Part Two. Diagnostic Test

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1 Part Two Diagnostic Test

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3 AP Calculus AB and BC Diagnostic Tests Take a moment to gauge your readiness for the AP Calculus eam by taking either the AB diagnostic test or the BC diagnostic test, depending on which test you plan to take. The questions in this diagnostic are designed to cover most of the topics you will encounter on the AP Calculus eam. After you take it, you can use the results to give yourself a general idea of the subjects in which you are strong and the topics you need to review more. You can use this information to tailor your approach to the following review chapters. Ideally you ll still have time to read all the chapters, but if you re pressed for time, you can start with the chapters and subjects you really need to work on. Important note for students planning to take the BC eam: The BC eam consists of both AB and BC material. If you plan to take the BC eam, you should test yourself on BOTH diagnostic tests. The BC diagnostic focuses on BC material in an effort to give you the greatest opportunity to test your understanding of BC topics. The real BC eam will not emphasize BC topics as heavily, so you should take the AB diagnostic as well to ensure that you have a good grasp of both the AB and the BC material. Give yourself 0 minutes for the 0 multiple-choice questions and 0 minutes for the freeresponse question. Time yourself, and take the entire test without interruption you can always call your friend back after you finish. Also, no TV or music. You won t have either luury while taking the real AP Calculus eam, so you may as well get used to it now. Be sure to read the eplanations for all questions, even those you answered correctly. Even if you got the problem right, reading another person s answer can give you insights that will prove helpful on the real eam. Good luck on the diagnostic!

4 Diagnostic Test AB Answer Grid To compute your score for this diagnostic test, calculate the number of questions you got right, then divide by 0 to get the percentage of questions that you answered correctly. The approimate score range is as follows: % (etremely well qualified) 60 79% (well qualified) 50 59% (qualified) 0 9% (possibly qualified) 0 9% (no recommendation) A score of 9% is a, so you can definitely do better. If your score is low, keep on studying to improve your chances of getting credit for the AP Calculus AB eam

5 Diagnostic Test AB Directions: Solve the following problems, using available space for scratchwork. After eamining the form of the choices, decide which one is the best of the choices given and fill in the corresponding oval on the answer sheet. No credit will be given for anything written in the test book. Do not spend too much time on any one problem. Note: For the actual test, no scrap paper is provided. In this test: () The domain of a function f is the set of all real numbers for which f () is a real number, unless otherwise specified. () The inverse of a trigonometric function f may be indicated using the inverse function notation f - or with the prefi arc (e.g., sin - arcsin ). e e. Evaluate lim. h (A) e h h (B) e (C) e (D) e (E) The limit does not eist. π. 0 sin cos d (A) 0 (B) (C) π (D) (E) π. The graph of y + + is concave down for (A) < 7 (B) 7 < < 7 (C) all (D) > 7 (E) < 7 and > If F( ) ln( ), then F ( e) (A) e ( e + ) (B) e( e + ) (C) e (D) e( e + ) (E) e GO ON TO THE NEXT PAGE

6 Part two: Diagnostic test AP Calculus AB Diagnostic Test 5. What are the values for which the function f ( ) + is increasing? (A) < < (B) < (C) < and > (D) 0 < < (E) > 8. Evaluate lim sin ( ). 0 (A) (B) 0 (C) 6 (D) (E) The limit does not eist. 6. If f ( ) cos (sin( )), then f π 8 (A) + cos (B) sin (C) cos sin (D) cos sin (E) cos 7. The graph of f is given below. Which of the following statements is true about f? (A) lim f ( ) 0 (B) lim f ( ) 0 (C) lim f ( ) + (D) lim f ( ) (E) lim f ( ) undefined GO ON TO THE NEXT PAGE

7 Part two: Diagnostic Test AP Calculus AB Diagnostic Test 5 0. Evaluate lim (A) 0 (B) - (C) 9. The graph of f is given above. Which graph below could represent the graph of f? (A) (B) (C) (D) (D) (E) The limit does not eist.. Evaluate lim +. (A) - 5 (B) 0 (C) (D) - (E) The limit does not eist.. The solution to the differential equation dy, where y( d ) is y (A) y 5 (B) y (C) y + 5 (D) y + (E) y + 5. The slope of the tangent to the curve y 5 y 8 at the point (,) is (E) (A) - 5 (B) - (C) 0 (D) - (E) - GO ON TO THE NEXT PAGE

