t £ 3.1 ? Justify your answer. -7- GO ON TO THE NEXT PAGE. Do not write beyond this border. Do not write beyond this border. (c) The laser is farthest to the right at time t = 1.253 seconds. Find the x-coordinate of the laser’s rightmost position. END OF PART A IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON PART A ONLY. DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO. -8- CALCULUS BC SECTION II, Part B Time—1 hour Number of questions—4 NO CALCULATOR IS ALLOWED FOR THESE QUESTIONS. DO NOT BREAK THE SEALS UNTIL YOU ARE TOLD TO DO SO. -13- 3 3 3 3 3 3 3 3 3 3 NO CALCULATOR ALLOWED    9 − x2   f (x ) =  px  −x + 3 cos   2   3. Let f be the function defined above. ( ) for −3 £ x £ 0 for 0 < x £ 4 Do not write beyond this border. Do not write beyond this border. (a) Find the average rate of change of f on the interval − 3 £ x £ 4. (b) Write an equation for the line tangent to the graph of f at x = 3. -14- Continue question 3 on page 15. 3 3 3 3 3 3 3 3 3 3 NO CALCULATOR ALLOWED (d) Must there be a value of x at which f (x ) attains an absolute maximum on the closed interval − 3 £ x £ 4 ? Justify your answer. -15- GO ON TO THE NEXT PAGE. Do not write beyond this border. Do not write beyond this border. (c) Find the average value of f on the interval − 3 £ x £ 4. 4 4 4 4 4 4 4 4 4 4 4. The continuous function f is defined for −4 £ x £ 4. The graph of f, shown above, consists of two line 1 1 5 segments and portions of three parabolas. The graph has horizontal tangents at x = − , x = , and x = . 2 2 2 It is known that f (x ) = −x 2 + 5x − 4 for 1 £ x £ 4 . The areas of regions A and B bounded by the graph of f and the x-axis are 3 and 5, respectively. Let g be the function defined by g(x ) = x ∫− 4 f (t) dt . (a) Find g(0) and g(4). -16- Continue question 4 on page 17. Do not write beyond this border. Do not write beyond this border. NO CALCULATOR ALLOWED 4 4 4 4 4 4 4 4 4 4 NO CALCULATOR ALLOWED (c) Find all intervals on which the graph of g is concave down. Give a reason for your answer. -17- GO ON TO THE NEXT PAGE. Do not write beyond this border. Do not write beyond this border. (b) Find the absolute minimum value of g on the closed interval [−4, 4]. Justify your answer. 5 5 5 5 5 5 5 5 5 5 NO CALCULATOR ALLOWED (a) Show that dy −4x = . dx 3(y + 1) (b) Using the information from part (a), find Do not write beyond this border. Do not write beyond this border. 5. The graph of the curve C, given by 4x 2 + 3y 2 + 6y = 9, is shown in the figure above. d 2y dx 2 in terms of x and y. -18- Continue question 5 on page 19. 5 5 5 5 5 5 5 5 5 5 NO CALCULATOR ALLOWED (c) In polar coordinates, the curve C is given by r = 3 dr for 0 £ q £ 2p . Find . 2 + sin q dq As q increases, on what intervals is the distance between the origin and the point (r, q) increasing? (d) Let S be the region inside curve C, as defined in part (c), but outside the curve r = 2 . Write, but do not evaluate, an integral expression for the area of S. -19- GO ON TO THE NEXT PAGE. Do not write beyond this border. Do not write beyond this border. Give a reason for your answer. 6 6 6 6 6 6 6 6 6 6 NO CALCULATOR ALLOWED ∞ (−1)n+1(x − 3)n 6. Consider the series ∑ , where p is a constant and p > 0. 5n n p n=1 ◊ (b) For p = 1 and x = 8, does the series converge absolutely, converge conditionally, or diverge? Explain your reasoning. -20- Continue question 6 on page 21. Do not write beyond this border. Do not write beyond this border. (a) For p = 3 and x = 8, does the series converge absolutely, converge conditionally, or diverge? Explain your reasoning. 6 6 6 6 6 6 6 6 6 6 NO CALCULATOR ALLOWED (d) When p = 1 and x = 3.1 , the series converges to a value S. Use the first two terms of the series to approximate S. Use the alternating series error bound to show that this approximation differs from S by less than 1 . 300,000 -21- GO ON TO THE NEXT PAGE. Do not write beyond this border. Do not write beyond this border. (c) When x = −2, for what values of p does the series converge? Explain your reasoning. STOP END OF EXAM ____________________________________ THE FOLLOWING INSTRUCTIONS APPLY TO THE COVERS OF THE SECTION II BOOKLET. • MAKE SURE YOU HAVE COMPLETED THE IDENTIFICATION INFORMATION AS REQUESTED ON THE FRONT AND BACK COVERS OF THE SECTION II BOOKLET. • CHECK TO SEE THAT YOUR AP NUMBER LABEL APPEARS IN THE BOX ON THE FRONT COVER. • MAKE SURE YOU HAVE USED THE SAME SET OF AP NUMBER LABELS ON ALL AP EXAMS YOU HAVE TAKEN THIS YEAR. -22- I P 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q , 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q 0 S RF F R R 4 − Topic FUN-6.B The Fundamental Theorem of Calculus and Definite Integrals ) = − =( − )− ( − )= − = Incorrect. This response would result if the antidifferentiation was not done and the endpoints were substituted directly into the integrand, as follows. − =( − )−( − )= − = Incorrect. This response would result if the integrand was differentiated rather than antidifferentiated, as follows. − (D) Learning Objective Correct. This question involves using the basic power rule for antidifferentiation and correctly substituting the endpoints and evaluating, as follows. ∫( (C) R Q 1.E (B) F R Skill (A) R =( − )−( − )= − = Incorrect. This response would result if the powers of were not divided by the new exponent when the power rule for antiderivatives was applied, as follows. − =( − )−( − ) = − = 4 Q Skill 3.D (A) Learning Objective Topic FUN-4.A Using the First Derivative Test to Find Relative (Local) Extrema Correct. A local minimum for will occur where the derivative changes sign from negative to positive. ′( ) = − − = ( − − )= ′ ( + )( − ) The zeros of ′ are at = − and = A sign chart shows that ′ changes from negative to positive at = Another way to see this is to observe that the graph of ′ is a parabola opening up so that the graph crosses the -axis from negative to positive at the larger zero, = (B) Incorrect. This is a zero of the second derivative ′′( ) = − = ( − ) where ′′ changes sign from negative to positive. Therefore, = is a point of inflection of the graph of not the location of a local minimum. This response might also have been chosen if the derivative had been incorrectly factored as ( − )( + ) Then = is a zero of this expression where the sign changes from negative to positive. (C) Incorrect. This is a zero of the derivative ′ where ′ changes sign from positive to negative. Therefore, = − is the location of a local maximum of not a local minimum. (D) Incorrect. This response might have been chosen if the derivative had been incorrectly factored as ( − )( + ) and the zero where the derivative changes sign from positive to negative was selected. 