AP Calculus BC Chapter 4 AP Exam Problems. Answers

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1 AP Calculus BC Chapter 4 AP Exam Problems Answers. A 988 AB # 48%. D 998 AB #4 5%. E 998 BC # % 5. C 99 AB # % 6. B 998 AB #80 48% 7. C 99 AB #7 65% 8. C 998 AB # 69% 9. B 99 BC # 75% 0. C 998 BC # 80%. B 985 AB #6 7%. A 985 BC # 65%. B 99 BC #4 75% 4. C 99 AB #44 8% 5. B 998 AB #89 8% 6. B 988 BC #9 48% 7. E 988 AB #4 60% 8. C 99 AB # 40% 9. D 998 AB # 78% 0. C 998 AB #9 8%. B 988 BC #7 9%. C 998 BC #6 7%. D 99 BC #9 70% 4. C 988 AB #0 4% 0. B 988 AB #8 59%. D 998 AB #7 %. A 998 AB #76 6%. A 998 AB #79 47% 4. C 985 BC #0 79% 5. E 985 BC #4 76% 7. D 99 AB #8 40% 8. B 985 BC # 5% 9. D 988 BC #4 55% 4. B 985 AB # 55% 4. A 998 AB # 68% 4. D 988 BC #9 64% 44. E 998 BC #6 8% 45. C 998 BC #88 75% 46. D 988 AB #45 4% 47. B 988 BC #45 44% 48. E 99 BC #6 44% 49. D 998 BC #9 65% 50. B 99 AB #45 47% 5. B 988 AB #40 50% 5. D 99 AB #4 8% 5. E 99 BC #4 8% 54. D 988 BC #7 4% 55. D 998 AB #90 6% 56. D 99 AB #9 4% 57. B 998 AB #78 46% 58. A 985 BC # 69% 59. C 985 AB # 5% 5

2 DETAILED ANSWERS to CHAPTER 4. A Check the critical points and the endpoints. f ( x) = x 6x=x(x ) so the critical points are 0 and. x 0 4 f ( x) Absolute maximum is at x = 4.. D The maximum acceleration will occur when its derivative changes from positive to negative or at an endpoint of the interval. at ()= v () t = t 6t+ =(t t+4) which is always positive. Thus the acceleration is always increasing. The maximum must occur at t = where a () =. E I. False. The relative maximum could be at a cusp. II. True. There is a critical point at x = c where f (c) exists III. True. If f ( c) > 0, then there would be a relative minimum, not maximum AB4/BC Solution x = (a) f ( ) cos x; + sinx = x= π π, π, cos x ; In [ ] 0 when x π π π f ( ) ( ) () ( x) ln = 0.69 ln ln = 0 absolute maximum value is ln absolute minimum value is 0

3 x = (b) f ( ) ( ( sin x) + sin x) cos xcos x ( + sinx) = sin x ; ( + sinx) f ( x) = 0 when sin x= 7 x = π π, 6 6 sign of f concavity π down + up down 7π π π 6 6 7π π x = a nd since concavity changes as indicated at these points C A quick way to do this problem is to use the effect of the multiplicity of the zeros of f on the graph of y = f( x ). There is point of inflection and a horizontal tangent at x =. There is a horizontal tangent and turning point at x =. There is a horizontal tangent on the interval 4 (,). Thus, there must be critical points. Also, f ( x)= (x ) ( x+ ) (9x 7). 6. B Look at the graph of f (x) on the interval (0,0) and count the number of x-intercepts in the interval. 7. C f ( x) = x +>0. Thus f is increasing for all x. 8. C f is increasing on any interval where f ( x)> 0. f ( x)= 4x + x= x( x +)>0. Since ( x + ) > 0 for all x, f ( x) > 0 whenever x > B f ( x) = x x x e +xe = xe x ( x+ ); f (x) < 0 for < x< C f will be increasing when its derivative is positive. f ( x) = x +6x 9=( x +x ) f ( x) = ( x+ )( x ) > 0 for x< or x>.. 6. B f ( x) x = 6x=x(x ) changes sign from positive to negative only at x = 0.