8 Part two: Diagnostic test AP Calculus AB Diagnostic Test. Which of the following is a slope field for the differential equation dy? d y (A) (B) (C) (D) (E) 5. The area of the region enclosed by the graph of y + and the line y + 5 is (A) 5 (B) 7 (C) (D) 9 (E) 6. At what point on the graph of y is the tangent line perpendicular to the line + y 6? (A) (0, 0) (B) (C) (D) (E) ( ), 6,, 6 ( ), 6 7. If the region enclosed by the -ais, the line, the line, and the curve y is revolved around the -ais, the volume of the solid is (A) 0π 5 (B) π (C) π (D) w 00 (E) π GO ON TO THE NEXT PAGE

9 Part two: Diagnostic Test AP Calculus AB Diagnostic Test 5 8. An equation of the line tangent to the graph of π y sin( + ) at π is 8 π (A) y π (B) y + π (C) y + π (D) y + (E) y + π 9. Find the derivative of f ( ) cos( t ) dt. (A) cos( ) (B) cos( ) (C) cos( ) (D) cos( ) (E) cos( ) 0 0. Suppose that f is a continuous function and differentiable everywhere. Suppose also that f ( 0), f ( 5), f ( 5). Which of the following statements must be true about f? I. f has eactly two zeros. II. f has at least two zeros. III. f must have a zero between 0 and -5. IV. There is not enough information to determine anything about the zeros of f. (A) I only (B) II only (C) I and III only (D) II and III only (E) IV GO ON TO THE NEXT PAGE

10 6 Part two: Diagnostic test AP Calculus AB Diagnostic Test Free-Response Question Directions: Solve the following problem, using available space for scratchwork. Show how you arrived at your answer. You should write out all your work for each part. On the actual test, you will do this in the space provided in the test booklet. Be sure to write clearly and legibly. If you make a mistake, you can save time by crossing it out rather than trying to erase it. Erased or crossed-out work will not be graded. Show all your work. Clearly label any functions, graphs, tables, or other objects that you use. On the actual eam, you will be graded on the correctness and completeness of your methods as well as your answers. Answers without any supporting work may not receive credit. Justifications (i.e., the request that you justify your answer ) require that you give mathematical (non-calculator) reasons. Work must be epressed in standard mathematical notation, not calculator synta. Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If you use decimal approimations in calculations, the readers of the actual eam will grade you on accuracy. Unless otherwise specified, your final answers should be accurate to three places after the decimal point. Unless otherwise specified, the domain of function f is the set of all real numbers for which f() is a real number.. A particle moves along the -ais so that its velocity at any time t > 0 is given by v(t) 5t - t + 7. The position of the particle, ( t ), is 8 for t. (a) Write a polynomial for the position of the particle at any time t 0. (b) Find the total distance traveled by the particle from time t 0 until time t. (c) Does the particle achieve a minimum velocity? And if so what is the position of the particle at this time? GO ON TO THE NEXT PAGE

11 Part two: Diagnostic Test AP Calculus AB Diagnostic Test 7 GO ON TO THE NEXT PAGE

12 8 Part two: Diagnostic test AP Calculus AB Diagnostic Test If you finish before time is called, you may check your work on this section only. Do not turn to any other section in the test. Stop

13 Part two: Diagnostic Test Diagnostic Test AB answers and eplanations 9 answers and eplanations. C f h f a This problem is a direct application of the definition of the derivative. Remember: f ( a) lim ( ) ( ). h a h a Here f ( ) e and a. We know that f e e e f eh d e d ( ) and ( ) lim h 0 h. So the result follows.. D Let u sin ; du cos d. We can rewrite the integral as: π sin cos d u du u (sin ) ( ) 0 0 π 0 ( ).. A The function f is concave down for all such that f ( ) < 0. By direct calculation, y + - and y 6 +. So f is concave down when 6 + < 0, that is, when < 7.. B To calculate the derivative of a quotient we must use the quotient rule: f gf fg g. With f ( ) + 5 and g( ) ln( ), g [ln( )][ + 5 ] [ + 5][ ] we get F ( ) [ln( )]. And since ln(e), we get, after much reducing, F ( e ) e( e + ). 5. C The function f is increasing for all such that f ( ) > 0. By direct calculation, f () - - ( - - ) ( - )( + ). Therefore, f '( ) > 0 when ( - ) and ( + ) are either both positive or both negative. This occurs when < and >.