4 Q Skill Learning Objective Topic 1.E FUN-3.C The Chain Rule (A) Incorrect. This response might come from incorrectly applying the chain rule twice as ( (B) ( ( ) = ⋅( ⋅ ( ) )= ( ⋅ ⋅ ( ) )= ( ) ) )= ⋅ ( )⋅ ( )= ⋅ ( )⋅ Incorrect. This response might come from using the chain rule only once, with the innermost “inside” function, as follows. ( (D) ) Incorrect. This response might come from correctly applying the chain rule the first time but not the second, as follows. ( (C) ( ( ( ) ) ) = ′( ′( ) ) as follows. ( ) )= ⋅ ( )⋅ ( )= ⋅ ( )⋅ Correct. The chain rule must be used twice for this composition of three functions. ( ( ) )= = ⋅ ( )⋅ = ⋅ ( )⋅ = ⋅ ( ( )⋅ ⋅ ⋅ ( ( )) ( ) ) 4 Q Skill 1.E (A) (B) Learning Objective Topic FUN-8.B (BC ONLY) Solving Motion Problems Using Parametric and Vector-Valued Functions Incorrect. This response would result if the chain rule was not used during the differentiation of the ( ) component of the position vector resulting in the loss of the negative sign, as follows. ()= ( + ) − = ()= ( + ) − = ()= ( + ) − = ()= ( + ) − = − ⇒ ( )= + Correct. The components of the velocity vector are the derivatives of the components of the position vector. − (C) ⇒ ( )= − + Incorrect. This response would result if the chain rule was not used during the differentiation of both components of the position vector, as follows. (D) − ⇒ ( )= + Incorrect. This response would result if the chain rule was not used during the differentiation of the ( ) component of the position vector, as follows. − + − − ⇒ ( )= − 4 Q Skill Learning Objective Topic 3.D LIM-7.A (BC ONLY) Working with Geometric Series (A) Incorrect. This response might be chosen if the series is correctly identified as a convergent geometric series with common ratio = in (B) π but the first term was taken to be − = − = π π π − = converges to π = − Since this ratio is less than π − = π = π and the series π − Incorrect. The sum of the series might have been taken to be the value of the definite integral used in performing the integral test for ( )= convergence. Let ∫ ∞ ( ) =⌠  ⌡ ∞ (π ) (π ) = ( ) ⌠  →∞ ⌡ π ( ) ( ) ( ) (D) resulting π Correct. The series is a geometric series with first term common ratio (C) rather than = →∞ (π ) (π )    π  π = − π = − π = =  →∞  π  − π   π π π π   Incorrect. This response might be chosen if the series is correctly ( ) ( ) identified as a geometric series with common ratio ratio is thought to be greater than ( ) = π but the 4 Q Skill Learning Objective Topic 1.E FUN-3.B The Quotient Rule (A) Correct. The derivative of is found by using the quotient rule. ′′( ) ( ) − ′( ) ′( ) ′( ) = ( ( )) The values for the functions and derivatives at = are obtained from the table. ′′( ) ( ) − ′( ) ′( ) ( )( − ) − ( − )( ) − + ′( ) = = = = (− ) ( ( )) (B) Incorrect. This response would result if the derivative of a quotient was taken to be the quotient of the derivatives, as follows. ′′( ) ′′( ) ′( ) = ⇒ ′( ) = = ′( ) ′( ) (C) Incorrect. This response comes from only differentiating the numerator in the quotient, as follows. ′′( ) ′′( ) ′( ) = ⇒ ′( ) = = =− − ( ) ( ) (D) Incorrect. This response would result if the terms in the numerator were added rather than subtracted in the quotient rule, as follows. ′′( ) ( ) + ′( ) ′( ) ( )( − ) + ( − )( ) − − ′( ) = ⇒ ′( ) = = =− (− ) ( ( )) 4 Q Skill Learning Objective Topic 2.C LIM-8.G (BC ONLY) Representing Functions as Power Series (A) Correct. Beginning with the Maclaurin series for ( ) obtain the Maclaurin series for is substituted to This series is then multiplied term- by-term by = − ( )= ( )= (B) + − − ( )   −  + + − ( ) +L = −  + L =  − −  =  −  − + + − +L − +L − + + ( ) =( )−  + L =  − ( )=    − ( ) − − + − +L was +L + ( ) + − − ( ) +L =  + L =  − − + + − − − = + − +L   −  + −  + L =  − + − +L +L Incorrect. This response would result if the Maclaurin series for used, and the series was multiplied by = was +L Incorrect. This response would result if the Maclaurin series for used instead of the series for = (D) ( ) + Incorrect. This response would result if the Maclaurin series for used, but there was no substitution of before multiplication by = − (C) +L +L was 4 Q Skill 2.B (A) Learning Objective Topic LIM-2.D Connecting Limits at Infinity and Horizontal Asymptotes Incorrect. This response might come from treating the problem like the limit of a rational function as goes to infinity when the numerator and denominator are polynomials of the same degree. If only the coefficients of the term and the term are considered, it might be thought that the limit would be (B) Correct. The numerator of − + − =− is a translated power function and the denominator is a translated exponential function. Since the exponential function grows faster than the power function the relative magnitude of the denominator compared to the numerator will result in this expression converging to as goes to infinity. (C) Incorrect. This response might come from treating the problem like the limit of a rational function as goes to If only the constant terms are considered, it might be thought that the limit would be = (D) Incorrect. It might be thought that the limit is nonexistent since the numerator goes to −∞ and the denominator goes to +∞ as goes to infinity, but this does not take into account the relative magnitude of the exponential function in the denominator compared to the quadratic term in the numerator as gets larger. 4 Q Skill 3.B Learning Objective Topic CHA-2.B Defining Average and Instantaneous Rates of Change at a Point (A) Incorrect. This statement could be false. A function can be continuous at a point without it being differentiable at that point. For example, ( ) = − is continuous at = but not differentiable there. (B) Incorrect. This statement could be false. For example, if then is not differentiable at = but ( )= − ( )= → (C) (D) Correct. The expression ( )− ( ) is one form of the − definition of the derivative of at = Since is not differentiable at = however, this limit does not exist. → Incorrect. This statement could be false. The definite integral can be found if is continuous even if is not differentiable at one of the endpoints. For example, if ( ) = − , then is not differentiable at = but ∫ ( ) = ( )( ) = 4 Q Skill 2.B Learning Objective Topic FUN-4.A Determining Concavity of Functions over Their Domains (A) Incorrect. This response would result if the change of sign of the second derivative ′′ at = is detected, but the change in sign at = is overlooked. (B) Correct. A point of inflection occurs where the second derivative ′′ changes sign. The zeros of ′′( ) occur at = = and = Constructing a sign chart with the three factors ( − ) and ( − ) shows that ′′( ) is positive for < negative for < < positive for < < and positive for > Therefore, the graph of has a point of inflection only at = and = where ′′( ) changes from positive to negative and then from negative to positive, respectively. (C) Incorrect. The second derivative ′′ is positive both to the left and to the right of = so there is no sign change in ′′( ) at = Therefore, the graph of does not have a point of inflection at = There is a sign change in ′′( ) from positive to negative at = so the graph of does have a point of inflection at = in addition to the one at = (D) Incorrect. These are the three zeros of the second derivative ′′( ) only changes sign at = and = ′′ but 4 Q Skill Learning Objective Topic 1.E FUN-6.D Integrating Using Substitution (A) Incorrect. This response would result if the substitution an attempt to evaluate the definite integral. = − ⇒ = − ⇒ When = π = π −π Only the substitutions for − however, and not for ⌠  ⌡ (B) π − − = ∫ π −π − ⇒ = =π When = = − = and for the limits of integration were made, as follows. π −π = ⌠  ⌡ π − − − ⇒ π = − Substituting for = ( π − π) − When − = = − = −π for π −π =⌠ = and for the limits of integration gives π −π = ⌡ ( π − π) − = (π = ) −π − Incorrect. This response would result if the integrand was incorrectly simplified − as or ⌠  ⌡ (D) When − was used in − Incorrect. This response would result if the substitution = − was correctly used in an attempt to evaluate the definite integral, but the evaluation used = rather than = = (C) = = and then either the antiderivative was only evaluated at the upper limit was taken