4 4. A f ( x) = 5x 5x = 5 x ( x ) = 5 x ( x ) (x+ ), changes sign from positive to negative only at x =. So f has a relative maximum at x = only.. B The only place that f (x) changes sign from positive to negative is at x =. 4. C 4. f ( x) = lnx+x ; f ( x ) changes sign from negative to positive only at x = e. x ( e f e )= e =. 5. B The graph of y = x 4 is a parabola that changes from positive to negative at x = and from negative to positive at x =. Since g is always negative, f changes sign opposite to the way y x = 4 does. Thus f has a relative minimum at x = and a relative maximum at x =. 6. B Make a sketch. x < one zero, < x < 5 no zeros, x > 5 one zero for a total of zeros 7. E Students should know what the graph looks like without a calculator and choose option E. Or y = 5( x ) ; y = 5( x ) ; y = 0( x ). y < 0 for x >. 8. C y = x x ; y = x + x 4 ; y = 6x 4 x = 6x 5 ( x ). The only domain value at which there is a sign change in y is x =. Inflection point at x =. 9.. D y = x +0 x; y =x+0; y changes sign at x= 5 0. C Points of inflection occur where f changes sign. This is only at x = 0 and x =. There is no sign change at x =. 7.. B y ( x) = x +ax+b, y (x) = 6x+a, y () = 0 a= y() = 6 so, 6 =+ a+ b 4 6 = + b 4 b=0 4.. C Inflection point will occur when f changes sign. f ( x) = 5x 0x f ( x) = 60x 60x = 60x ( x ). The only sign change is at x =.. D f ( x)= x /. This does not exist at x = 0. D is false, all others are true.

5 4. C Consider the cases: I. false if f x ( ) = II. This is true by the Mean Value Theorem III. false if the graph of f is a parabola with vertex at Only II must be true. x = a+ b AB4 Solution (a) y = 0 x x ( ) (b) f ( x) = xe + e =e x ( x) There is a critical point at x = where f (x) has a relative maximum since f ( x)> 0 for x < and f ( x) < 0 for x >. x (c) f ( x) = e ( ) + ( e x x )( x) = e (x ) (d) The graph of f is concave down when: x e ( x ) < 0 x < 0 x <

6 6. 99 AB Solution (a) ( ) f x =x x 0 + ( x )( x ) = f ( x) = 0 at x= and x= s ign of f + + f is increasing on, [, ) (b) f ( x)= 6x 0 sign of f + 5 The graph is concave down on, 5 (c) From (a), f has its relative minimum at x =, so f =() 5() +() () + k = 9+k = k = 0

7 7. 99 AB4/BC Solution x = (a) f ( ) cos x; + sinx = x= π π, π, cos x ; In [ ] 0 when x π π π f ( ) ( ) () ( x) ln = 0.69 ln ln = 0 absolute maximum value is ln absolute minimum value is 0 x = (b) f ( ) ( ( sin x) + sin x) cos xcos x ( + sinx) sin x = ; ( + sinx) f ( x) = 0 when sin x= 7 x = π π, 6 6 sign of f concavity π down + up down 7π π π 6 6 7π π x = a nd since concavity changes as indicated at these points 6 6

8 Calculus AB Scoring Guidelines. Let f be the function given by f(x) = xe x. (a) Find lim f(x) and lim f(x). x x (b) Find the absolute minimum value of f. Justify that your answer is an absolute minimum. (c) What is the range of f? (d) Consider the family of functions defined by y = bxe bx, where b is a nonzero constant. Show that the absolute minimum value of bxe bx is the same for all nonzero values of b. (a) lim x xex = 0 lim x xex = or DNE { : 0 as x : or DNE as x (b) f (x) = e x + x e x = e x ( + x) = 0 if x = / f( /) = /e or 0.68 or 0.67 /e is an absolute minimum value because: (i) f (x) < 0 for all x < / and f (x) > 0 for all x > / or f + (ii) (x) / : solves f (x) = 0 : evaluates f at student s critical point 0/ if not local minimum from student s derivative : justifies absolute minimum value 0/ for a local argument 0/ without explicit symbolic derivative Note: 0/ if no absolute minimum based on student s derivative and x = / is the only critical number (c) Range of f = [ /e, ) or [ 0.67, ) or [ 0.68, ) (d) y = be bx + b xe bx = be bx ( + bx) = 0 if x = /b At x = /b, y = /e y has an absolute minimum value of /e for all nonzero b : answer Note: must include the left hand endpoint; exclude the right hand endpoint : sets y = be bx ( + bx) = 0 : solves student s y = 0 : evaluates y at a critical number and gets a value independent of b Note: 0/ if only considering specific values of b

9 BC Solution x x (a) f ( x)=(x ) e + xe =e x (x+)(x ) f ( x ) = 0 for x=,. Therefore f is increasing for x < and x >. x x x (b) f ( x) = (x +x)e +(x+) e =e ( x +4x ) 4± 0 f ( x) = 0 for x= The points of inflection occur at x = ± 5. (c) f ( x) has a relative minimum at x =; f() = e. f ( x) has a relative maximum at x = and limit f( x) = 0 x So, f has an absolute minimum at x =, y = e.

10 0. B dy dx > 0 y is increasing; d y < 0 graph is concave down dx. This is only on b < x < c.. D From the graph f () = 0. Since f () represents the slope of the graph at x =, f () > 0. Also, since f () represents the concavity of the graph at x =, f () <0.. A From the graph it is clear that f is not continuous at x = a. All others are true.. A The graph of the derivative would have to change from positive to negative. This is only true for the graph of f. 4. C Look for concavity changes, there are. 5. E Graphs A and B contradict f < 0. Graph C contradicts f (0) does not exist. Graph D contradicts continuity on the interval [,]. Graph E meets all given conditions.

11 AB6 Solution (a) f has a relative maximum at x = because: f changes from positive to negative at x = or f changes from increasing to decreasing at x = or f ( ) =0 and f ( ) <0 f has a relative minimum at x = 0 because: f changes from negative to positive at x = 0. or f changes from decreasing to increasing at x = 0. or f (0) = 0 an df (0 ) > 0 (b) f is concave up on (,) and (,) because: f is increasing on those intervals or f > 0 on those intervals (c)

12 π π π π f f sin sin D Want c so that f ( c) = = =. π π π π c f ()= c cos = 0 c=π 8. B f (0) = 0, f() =0, f ( x) = x 6 x ; by the Mean Value Theorem, f ( c)= f () f(0) = 0 for c (0,). So, 0= c 6 c=c c. The only value in the open interval is. ( ) 9. D f ( x) = x 4x, f(0) = 0 and f() = 0. By the Mean Value Theorem, f() f(0) 0= = f ( c) = c 4c for c ( 0, ). So, c = AB Solution f x x (a) ( ) = x ( x )(x )( = x+ x=, x=, x= f x = x 7 f = 4, f = (b) ( ) ( ) ( ) y = 4( x+ ) or 4x+ y = 8 or y = 4x+ 8 ) (c) f f ) 0 = = 6 c 7= f c = 6 () ( c = ( )

13 4. B. f changes sign from positive to negative at x = and therefore f changes from increasing to decreasing at x =. Or f changes sign from positive to negative at x = and from negative to positive at x =. Therefore f has a local maximum at x = and a local minimum at x =. 4. A The graph shows that f is increasing on an interval ( ac), and decreasing on the interval (, cb), where a< c<b. This means the graph of the derivative of f is positive on the interval ( ac), and negative on the interval (, cb), so the answer is (A) or (E). The derivative is not (E), however, since then the graph of f would be concave down for the entire interval. 4. D II does not work since the slope of f at x = 0 is not equal to f (0). Both I and III could work. For example, f ( x)= e x in I and f ( x) = sin x in III. 44. E The graph of h has turning points and one point of inflection. The graph of h will have x-intercepts and one turning point. Only (C) and (E) are possible answers. Since the first turning point on the graph of h is a relative maximum, the first zero of h must be a place where the sign changes from positive to negative. This is option (E). 45. C From the given information, f is the derivative of g. We want a graph for f that represents the slopes of the graph g. The slope of g is zero at a and b. Also the slope of g changes from positive to negative at one point between a and b. This is true only for figure (C). 46. D Volume =πr h=6π h=6r. A= πrh+πr = π(6r +r ) da = π 6 r ( + r ) = 4πr ( 8+ r ); dr da < 0 dr for 0< r < and da dr > 0 for r > The minimum surface area of the can is when r = h = 4.