14 0 Part two: Diagnostic test Diagnostic Test AB answers and eplanations 6. D In this problem we have to use the chain rule three times. Perhaps it helps to rewrite f as f () [cos(sin())]. Then by applying the chain rule (three times) we get f ( ) [ cos(sin( )) ] [ sin(sin( )) ] cos( ) cos(sin( ))sin(sin( ))cos( ). Now by plugging in the value π 8 and using the fact that sin π π π cos, we get f cos sin ( 8 ) ( ) ( ). 7. C We can easily calculate each limit and check which graphs are plausible. Choice (C) is correct because as approaches from the right, f () approaches. The limit in (A) is undefined because the left and right side limits are not equal. For (B), by eamining the graph, the limit of f as approaches 0 from the left is, not. For (D), the left hand and right hand limits are not equal so the limit does not eist. For (E), the left hand limit of the function is defined. In fact, we get lim f ( ). 8. C Students are epected to know that lim sin. For this problem, the limit can be rewritten: 0 lim sin ( ) lim sin sin lim lim 0 sin 6. Note that as 0, 0, so the limit is still. 9. C Notice that the graph of f has a point of inflection at 0; i.e., it changes concavity at 0. Points of inflection can only occur when f 0 or is undefined. Since f is the rate of change of f, the slope of the tangent line to the graph of f must be 0 at the point of inflection. We can therefore eliminate all but (C). 0. D Because the degree of the numerator polynomial is equal to the degree of the denominator polynomial, the limit as approaches infinity is simply the ratio of the coefficients of the highest-order terms. Note that lim lim A The limit is indeterminate because if we plug in the value we get 0. However, by factoring the numerator 0 and the denominator we can reduce the argument and easily plug in the value to evaluate the limit. See + that lim lim ( + )( ) lim ( + ) 5 ( )( + ) +.

15 Part two: Diagnostic Test Diagnostic Test AB answers and eplanations. B First, cross-multiply and integrate both sides to get y dy d. Thus y + C and so, solving for y, we get y + C. Now using the initial condition we can figure out what C should be. We know that + C ; solving for C we get C 5. Therefore, we can conclude that y D Differentiate the epression term by term, keeping in mind the product rule and attaching a term d or dy each time we differentiate or y, respectively. Then solve for dy. We get d ( y ) ( 5 y) 8 ( ydy + 6 y d) ( 0yd + 5 dy) 0 ydy 5 dy 0yd 6 y d ( y 5 ) dy ( 0y 6 y ) d dy ( 0y 6 y) d ( y 5. ) Now evaluate dy at the point (,) to get the slope of the tangent line to the curve. We get d dy [ 0( )( ) 6( ) ( ) ] d (, ) [ ( )( ) 5( ) ].. B Cross-multiply and integrate to get y dy d. Thus, y + C. Solving for y, we see that y ± + C. For eample, when C 0, we have y ±. These integral curves are given in (B). 5. D To calculate the area between the curves y + and y + 5, we must evaluate the integral b ( + 5) ( + ) d. To determine which values to use for a and b as the limits of the integral, we a calculate the values where the two curves intersect. Solve by factoring to get and -. Set a -, b. The enclosed area, A, is therefore given by the equation A ( + 5) ( + ) d + + d