to equal π − −  =⌠ π ⌡ as follows. (− ) π =( − ) =π − π Correct. This integral can be evaluated by using substitution of variables with = − = − ⇒ = =π When ⌠  ⌡ = − − ( π − π) π = − Substituting for π − ⇒ =⌠ ⌡ π −π = When − = = − = −π for and for the limits of integration gives = π −π = ( π −π) − = (π ) −π − 4 Q Skill Learning Objective Topic 3.D FUN-1.B Using the Mean Value Theorem (A) Incorrect. This interval might be chosen because of an error in computing the average rate of change over the interval as ( )− ( ) − = = rather than ( )− ( ) − = − − =− (B) Incorrect. This interval might be chosen because of an error in computing the average rate of change over the interval as ( )− ( ) − − = = = = rather than − − ( )− ( ) (C) Correct. The function is continuous on the closed interval [ ] and differentiable on the open interval ( ) By the Mean Value Theorem, there is a number in the interval ( ) such that ( )− ( ) − ′( ) = = = − Incorrect. This response would result if the Intermediate Value Theorem was used instead of the Mean Value Theorem to select the open interval ( ) where ( ) = for some number in the interval. (D) 4 Q Skill Learning Objective Topic 1.E FUN-7.C (BC ONLY) Approximating Solutions Using Euler's Method (A) Correct. Euler’s method for the differential equation written as + = ( )= + can be is an ) Δ , where Δ is the step size. Then − − = − 0.2, since there ) Here the step size is Δ = ( + approximation for = ( are two steps of equal length. = = − =− = + (B) = = ( + ) Δ = 2 + ( 0 + 2 ( 2 ) ) ( −0.2 ) = 2 + ( 4 )( −0.2 ) = 1.2 ) Δ = 1.2 + ( −0.2 + 2 (1.2 ) ) ( −0.2 ) = 1.2 + ( 2.2 )( −0.2 ) = 0.76 ( Incorrect. This response would result if only the first step was done using Euler’s method for the differential equation = = − (C) ( )= + = = ( + ) Δ = 2 + ( 0 + 2 ( 2 ) ) ( −0.2 ) = 2 + ( 4 )( −0.2 ) = 1.2 Incorrect. This response would result if the step size was taken to be Δ = −0.1 rather than Δ = −0.2. As a result, this response was the approximation for ( − ) rather than for ( − ) = = − =− = + (D) = = = ( ( + ) Δ = 2 + ( 0 + 2 ( 2 ) ) ( −0.1) = 2 + ( 4 )( − 0.1) = 1.6 ) Δ = 1.6 + ( −0.1 + 2 (1.6 ) ) ( − 0.1) = 1.6 + ( 3.1)( − 0.1) = 1.29 Incorrect. This response would result if the step size was taken to be Δ = 0.2 rather than Δ = −0.2. As a result, this response was the approximation for ( ) rather than for ( − ) = = = = + = = + ( ) Δ = 2.8 + ( 0.2 + 2 ( 2.8 ) ) ( 0.2 ) = 2.8 + ( 5.8 )( 0.2 ) = 3.96 ( ) Δ = 2 + ( 0 + 2 ( 2 ) ) ( 0.2 ) = 2 + ( 4 )( 0.2 ) = 2.8 4 Q Skill Learning Objective Topic 1.E FUN-3.D Implicit Differentiation (A) Incorrect. This response would result if the equation obtained by implicit differentiation was solved incorrectly for + ( ) = =− ⇒ =− =− =− as follows. It would also be obtained if the expression for had been correctly found but the - and -values were reversed when substituting into the first derivative, as follows. + ( ) (B) = =− ⇒ =− =− =− Correct. The slope of the tangent line is the value of at the point ( ) During the implicit differentiation to find both the power rule and the chain rule are needed. + The point = ( ) is on the curve, since equation + + () () = = = and = satisfy the At this point, ⇒ + = ⇒ =− (C) Incorrect. This response might come from observing the symmetry of the expressions for and in the equation of the curve and concluding that the derivative expressions will be symmetric, leading to a slope of without considering that the - and -coordinates of the given point are not the same. (D) Incorrect. This response would result if the method of implicit differentiation was incorrectly applied by taking the derivative of the expression without consideration of the chain rule and + setting the result equal to = + ⇒ as follows. ( ) = + = + = 4 Q Skill Learning Objective Topic 1.E LIM-6.A (BC ONLY) Evaluating Improper Integrals (A) Incorrect. This response would result if the integrand was just evaluated at the lower limit, = rather than antidifferentiated, as follows. = ( + ) (B) (C) (D) = = = Incorrect. This response would result if in applying the power rule for antiderivatives, the power of ( + ) was not divided by the new exponent, as follows.   ∞ ⌠ ⌠   = = −   →∞  →∞  ⌡ ( + ) ⌡ ( + )  ( + )      − = + = = + →∞    ( + )  Correct. This is an improper integral, since the region over which the integrand is being integrated is unbounded. The antidifferentiation is an application of the power rule.   ∞ ⌠ ⌠   = = −   →∞  →∞  ⌡ ( + ) ⌡ ( + )  ( + )      − = + = + = →∞   + )  (  Incorrect. This response might come from assuming that the definite integral of a function integrated over an unbounded region will always diverge. 4 Q Skill 1.E (A) Topic FUN-7.D Finding General Solutions Using Separation of Variables Incorrect. This response would result if a chain rule error was made during the antidifferentiation of the term. = − ⇒ ⌠  ⌡ − + − Since − ⇒ ⇒ = − − = − at the initial value > − − = or ⇒ ⌠  ⌡ − = − = − − Since = the solution would be − − ∫ + ⇒− ⇒ = − = − − − ⇒ > = + = − − =− + ⇒ − = − at the initial value = the solution to the − or = − differential equation is = − − Incorrect. This response would result if an arbitrary constant was not included during the antidifferentiation. = − ⇒ ⌠  ⌡ − = Since − (D) + Correct. The differential equation can be solved using separation of variables and the initial condition to determine the appropriate value for the arbitrary constant. = (C) = = − − ⇒ − = = − ∫ = = (B) Learning Objective − ∫ ⇒− − = at the initial value > − = = − or ⇒ = − = − the solution would be − = − Incorrect. This response would result if an arbitrary constant was not included during the antidifferentiation and the incorrect sign was taken for the absolute value when solving for = − ⌠  ⌡ − − ⇒ = =− − = − ∫ ⇒− ⇒ = − + − = ⇒ − = − 4 Q Skill 3.D (A) Learning Objective Topic LIM-7.A (BC ONLY) Alternating Series Test for Convergence Incorrect. This series diverges by the − = ≠ (− ) − = term test since Incorrect. This series diverges by the + = ≠ (− ) + = term test since Incorrect. This series diverges by the term test since →∞ (B) →∞ →∞ (C) →∞ (D) →∞ (− ) = Correct. The series →∞ ∞ ∑ ( − )  = does not exist. =   ∞ ∑ (− ) = three conditions: (1) the series is alternating, (2) ≥ →∞ = and < for all + Therefore, the series converges by the alternating series test. (3) the terms = ( ) satisfies the are decreasing since 4 Q Skill 1.C (A) π ( ) − ( )) ( ( )− = ∫− ( ( ) − ( )) ( )) = ∫ π ( ( ) − ( )) Incorrect. The square of the difference between the two polar curves was integrated rather than the difference of the squares. ∫ (D) CHA-5.D (BC ONLY) Finding the Area of the Region Bounded by Two Polar Curves Incorrect. This response comes from using a rectangular form of the area instead of a polar form for area, as if the curves were treated as functions of in terms of but expressing the limits in terms of the polar angle θ ∫ (C) Topic Incorrect. This response comes from using a rectangular form for area instead of a polar form for area, as if the curves were treated as functions of in terms of The region appears to lie above the interval [ − ] on the -axis. ∫− ( (B) Learning Objective π ( (θ ) − (θ ) ) ∫ θ = π ( (θ ) − (θ ) ) θ Correct. The area bounded by the two polar curves can be found with a definite integral. Let be the smaller radius, and let be the larger radius. The graphs of the two polar curves bound the region the domain ≤ θ ≤ π Then the area of is given by ∫ = π ( (θ ) ) ∫ (( π θ − ∫ π ( (θ ) ) (θ ) ) − ( (θ ) ) ) θ θ = ∫ (( π (θ ) ) − ( (θ ) ) over ) θ 4 Q Skill 3.B (A) Learning Objective Topic LIM-7.A (BC ONLY) Alternating Series Test for Convergence Incorrect. The series + ∞ ∑= is the divergent harmonic series. However, so this inequality goes the wrong way to use the < comparison test to show that the series ∞ ∑ addition, the divergence of the positive series that the series (B) ∞ ∑= + = that the series ∞ ∑= →∞ = + ∞ ∑= + the limit comparison test shows also diverges. However, it cannot be divergence of the positive series (C) ∞ ∑ = does not imply = is the divergent harmonic series. Since concluded from that the alternating series series ∞ ∑ diverges. Incorrect. The series →∞ diverges. In + = ∞ ∑ = ∞ − ( ) ∑= + diverges. The does not imply that the diverges. Incorrect. The series ∞ ∑ is a convergent = -series with for all > so this inequality + goes the wrong way to use the comparison test to show that the series = > ∞ ∑ + convergent. = However, < converges and that therefore ∞ ∑ = (− ) + is absolutely 4 (D) Q Q Q Correct. The series ∞ ∑= (− ) series is alternating, (2) + →∞ satisfies the three conditions: (1) the and (3) the terms = + = are decreasing. Therefore, the series converges by the + alternating series test. = To verify that the terms following. + < ⇔ ⇔ ⇔ + < + < ( + ) + + + < + + + + + ≥ The last inequality is true for all Alternatively, if ( )= is decreasing because for > are decreasing, consider the + then ′( ) = ( = )− ( + ) + ( ) and the function = ( − + ) < 4 Q Skill Learning Objective Topic 1.C FUN-6.A Applying Properties of Definite Integrals (A) Incorrect. The value of this integral can be determined using the properties of the definite integral, as follows. ∫ (B) = ⋅∫ ( ) = − (− ) = ( ) = ⋅ = Correct. It is not true in general that ∫ ( ) ( ) ∫ values of ∫ ( )=− ∫ and ∫ ∫ ∫ and ∫ then ⋅∫ ( ) ∫ ( ) ( ) ( ) ( ) ( )= = ( ) the value of (D) ( ) Incorrect. The value of this integral can be determined using the properties of the definite integral, as follows. ∫ (C) = −∫ ( ) ( ) ( ) ( ) so the individual cannot be used to determine For example, if ∫ = =⌠ − ( ) =− ( − ) and ( ) = − ( − ) then ( ) ( ) ( ) and and However, if =− ⌡ ( )= ∫ ( ) = = − as before, but now =⌠ − ( − ) =− ⌡ Incorrect. The value of this integral can be determined using the properties of the definite integral, as follows. ∫ = ( ( ) + ( )) ⋅∫ ( ) + = ∫ ∫ ( ) ( ) = + ⋅ ∫ ( ) + (− ) = 4 Q Skill Learning Objective Topic 3.D LIM-8.D (BC ONLY) Radius and Interval of Convergence of Power Series (A) Correct. The ratio test can be used to determine the interval of convergence. + →∞ ⇒ + = < + ( + ) →∞ ( + ) ⋅ ⇒− < ⋅ + ⋅ + ( + ) =− At = the series is ∞ (− ) ∑= ∞ ∑ (− = ⋅ + = multiple of the alternating harmonic series. the series is ∞ ∑= + ⋅ = ∞ ∑= + = + < ) which converges since it is a , which diverges since it is a multiple of the harmonic series. Therefore, the interval of convergence is − ≤ (B) ⋅ + →∞ < Now check the endpoints of the interval. At = < Incorrect. The ratio test might have been used to correctly determine that the radius of convergence is but the interval was taken to be centered at + →∞ At =− At = = →∞ + ( + )⋅ the series is the series is ∞ ∑= (− ) + ⋅ ∞ ∑= ⋅ ⋅ + ⋅ + = →∞ + ⋅ = ⇒ = which converges by the alternating series test. which diverges by the term test. (C) Incorrect. This response would result if the ratio test was not used to determine the radius of convergence. Looking at the form of the general term, the center was thought to be at = and the radius of convergence was taken to be It was assumed that the series must converge at one endpoint and diverge at the other. (D) Incorrect. Errors in applying the ratio test might have led to the conclusion that the radius of convergence was infinite. 4 Q Skill 1.E (A) Learning Objective Topic FUN-3.G (BC ONLY) Defining Polar Coordinates and Differentiating in Polar Form Incorrect. This response would result if the slope was taken to be θ = θ = θ = θ θ ⇒ = θ = θ θ ⇒ θ= π = θ = θ θ= − θ θ θ= θ= π π π =− θ− θ = θ θ =π = − θ+ θ = θ θ =π = π π (B) Correct. The slope of the polar curve is θ = θ = θ = θ θ ⇒ = θ = θ θ ⇒ θ= (C) = θ = θ θ= π − θ θ π θ= θ= π π =− θ− θ = = θ θ =π = − θ+ θ π θ θ =π = π Incorrect. This response would result if the product rule was applied incorrectly by taking the product of the derivatives. = θ = θ θ ⇒ = θ = θ θ ⇒ θ= (D) π π π = θ = θ θ= π − θ θ =− = θ⇒ θ ⇒ θ θ θ= θ= π π =− = θ θ =π = − θ θ =π = = Incorrect. This response would result if the slope was taken to be rather than θ θ= π = θ= π = θ 4 Q Skill Learning Objective Topic 1.F LIM-5.A Approximating Areas with Riemann Sums (A) Incorrect. It was correctly determined that the left Riemann sum is an underestimate and the right Riemann sum is an overestimate of the definite integral because the function ( ) = is increasing on the interval [ is concave down, ] Because the graph of however, the trapezoidal sum is an underestimate, not an overestimate. (B) Correct. Because the function ( ) = is increasing on the interval [ ] the left Riemann sum is an underestimate and the right Riemann sum is an overestimate of the definite integral, so < ∫ < Since the graph of is concave down, the trapezoidal sum is also an underestimate, but it is a closer approximation to the definite integral than the left Riemann sum. Therefore, < < ∫ < (C) Incorrect. This would be the correct response for the graph of a function that is decreasing and concave up. The graph of however, is increasing and concave down, so all the ( ) = inequalities are going in the wrong direction. (D) Incorrect. It was correctly determined that the trapezoidal sum is an underestimate of the definite integral since the graph of ( ) = is concave down on the interval [ is ] Because the function increasing, however, the left Riemann sum is an underestimate and the right Riemann sum is an overestimate of the definite integral. This response has reversed those two inequalities. 4 Q Skill 3.D (A) ′( ) = at = ( ) − ( ) = (− ) − and = =− where the values of are obtained from the graph. ( ) = ( )( ) − ( )( ) − ( )( ) = − Incorrect. This response would result if the Fundamental Theorem of Calculus was incorrectly applied, as follows. ∫ (D) FUN-6.B The Fundamental Theorem of Calculus and Definite Integrals Incorrect. This response would result if the function was integrated over the interval [ rather than as follows. ′ ] ∫ (C) Topic Correct. By the Fundamental Theorem of Calculus, ∫ (B) Learning Objective ′( ) = ′( ) − ′( ) = ( − ) − ( − ) = Incorrect. This response is the total area bounded by the graph of and the -axis over the interval [ ] ∫ ( ) = ( )( ) + ( )( ) + ( )( ) = + + = 4 Q Skill Learning Objective Topic 1.E FUN-6.E (BC ONLY) Using Integration by Parts (A) Correct. This definite integral can be found using integration by parts, where ∫ = = ∫ (B) = = = ′( ) ( ) −∫ ⇒ ( ) = ( ) = ( )− = − = Incorrect. This response would result if the technique of integration by parts was applied incorrectly as follows. = ⇒ = = ′( ) ⇒ = ( ) ⋅ ′( ) = ( ) − ( ) =( ( ) − ) − ( ( ) − ( )) = − Incorrect. This response would result if each factor in the integrand was antidifferentiated separately, as follows. ∫ (D) ⇒ ∫ ⋅ ′( ) ∫ (C) − ⋅ ′( ) = ( ) ( ) = ( )− = Incorrect. This response would result if an error was made in the integration by parts when addition was used instead of subtraction, resulting in ∫ ∫ = = ⇒ ⋅ ′( ) + ∫ = = = ′( ) ( ) +∫ ⇒ ( ) = = ( ) ( )+ = + = = 4 Q Skill 3.D (A) Learning Objective Topic