14 B A= (x)( y) = 4xy and y = 4 4 x. 9 So A= 8x x. 9 A = 8 x + x x x = x 9 9 (9 x ) A = 0 at x =±, the largest area is. The maximum area occurs when A= 4xy = 4 = x = and y =. The value of 48. E v=πr h and h+ πr =0 v= π(5 r πr ) for 0 < r < 5 ; dv = 6πr( 0 πr ). The π dr maximum volume is when r = 0 because dv > 0 on 0, 0 and dv <0 on 0,. π dr π dr π π 49. D f ( x) = x, f () =, and f() =, so an equation for the tangent line is y = x. The difference between the function and the tangent line is represented by ( x ). Solve ( x ) <0.5. This inequality is satisfied for all x such that 0.5 < x < This is the same as.9 < x <.707. Thus the largest value in the list that satisfies the inequality is B Let f ( x) = x + x. Then Newton s method (which is no longer part of the AP Course Description) gives f x n + x n = x n f x n xn + xn = ( n) x n + x ( ) x + = = + 4 x = = =

15 5. B x + y = z, take the derivative of both sides with respect to t. x dx + y dy = z dz dt dt dt Divide by and substitute: 4 dx dx dx + = 5 = dt dt dt 5. D Let x and y represent the horizontal and vertical sides of the triangle formed by the ladder, the wall, and the ground. + y = 5; x dy dx dx x + y = 0; (4) + (7)( ) =0; = 7. dt dt dt dt 8 5. E Let y = PR and x=rq. + y dx dy dy dy x = 40, x + y = 0, x + y = 0 y= x. dt dt 4 dt dt Substitute into x + y =40. x + x = 40, 6 6 x = 40, x= x+ S S 54. D = x = S 8 dx ds 4 = = 4 = dt dt D The area of a triangle is given by A= bh. Taking the derivative with respect to t of both da db dh sides of the equation yields = h+ b. Substitute the given rates to get dt dt dt da = (h b) = (h b). The area will be decreasing whenever da < 0. This is true dt dt whenever b> h. 56. D A=πr and C= π r; da =πr dr and dc =π dr dt dt dt dt. For da dt = dc, r =. dt 57. B A=πr da =πr dr dt dt. However, C dr = πr and dt = 0.. Thus da = 0.C. dt

16 58. A A=πr, A= 64 π when r = 8. Take the derivative with respect to t. da dr = πr ; 96 π= π( 8) dr dr =6 dt dt dt dt 59. C V = πr h, dv rh dr +r dh (6)(9) 4 dt = π = π dt dt +6 = π AB5/BC Solution (a) 4 V =πr h+ πr 44π=π() h + 4 π() h = At this instant, the height is centimeters. (b) dv dh =πr dr + πr h + 4πr dr dt dt dt dt dh 6π=π() +π ()()() +4π() () dt dh = 5 dt At this instant, the height is increasing at the rate of 5 centimeters per minute.

17 BC B x y A θ 00 C The figure above represents an observer at point A watching balloon B as it rises from point C. The balloon is rising at a constant rate of meters per second and the observer is 00 meters from point C. (a) Find the rate of change in x at the instant when y = 50. (b) Find the rate of change in the area of right triangle BCA at the instant when y = 50. (c) Find the rate of change in θ at the instant when y = 50. (a) x = y +00 x dx = y dy dt dt At y = 50, x = 50 5 and dx dt = 50 = m/s 5 Explicitly: x= y dx y dy + 00 = dt y + 00 dt = = 50 () m/s 5 (b) 00y A= = 50y da dy = 50 = 50 = 50 m/s dt dt (c) tan θ= y 00 sec dθ dy θ = = dt 00 dt 00 dθ = cos θ dt At y = 50, cos θ= and therefore 50 5 dθ = dt 00 5 = 5 radians/sec.