16 Part two: Diagnostic test Diagnostic Test AB answers and eplanations 6. D Because y, we get y. So the slope of the tangent at is. + y 6 implies y +, so the slope of the line given is. Recall that perpendicular lines have negative reciprocal slopes. Thus we must solve the equation for. We get. Therefore, is the point where the slope of the tangent is perpendicular to the line + y A b The equation to use is V π[ f ( )] d, where f () radius of a cross section given by the function a and a and b are the left and right bounds of the surface of revolution. The π[ f ( )] term measures the b area of a cross-sectional disk and the d term adds them up to give the entire volume of the solid. So, a π V π[ ] d π d π 0π 8. A To get the equation of the tangent line, we need to know the slope of the tangent line and a point on the tangent line. To calculate the slope of the tangent at π 8, find y π ( 8 ). ( ) ( + ) ( + ) ( + ) sin( + π ), π, y ( π ) (, ) is a point on the tangent line. Using π π π π π 8 Using the chain rule, y cos cos cos. Because the line ( ) π 8 0 we want is tangent to the curve y 8 8 the point-slope form we conclude that the equation of the tangent line is y π. 9. E Set f ( ) cos( t ) dt. Therefore, f ( ) cos( t ) dt 0. From the Fundamental Theorem of Calculus we 0 know that f ( ) cos( ). Now applying the chain rule we see that d d cos( t ) dt d d f ( ) 0 f ( ) cos(( ) ) cos( ). 0. D This question is a direct application of the intermediate value theorem, which states that if f is a continuous function on the closed interval [ a, b ] and d is a real number between f ( a) and f ( b), then there eists a c in [ a, b ] such that f ( c) d. Because f ( 0), f ( 5), by the Intermediate Value Theorem there clearly eists a c between 0 and 5 such that f ( c) 0. Similarly, because f ( 0), f ( 5), there eists a c between 0 and -5 such that f ( c) 0. Therefore, c and c are at least two zeros of f and by their location, (D) must be the correct answer.

17 Part two: Diagnostic Test Diagnostic Test AB answers and eplanations Free-Response. (a) The position of the particle is given by ( t) v( t) dt t t + dt t t + 7t + C. Solve for C by substituting 5 ( ) 8: ( ) ( ) + + C + + C + C. Therefore, C -0 and the position of the particle can be written 5 ( t) t t + 7t 0. (b) Total distance traveled is given by v( t) dt. 0 v( t) is positive everywhere, so this is just 5 ( ) ( 0) (c) We find the etrema by setting v ( t) 0. v ( t) 0t 0 when t. Note that v t 5 ( ) 0 > 0, so this value is a minimum. The position is, then, Note: Any question that asks you to compute total distance is asking for v( s) ds, where v is the velocity. Our problem is easier, since v( s) v( s) for our function. For part (c), it isn t enough to just compute the zero of v. You also need to eplain, however briefly, why v actually attains a minimum (and doesn t just continue to decrease to ). In the solution, this is because we stated that v is positive everywhere, and so can t go to.

18 Part two: Diagnostic test Diagnostic Test AB: Correlation Chart Diagnostic Test AB: Correlation Chart Use the results of your test to determine which topics you should spend the most time reviewing. Question Topic Number Definition of Derivative Integration of Trigonometric Functions Relationship Between Concavity of f and the Sign of the Second Derivative Quotient Rule, Chain Rule, and Logarithmic Derivatives 5 Relationship Between Increasing/Decreasing Property of f and the First Derivative 6 Chain Rule and Trigonometric Derivatives 7 Estimating Limits of a Function from the Graph 8 Trigonometric Limits 9 Graphical Relationship Between Increasing/Decreasing Nature of f and the Sign of f 0 Evaluation of Limits As Approaches Infinity Evaluating Indeterminate Limits Using Algebraic Techniques Solving Differential Equations with the Initial Condition Implicit Differentiation Graphical Interpretation of Solutions to Differential Equations 5 Integration Within Specified Boundaries 6 Derivatives As Slope of Tangent 7 Integration, Volume of a Solid with Known Cross-Section 8 Derivatives, Chain Rule, Derivatives As Slope of Tangent 9 Fundamental Theorem of Calculus and Chain Rule 0 Intermediate Value Theorem (a) Integration Using Initial Value Data (b) Integration to Determine Total Displacement (c) Differentiation to Find Etremum Value