LIM-7.A (BC ONLY) Determining Absolute or Conditional Convergence Incorrect. Series I is absolutely convergent, not conditionally convergent, since (B) ∞ ∑= converges by the ratio test. Incorrect. Series II was correctly identified as being conditionally convergent. Series III is also conditionally convergent, since ∞ − ( ∑= ) converges by the alternating series test but + ∞ ∑= + diverges by the limit comparison test with the harmonic series (C) Correct. A series ∑ ∞ ∑= is conditionally convergent if the series converges but the series of absolute terms ∑ diverges. Each of the three series in this problem converges by the alternating series test. The series ∞ ∑= (− ) is not conditionally convergent, since ∞ ∑= converges by the ratio test (so this series is absolutely convergent). The series ∞ ∑ The series ∞ = series (D) ∑ (− ) = is conditionally convergent because the series diverges, since it is a = ∑ ∞ ∞ − ( ∑ = + ) = -series with < is conditionally convergent because the series diverges by the limit comparison test with the harmonic + ∞ ∑ = Incorrect. This response might be chosen because all three series converge by the alternating series test. To determine whether they are conditionally convergent, however, it is also necessary to consider the series of absolute terms ∑ Series I is absolutely convergent, not conditionally convergent, since the ratio test. ∞ ∑ = converges by 4 Q Skill 1.C (A) Learning Objective Topic CHA-5.C Volume with Washer Method - Revolving Around - or -axis Incorrect. This response is the volume of the solid generated by revolving about the -axis rather than the -axis. A typical slice rotated about the -axis would form a disk of radius = = − ( )=π with cross-sectional area solid would be (B) ∫ = π ( − ) The volume of the π( − ) Incorrect. This response is the volume obtained by rotating about the -axis the region bounded by the graph of = the − horizontal line = and the vertical lines = and = A typical slice rotated about the -axis would form a washer with −π cross-sectional area ( ) = π = = − and the outer radius is (C) Incorrect. For this response, the cross-sectional area of the washer obtained by rotating a typical slice about the -axis was taken to be π ( − ) rather than π = (D) where the inner radius is −π where the inner radius is + and the outer radius is = Correct. Rotating a typical slice about the -axis will form a washer with inner radius and outer radius where each radius must be expressed in terms of Since = the inner radius is − = = + Since the region is bounded on the right by the vertical line = the outer radius is = The cross-sectional area of a typical slice rotated about the -axis is therefore ( )=π −π =π ( − ( + ) ) The volume of the solid is found by using the definite integral of the cross-sectional area for between and as follows. − = ⌠ π  ⌡ ( − ( + ) ) ⌠ = π ⌡ ( − ( + ) ) 4 Q Skill 1.E (A) Learning Objective Topic CHA-3.G (BC ONLY) Second Derivatives of Parametric Equations Incorrect. This response is the value of =    =   ( ( )) not the value of The chain rule must be used to find =       and then used again to find (B) = Incorrect. The chain rule was correctly used to find ( )  however. Another application of the     by chain rule would be needed to find by dividing     This response is just    (C) Correct. By the chain rule, = = ( ) The chain rule is needed again to find the second derivative, as follows. =    =         = ( ) = ( ) (D) Incorrect. This response comes from thinking that Since = the quotient would be undefined. = 4 Q Skill 2.C (A) Learning Objective Topic LIM-5.C Riemann Sums, Summation Notation, and Definite Integral Notation Incorrect. The sum inside the limit can be interpreted as a right Riemann     = sum in the form ∑   = ∑ ( + Δ ) Δ , where ( ) =  + =    and Δ = . The value of Δ corresponds to an interval of length Since Δ = 2 + for the initial value = the interval starts at Therefore, the limit of the Riemann sum would be equal to ∫ (B) ( ) =⌠ not ⌠ ⌡ ⌡ Correct. The sum inside the limit can be interpreted as a right Riemann     = sum in the form ∑   = ∑ ( Δ ) Δ , where ( ) + =  + =    and Δ = . The value of Δ corresponds to an interval of length Since Δ = for the initial value = the interval starts at Therefore, the limit of the Riemann sum is equal to the definite integral ∫ (C) ( ) =⌠ ⌡ + Incorrect. The sum inside the limit can be interpreted as a right Riemann     = sum in the form ∑   = ∑ ( Δ ) Δ , where ( ) + =  + =    and Δ = . The value of Δ corresponds to an interval of length Since Δ = for the initial value = the interval starts at therefore the limits on the definite integral should go from to (D) to and not Incorrect. The sum inside the limit was correctly interpreted as a right Riemann sum on an interval of length starting at = With that choice of interval, however, the Riemann sum would be written as     = The right Riemann ∑  = ∑ ( + Δ ) Δ , where ( ) =  + =    sum using = not ( )= = + would correspond to an interval starting at 4 Q Skill 2.B (A) Learning Objective Topic LIM-8.A (BC ONLY) Finding Taylor Polynomial Approximations of Functions Incorrect. This response would result if the chain rule was never used in finding the first and second derivatives of and ′′( ( ) )= series for (B) in the Maclaurin ( ) ′′( ) = was computed as = Incorrect. This response would result if the coefficient of in the Maclaurin series for was taken to be ′′( ) without dividing by In addition, the chain rule was never used in finding the first and second derivatives of (C) As a result, the coefficient of ( ) ′( ) = leading to ( ) ′′( ) = leading to = Incorrect. This response would result if the chain rule was correctly used in finding ′( ) but was not used again when the product rule was used during the computation of the second derivative, as follows. ′( ) = ′( ) ( ) ′′( ) = ′′( ) ( ) + ′( ) ( ) ′′( ) = ′′( ) ( ) + ′( ) ( ) = The coefficient of + = in the Maclaurin series for was found to be ′′( ) = (D) Correct. The coefficient of in the Maclaurin series for ( ) Since ( ) = the chain rule gives ′( product rule and the chain rule again gives ′′( ) = ′′( ) ( ) + ′( ) ( The values of ( ) ′( ) and Maclaurin series for = ( )= ′( ) = = ′′( ) = Therefore, ⋅ ( ′′( ) = the coefficient of ′( ) )= ( ( ) ( ) ) = ′( ) )=( ′′( ) is Using the ′′( ) + ( ′( ) ) ) ( ) ′′( ) can be determined from the ⋅ = ′′( ) + ( ′( ) ) ) ( ) =( + ) in the Maclaurin series for is and so = ( )= 4 Skill 2.E Q Learning Objective Topic FUN-4.A Determining Intervals on Which a Function Is Increasing or Decreasing (A) Incorrect. The graph of is concave up where ′ is increasing. This response might come from switching the roles of the function and its derivative and thinking that is increasing where the graph of ′ is concave up. The graph of ′ is concave up on the intervals ( ) and ( ) (B) Incorrect. This response might come from treating the given graph as the graph of rather than the graph of ′ These are the two intervals where ′ is increasing. (C) Incorrect. These are the intervals where both increasing. (D) Correct. The function is increasing on closed intervals where ′ is positive on the corresponding open intervals. The graph indicates that ′( ) > on the intervals ( is ) and ( ) so increasing on the intervals [ ] and [ ] and ′ are 4 Q Skill 1.E (A) Learning Objective Topic CHA-4.D Using Accumulation Functions and Definite Integrals in Applied Contexts changes from = to Incorrect. This response is how much The amount by which = that is, ′( ) − ′( ) = changes from = to = is ( ) − ( ) which can be computed by using the Fundamental Theorem of Calculus. (B) Incorrect. This response is the approximation to the change in along the line tangent to the graph of at = It can also be interpreted as the approximation to the integral ∫ ′( ) = ( ) − ( ) by using a left Riemann sum with one interval of length Δ = 5. Δ ≈ ′(1) Δ = ′(1) ⋅ 5 = 8.341 (C) Incorrect. This response is the approximation to the integral ∫ ′( ) = ( ) − ( ) by using the trapezoidal sum approximation over one interval of length Δ = 5.  ′( ) + ′( )  ⋅ Δ =  ′( ) + ′( )  ⋅ 5 = 18.017         (D) Correct. The change in from = to Fundamental Theorem of Calculus, ( )− ( )= ∫ ′( ) = ∫ − = (− ) is ( ) − ( ) By the = numerical integration is done with the calculator. where the 4 Q Skill 2.D (A) Learning Objective Topic LIM-2.D Connecting Infinite Limits and Vertical Asymptotes Incorrect. This graph displays the appropriate behavior as it approaches the horizontal asymptote at = but in this graph ( ) = − ∞ and ( ) = +∞ which is the opposite → − → + behavior for what the graph of the vertical asymptote at = should be doing as it approaches (B) Incorrect. This graph displays the appropriate behavior as it approaches the vertical asymptote at = but it has = − as a horizontal asymptote rather than = (C) Correct. Since → − ( ) = +∞ and → + approaches the vertical asymptote at ( ) = −∞ the graph of = in the upward direction as approaches from the left and approaches the vertical asymptote in the downward direction as approaches from the right. Since ( ) = and ( ) = the graph of →− ∞ →+ ∞ approaches the horizontal asymptote at = in both horizontal directions. This graph exhibits these properties and therefore could be the graph of (D) Incorrect. This graph has a vertical asymptote at = and a horizontal asymptote at = rather than a vertical asymptote at = and a horizontal asymptote at = 4 Q Skill 3.F Learning Objective Topic CHA-4.B Finding the Average Value of a Function on an Interval (A) Incorrect. The definite integral was not divided by the length of the interval [ ] over which the averaging is done. (B) Correct. The average value of a function is − ∫ ( ) Tara’s average heart rate from = is the average value of the function (D) − ∫ ] to = over the interval [ would therefore be given by the expression (C) over an interval [ ] and () Incorrect. This response is the average rate of change of Tara’s heart rate from = to = not the average of her heart rate over that interval. By the Fundamental Theorem of Calculus, this ( )− ( ) expression is equal to − Incorrect. This response is the average of the rate of change of Tara’s heart rate at the two times = and = not the average of her heart rate over the interval from = to = 4 Q Skill 1.E (A) Topic LIM-8.B (BC ONLY) Finding Taylor Polynomial Approximations of Functions Incorrect. This response comes from only using three terms of the Taylor polynomial rather than the Taylor polynomial of degree (− + (B) Learning Objective ( − )+ ( − ) ) = =− Correct. The third-degree Taylor polynomial for following. ( ) = ( ) + ′( )( − ) + =− + ( − )+ ( − ) + about ′′( )( − ) + = is the ′′′( )( − ) ( − ) =− + ( − )+ ( − ) + ( − ) ( (C) )≈ ( )=− + ( ) + ( ) =− Incorrect. This response would result if the coefficient of the degree term was taken to be (− + (D) )+ ( ( ) ( − )+ ( − ) + ( − ) Incorrect. The coefficient of the degree rather than (− + ( ) ( ) ( ) rather than ) = ( ) =− term was taken to be ( ) ( − )+ ( − ) + ( − ) ) = =− ( ) ( ) 4 Q Skill 1.E (A) Topic CHA-6.A (BC ONLY) The Arc Length of a Smooth, Planar Curve and Distance Traveled Incorrect. This response comes from not taking the square root in the integrand of the definite integral for the length of a curve, as follows. ∫ ( (B) Learning Objective + ( ′( ) ) )  =⌠   + ⌡  ( + Correct. The length of the graph of by the definite integral ∫  =⌠    = ∫ ( = ( ) from = ( ) ) + (C) + = + is given = Incorrect. This response comes from not squaring the derivative in the integrand of the definite integral for the length of a curve, as follows. ∫ (D) + = = to ∫ ⌡ where the numerical integration is done with the calculator. + ( ′( ) ) ) + ′( ) = ∫ + ( + ) Incorrect. This response is the change in ( )− ( )= ∫ ′( ) = ∫ = = from + = to = 4 Q Skill Learning Objective Topic 2.D FUN-7.C Sketching Slope Fields (A) Correct. In the slope field for a differential equation of the form = ( ) the slope at a point ( ) depends only on the value of The line segments in the slope field at each point on a vertical line perpendicular to the -axis should therefore all have the same slope. The line segments in this slope field show that behavior and therefore this could be a slope field for a differential equation of the form (B) = ( ) Incorrect. In the slope field for a differential equation of the form = ( ) the slope at a point ( ) depends only on the value of The line segments in the slope field at each point on a vertical line perpendicular to the -axis should therefore all have the same slope. The line segments in this slope field do not exhibit that behavior. The slopes of the line segments depend on both the - and the -values. (C) Incorrect. In the slope field for a differential equation of the form = ( ) the slope at a point ( ) depends only on the value of The line segments in the slope field at each point on a vertical line perpendicular to the -axis should therefore all have the same slope. The line segments in this slope field do not exhibit that behavior. The slope of the line segment at a point depends only on the value of This could be a slope field for a differential equation of the form (D) = ( ) Incorrect. In the slope field for a differential equation of the form = ( ) the slope at a point ( ) depends only on the value of The line segments in the slope field at each point on a vertical line perpendicular to the -axis should therefore all have the same slope. The line segments in this slope field do not exhibit that behavior. The slopes of the line segments depend on both the - and the -values. 4 Q Skill Learning Objective Topic 3.D LIM-2.C Removing Discontinuities (A) Correct. The limit at limits are equal. = ( )= ( )⇒ exists if the left-hand and right-hand + = − The solution to this equation for > is = With this value of is ( ) exists and is equal to ( ) Therefore, → − → + → = continuous at (B) Incorrect. This response comes from trying to make the left-hand and right-hand limits of the derivative equal at = as follows. <   ′( ) =  > ( − )  → − ′( ) = → + ′( ) ⇒ = ( > The solution to this equation for (C) − ) Incorrect. In trying to set the left-hand and right-hand limits of equal at = the might have been substituted for the parameter rather than the variable as follows. + = − = The positive solution to this equation is (D) = is Incorrect. This response might come from errors that lead to an equation that has no positive solution. For example, it might come from trying to make the left-hand and right-hand limits of the derivative equal at = but also making a chain rule error in the derivative of the piece for > as follows.   ′( ) =    → − < ( ′( ) = − − → > ) + ′( ) ⇒ = This equation has no solution for ( − − ) 4 Q Skill 2.D Learning Objective Topic CHA-2.C Defining the Derivative of a Function and Using Derivative Notation (A) Incorrect. The slope of the line tangent to the graph of at = is ′( ) Therefore, ′( ) is negative. Since the tangent line goes through the point ( ( )) ( ) is also negative. Therefore, it cannot be true that ′( ) = − ( ) since both are negative. (B) Incorrect. The slope of the line tangent to the graph of at = is ′( ) and the tangent line goes through the point ( ( )) Therefore, ′( ) = − and ( ) = − so it is not true that ′( ) < ( ) (C) Correct. The slope of the line tangent to the graph of at = ′( ) and the tangent line goes through the point ( ( )) Therefore, ′( ) = − and ( ) = − so ′( ) = ( ) (D) Incorrect. The slope of the line tangent to the graph of at = is ′( ) and the tangent line goes through the point ( ( )) Therefore, ′( ) = − and ( ) = − so it is not true that ′( ) > ( ) 4 Q Skill 1.E (A) is Learning Objective Topic CHA-4.C Connecting Position, Velocity, and Acceleration Functions Using Integrals Incorrect. This response is the referee’s displacement over the time interval ≤ ≤ rather than the total distance the referee traveled. ∫ () =∫ ( − ) ( + ) = (B) Incorrect. This response is the absolute value of the referee’s change in velocity from time = to time = ( )− ( ) = (C) Correct. The referee’s total distance traveled on the time interval ≤ ≤ is ∫ () = ∫ ( − ) ( + ) = where the numerical integration is done with the calculator. (D) Incorrect. This response comes from averaging the referee’s initial and final velocities on the time interval ≤ ≤ then multiplying by the length of the time interval. ( )+ ( ) ( )+ ( ) ⋅Δ = ⋅ 4 = 29.156 4 Q Skill 2.B (A) Learning Objective Topic FUN-5.A Interpreting the Behavior of Accumulation Functions Involving Area Correct. Because is continuous, the Extreme Value Theorem guarantees the existence of an absolute maximum on the closed interval [ − ] and that the maximum will occur at a critical value or at one of the endpoints. Since is an = − and = Therefore, the antiderivative of ′( ) = ( ) = at candidates are = − =− = and = Evaluate at each candidate and select the largest. (− ) = (− ) = ∫ ∫ − − () = −∫ () = −∫ − − ( )= ∫ () = ( )= ∫ ( ) =− ( () () ( = − − ( )( ) + =− ( ) ( )( ) = − ( − + ) = ) ( )( ) = − ) ( )( ) + ( + ) = − ( + The maximum occurs at )=− =− (B) Incorrect. Both critical values might have been found, and the relative minimum was picked while not accounting for the endpoints. (C) Incorrect. Both critical values might have been found, and the relative maximum was picked while not accounting for the endpoints. (D) Incorrect. The four candidates might have been identified, but the computation of ( ) found the area of the region from = to = and did not account for the region being below the horizontal axis, as follows. ( )= ∫ () = ( ) ( )( ) + ( + ) = ( + )= 4 Q Skill 1.E (A) Learning Objective Topic FUN-8.B (BC ONLY) Solving Motion Problems Using Parametric and VectorValued Functions Incorrect. This response would result if the problem was treated like rectilinear motion. The velocity of the particle was taken to be the value of the derivative = = = = at time ( ) − ( = − − = ) − =− = Then the speed was taken to be the absolute value of velocity. (B) Incorrect. This response is the magnitude of the position vector at time = not the magnitude of the velocity vector at that time. ( ( )) + ( ( )) (C) ) − +( ( ⋅ Correct. The speed of the particle at time ′( ) velocity vector ( ′( ) ) + ( ′( ) ) (D) ( = = ′( ) = − = = (− − − )) = is the magnitude of the ( ) at time ) +( ( ⋅ )) = Incorrect. This response would result if the components of the vector ′( ) ′( ) = − − ( ) were not squared when finding the = magnitude of the vector at ′( ) + ′( ) = = (− − )+( ( ⋅ )) = 4 Q Skill 1.E Learning Objective Topic CHA-3.A Interpreting the Meaning of the Derivative in Context (A) Incorrect. This response might be chosen if the calculation of the average rate of change resulted in a value that was greater than or less than − It would also be chosen if the average rate of change but the instantaneous rate of was correctly found to be − change was taken to be the second derivative of not the first derivative. In either case, the resulting equation would have no solution in the interval [ ] (B) Incorrect. This response would be chosen if the average rate of change was correctly found to be − but the graph of not ′ was drawn to determine the number of intersection points with the horizontal line = − It would also be chosen if the instantaneous rate of change was correctly identified as the derivative of but the average rate of change over the interval [ ] was thought to be the average at the endpoints, ( )+ ( ) =− or the average value of the function over the interval, ∫ ( ) =− In all these cases, the resulting equation would have only one solution in the interval [ ] (C) Correct. The average rate of change of ( )− ( ) =− [ ] is − of change of is the derivative, ′( ) = on the closed interval The instantaneous rate − + − The graph of ′ produced using the calculator, intersects the horizontal line = − three times in the open interval ( ) (D) Incorrect. This response might be chosen because the function a polynomial of degree is 4 Q Skill 3.F Learning Objective Topic FUN-7.H (BC ONLY) Logistic Models with Differential Equations (A) Incorrect. This differential equation is for a rate of change that is equal to the difference between the carrying capacity and a term that is proportional to the size of the population. There is no joint proportionality with the size of the population and the difference between the size of the population and the carrying capacity. (B) Incorrect. This differential equation would be for a model where the population grows at a rate that is only proportional to the difference between the carrying capacity and the size of the population. There is no joint proportionality with the size of the population. (C) Incorrect. This differential equation would be for a model where the population grows at a rate that is jointly proportional to the reciprocal of the size of the population, not the population itself, and the difference between the carrying capacity and the size of the population. (D) Correct. The model for logistic growth that arises from the statement “The rate of change of a quantity is jointly proportional to the size of the quantity and the difference between the quantity and the carrying ( − ) where represents the quantity that is = capacity” is changing and is the carrying capacity. Here the population is changing according to the logistic growth model with carrying capacity Therefore, the differential equation for the model would be = ( − ) 4 Q Skill Learning Objective Topic 1.E CHA-3.E Solving Related Rates Problems (A) Incorrect. This response would result if the area of the region was taken to be = π ( − ) instead of π In addition, an −π error was made in the power rule when doing the differentiation, as follows. =π( − ) (B) ( − ) ( − ( − )) = π Incorrect. This response would result if the area of the region was taken to be = = π ( − ) instead of π −π = − (since the inner radius and the instant when = π( − ) (C) ) =π( − ( = and )= − = It is given that is decreasing). At this gave π ( − ) ( − ( − )) = π Incorrect. This response would result if the area of the region was correctly taken as = π Using the chain rule gave the −π rate of change of the area with respect to time as = π − π It is given that = but was taken to be rather than − by not taking into account that the inner radius is decreasing. At the instant when = and = this = π( ) gave (D) − π( ) = π ( ( ) − ( )) = π Correct. The area of the region is = π Using the chain −π rule gives the rate of change of the area with respect to time as = π − π (since the inner radius and = = π( ) It is given that = =− and is decreasing). At the instant when − π( ) = π ( ( ) − ( − )) = π = 3 6 25 1 4 6 BC 1 6 LRQ ′( ) = 7 V RI F R E ∫ F ∫ () P V = F V R R () 7R P RI V = () { = V RO PE R P = ≈ ( − )⋅ +( = ⋅ ( )+ ( ) − )⋅ + F V + + ⋅ OR I RP ( )+ ( ) + ( − )⋅ ( )+ ( ) V +( + ⋅ − )⋅ + ( )+ ( ) + ⋅     R OV P R P R O R + = ∫ () V PE R VE ∫ () 7R P = RI F V = 0 + V ∫ ROO () O OR 30 = = ROO V 9V FROO F I RP P = R     OP V V FR V V 3 6 25 1 6 1 6 4 LRQ ( ′( ) ) + ( ′( ) ) = E V 7 F V P V U UV F P ∫ ( ′( ) ) + ( ′( ) ) = 7 = UV V F F U V V F E )= +⌠ ( F V F = V F U V U U P P UV P V UVU ⇒ = V V U = P V U = = V ( )= ( 7 )= +⌠ =− ⌡ +⌠ = ⌡ U F E V F V V V 9V U V ( )− ( )= { V        = = ⌡ V U U U U V ( )= ( )= +⌠ { P = ⌡ 7 = U V UV V V V FU F U V U = V M V F 3 6 25 1 4 RI F 1 6 LRQ ( ) − (− ) − − = − (− ) = 6 BC V =− ( ) ( ) = − + FRV π = − E )R < ( ( ( π )) ⋅ π = − + ′( ) = − + − V T R IR F 7 O O − ( − ) ∫− ∫− ( ) RI ∫− O − ≤ ∫ = ∫ ( )  − + FRV π =⌠ → − ⌡ − ( ( ) ( )= ( )= = F V V FR V ≤ )( π = V π ( )) = OV RI − ≤ ≤ − ) ( ) ( ) OP ∫− + ∫− − ( − ) ∫− ( =− + − + V R ( )            π ( ) = ′( ) T R ( π )) ⋅ π ′( ) = − + − V <    = −  + V ( π ) V FR R V π O RI R ≤ ∫− − + FRV − RI ( ) π V =− (π− ) OP → + ( )= VR R V R [− ( P 9 O ] V EVRO P P P R [ − 9V 7 R P ] { FR V = M V IF R ≤ 3 6 25 1 4 ∫− () ( )= ∫− ( ) = ∫− () ( )= E = + − ∫ + + = + + −  = + + − ⋅  = + + 1 6 LRQ =      ( ) ∫ (− = 6 BC + + ( + O RI −   ⋅ − ⋅ ( − (− )) = ) − (− O ⋅ + ( ) R RI R ) − ⋅   ⋅ [−      ]F (− ) = ′( ) = ( ) I V =− = VF V V M V IF R ( )= EVRO F 7 P RI < V < P P O V FR F < < R [− RI R R E F V OV 9V ≤ ≤ () ′( ) = ( ) V IR − < < OV 7 V EVRO P P P R O RFF =− R = 7 R ) − + ∫ ( ) ] V (− ) = OV − < ′( ) = ( ) V <− F V R { OV VR 3 6 25 1 4 ( + ⇒ + + ⇒ ( + ) ⇒ = E = 1 6 LRQ { ( ) = − + ( + ) − )⋅ = ( ( + )) U > +V θ ( + V θ) > π 7 π ≤θ ≤ = U F E π ( + )−      + ( ( + )) ≤θ ≤ π ⌠  ⌡ π π    UP    θ     P V UP U V T U U U V θ F Vθ < V θ > ( θ ) V FU V U ⇒ +V θ = U π <θ < V U − = − = ⋅F Vθ θ ( + V θ) ) U P F U F =− − ⋅ ( + ) − (− = F )= + 6 +V θ =  − + V θ  9V ⇒ V θ =−   θ  ⇒ θ = π U Uθ = π F V U 3 6 25 1 6 1 6 4 7 V U V = S= ∞ − ( ∑ V ∑= + (− ) 7 U ∞ ∑= = U V S V U V ∞ − ( V U V ∞ − ( + ∑= ) ∑ (− ) ∞ = S= F ⋅ + ∞ ∑ = =       = F F > U V + ) = ⋅ ∞ − ( ∑= ) +       V U V V UP U ∞ − ( FV U V + ) ∑ V U V =− V U V F ⋅ = V ∞ − ( + ) ∑ F V UV ∞ ∑ = F U V EV F F V UV ∞ − ( ∑ = F U VF U V ∞ ∑= V S V U V U V U V S 7 U F F ∞ − ( ∑ ∞ − ( ∑= ) + ( ⋅ 8V ∞ − ( ∑= − ) = UPV ) + ( ) ⋅ ∞ − ( ∑= ≈ ∞ ∑ −S = ∞ = −∑      S = U V V + ) ( ( ) ⋅ U      ) = V U V UPV FU V P ≈ P U EV 7 U U V U V UU U E UU U E E − < U P ⋅ 9V U ⋅ = F V V UV U ∞ ∑= − U S> ⋅ − V = S> F S = S= S (− ) ⋅ = V U V (− ) U V + ) U VF ⋅ = 7 ∑ + ) = 7 U F 7 ∑= U V ∞ − ( U V EV ∞ − ( V U VE = S= F ⋅ = E 7 + ) ∑ V U V + ) = ∞ LRQ < UP V ⋅ U P V V U UP V U V UU U E S ) + F FW R 0 WS FR :R RF × 1 P R R W RI W FW R 4 W 6 W RQ 6 R R QRW R Q SR × W RQ R W RI 4 R QRW R Q × W RQ R W RI 4 R QRW R Q × W RQ R W RI 4 R QRW R Q × W RQ R W RI 4 R QRW R Q × W RQ R W RI 4 R QRW R Q × W RQ R W RI R QRW R Q 6 P W 6 W RQ 6 R R QRW R Q RPSR W FR W 6 W RQ 6 R 6 W RQ W 6 R RPSR W 6 R 5R Q WR Q D W R Q P 36 R RQ RQ D W D RPSR W 6 R 5DQ 36 R W F 4 FW R 0 WS W RQ FR RF × 1 P R R W RI W FW R 4 W 6 W RQ , 6 R R QRW R Q SR × W RQ R W RI 4 R QRW R Q × W RQ R W RI 4 R QRW R Q × W RQ 4 R W RI R QRW R Q 6 P W 6 W RQ ,, 6 R R QRW R Q RPSR W FR W 6 W RQ , 6 R W 6 W RQ ,, 6 R RPSR W 6 R 5R Q WR Q D W R Q P 36 R RQ RQ D W D 6 R RPSR W 6 R 5DQ 36 R FR :R W RQ F 0 Question Skill Learning Objective R DF DQ RF R PDQF DD RQ Topic The Fundamental Theorem of Calculus and Definite Integrals Using the First Derivative Test to Find Relative (Local) Extrema Key % Correct A 95 A 82 1 1.E FUN-6.B 2 3.D FUN-4.A 3 1.E FUN-3.C The Chain Rule D 85 FUN-8.B (BC ONLY) Solving Motion Problems Using Parametric and Vector-Valued Functions B 80 4 1.E 5 3.D LIM-7.A (BC ONLY) Working with Geometric Series B 66 6 1.E FUN-3.B The Quotient Rule A 88 7 2.C LIM-8.G (BC ONLY) Representing Functions as Power Series A 69 B 81 C 68 B 67 Integrating Using Substitution D 70 Connecting Limits at Infinity and Horizontal Asymptotes Defining Average and Instantaneous Rates of Change at a Point Determining Concavity of Functions over Their Domains 8 2.B LIM-2.D 9 3.B CHA-2.B 10 2.B FUN-4.A 11 1.E FUN-6.D 12 3.D FUN-1.B Using the Mean Value Theorem C 80 (BC ONLY) Approximating Solutions Using Euler's Method A 74 13 1.E FUN-7.C 14 1.E FUN-3.D Implicit Differentiation B 74 15 1.E LIM-6.A (BC ONLY) Evaluating Improper Integrals C 61 Finding General Solutions Using Separation of Variables B 46 16 1.E FUN-7.D 17 3.D LIM-7.A (BC ONLY) Alternating Series Test for Convergence D 57 D 65 18 1.C CHA-5.D (BC ONLY) Finding the Area of the Region Bounded by Two Polar Curves 19 3.B LIM-7.A (BC ONLY) Alternating Series Test for Convergence D 69 20 1.C FUN-6.A Applying Properties of Definite Integrals C 78 A 71 B 59 (BC ONLY) Radius and Interval of Convergence of Power Series (BC ONLY) Defining Polar Coordinates and Differentiating in Polar Form 21 3.D LIM-8.D 22 1.E FUN-3.G 23 1.F LIM-5.A Approximating Areas with Riemann Sums B 67 24 3.D FUN-6.B The Fundamental Theorem of Calculus and Definite Integrals A 65 25 1.E FUN-6.E (BC ONLY) Using Integration by Parts A 33 C 53 D 58 C 40 B 37 D 22 26 3.D LIM-7.A 27 1.C CHA-5.C 28 1.E CHA-3.G 29 2.C LIM-5.C 30 2.B LIM-8.A (BC ONLY) Determining Absolute or Conditional Convergence Volume with Washer Method - Revolving Around x- or y-axis (BC ONLY) Second Derivatives of Parametric Equations Riemann Sums, Summation Notation, and Definite Integral Notation (BC ONLY) Finding Taylor Polynomial Approximations of Functions RQ Question Skill F R DF DQ Learning Objective 76 2.E FUN-4.A 77 1.E CHA-4.D 78 2.D LIM-2.D 79 3.F CHA-4.B 80 1.E LIM-8.B 81 1.E CHA-6.A 82 2.D FUN-7.C 83 3.D LIM-2.C R PDQF DD Topic Key % Correct D 92 D 82 C 95 B 70 B 80 B 68 Sketching Slope Fields A 71 Removing Discontinuities A 72 C 63 C 62 A 49 C 72 Determining Intervals on Which a Function Is Increasing or Decreasing Using Accumulation Functions and Definite Integrals in Applied Contexts Connecting Infinite Limits and Vertical Asymptotes Finding the Average Value of a Function on an Interval (BC ONLY) Finding Taylor Polynomial Approximations of Functions (BC ONLY) The Arc Length of a Smooth, Planar Curve and Distance Traveled Defining the Derivative of a Function and Using Derivative Notation Connecting Position, Velocity, and Acceleration Functions Using Integrals Interpreting the Behavior of Accumulation Functions Involving Area (BC ONLY) Solving Motion Problems Using Parametric and Vector-Valued Functions 84 2.D CHA-2.C 85 1.E CHA-4.C 86 2.B FUN-5.A 87 1.E FUN-8.B 88 1.E CHA-3.A Interpreting the Meaning of the Derivative in Context C 42 D 66 D 60 89 3.F FUN-7.H (BC ONLY) Logistic Models with Differential Equations 90 1.E CHA-3.E Solving Related Rates Problems RQ RQ Question Skill Learning Objective Topic Mean Score 1 1.E|3.D|3.F|4.D|4.A|4.B|4.C|4.E CHA-2.D|CHA-4.E|LIM-5.A 2.3|8.3|6.2|8.3 7.33 2 1.D|1.E|3.B|3.D|4.D|4.E FUN-8.B 9.6 5.05 3 1.D|1.E|3.C|3.D|4.A CHA-2.A|CHA-2.C|CHA-4.B|FUN-1.C 2.1|2.2|8.1|5.2 3.03 4 1.D|1.E|2.B|2.E|4.A FUN-6.A|FUN-4.A 6.6|5.5|5.6 4.89 5 1.D|1.E |3.E|3.G|4.C FUN-3.D|FUN-4.E|FUN-3.G|CHA-5.D 3.2|5.12|9.7|9.9 5.04 6 1.E|3.B|3.D|3.E|4.A LIM-7.A|LIM-8.C 10.9|10.5|10.12 3.64

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