AP Calculus BC Chapter 4 – AP Exam Problems

15

Answers

1. A 1988 AB #33 48% 2. D 1998 AB #24 52% 3. E 1998 BC #23 31%

5. C 1993 AB #23 21% 6. B 1998 AB #80 48% 7. C 1993 AB #27 65% 8. C 1998 AB #22 69% 9. B 1993 BC #22 75% 10. C 1998 BC #1 80% 11. B 1985 AB #16 73% 12. A 1985 BC #2 65% 13. B 1993 BC #14 75% 14. C 1993 AB #44 18% 15. B 1998 AB #89 38% 16. B 1988 BC #19 48% 17. E 1988 AB #4 60% 18. C 1993 AB #21 40% 19. D 1998 AB #1 78% 20. C 1998 AB #19 28% 21. B 1988 BC #27 39% 22. C 1998 BC #16 37% 23. D 1993 BC #9 70% 24. C 1988 AB #20 41%

30. B 1988 AB #8 59% 31. D 1998 AB #17 33% 32. A 1998 AB #76 61%

33. A 1998 AB #79 47% 34. C 1985 BC #20 79% 35. E 1985 BC #43 76%

37. D 1993 AB #18 40% 38. B 1985 BC #13 51% 39. D 1988 BC #24 55%

41. B 1985 AB #33 55% 42. A 1998 AB #23 68% 43. D 1988 BC #9 64% 44. E 1998 BC #6 83% 45. C 1998 BC #88 75% 46. D 1988 AB #45 43% 47. B 1988 BC #45 44% 48. E 1993 BC #36 44% 49. D 1998 BC #92 65% 50. B 1993 AB #45 47% 51. B 1988 AB #40 50% 52. D 1993 AB #34 28% 53. E 1993 BC #34 18% 54. D 1988 BC #37 43% 55. D 1998 AB #90 36% 56. D 1993 AB #39 42% 57. B 1998 AB #78 46% 58. A 1985 BC #22 69% 59. C 1985 AB #31 52%

DETAILED ANSWERS to CHAPTER 4

1. A Check the critical points and the endpoints.′f x( ) 3x2= − 6x = 3x(x − 2) so the critical points are 0 and 2.

x( )2 0 2 48 12 8 28

xf

−−

Absolute maximum is at x = 4.

2. D The maximum acceleration will occur when its derivative changes from positive to negativeor at an endpoint of the interval. a t( ) = v′ t( ) 3t= − 6t 1+ 2 = 3(t2 2 − 2t + 4) which is always positive. Thus the acceleration is always increasing. The maximum must occur at t = 3 where (3a ) = 21

3. E I. False. The relative maximum could be at a cusp.II. True. There is a critical point at x = c where f ′(c) existsIII. True. If ′′(f c) > 0 , then there would be a relative minimum, not maximum

1993 AB4/BC3 Solution

(a) ( )

In [ ] 0 when = = 3x π ;2

f 1 cos ;x+2 sin x

, 2π π , cos x

′ x =

( )( )( )

ln 2 = 0.693

ln 2

ln 1 = 0

232

πππ

f x( )x

absolute maximum value is ln 2

absolute minimum value is 0

4.

5. C A quick way to do this problem is to use the effect of the multiplicity of the zeros of f on thegraph of =y f ( )x . There is point of inflection and a horizontal tangent at x = −2 . There is a horizontal tangent and turning point at x = 3 . There is a horizontal tangent on the interval −( 2,3) . Thus, there must be 3 critical points. Also, ′f x( ) = (x 3) (− +x 23 4) (9x 7− ) .

6. B Look at the graph of f ′(x) on the interval (0,10) and count the number of x-intercepts in theinterval.

(b) f ( ) ( (x)

+2 s( in x)

( )

2

2

sin x 2) sin+ c) os−x x cos+2 s( in

10 when sin2

7 ,116 6

′′f x x

x = π π

−′′ x =

= −2sin x 1 ;−

= = −

x

up

sign of f ′′

concavity

− −+

πdown down

76π 211

6π π

6 6x = a7π nd 11π since concavity changes as indicated at these points

7. C ′f x( ) 3x2= +12 > 0 . Thus f is increasing for all x.

8. C f is increasing on any interval where ′f x( ) > 0 . ′f ( ) = 4x x x2+ = 2 (x 23 2x +1) > 0 .

Since x2( 1+ >) f0 or all x, (′f x) > 0 whenever x > 0 .

22.2. f x( ) x2e= + 2xx xe = xe (x x 2+ ) ;′ ′f (x) 0< for 2− < x 0<9.

10.. f will be increasing when its derivative is positive.′f x( ) 3x= + 6x −9 = 3(2 2x + 2x −3) f ′( ) = 3(x x 3)(+ −x 1) 0> for x < 3− or x 1> .

B

C

11. 6. ′f x( ) 3x2= − 6x = 3x(x − 2) changes sign from positive to negative only at x = 0.B

12. A f ′( )x 15x 15= − =4 2x 15 (x x2 2 1)− 1= 5 (x x2 1)− (x 1)+ , changes sign from positive tonegative only at x = −1 . So f has a relative maximum at x = −1 only.

13. B The only place that f ′(x) changes sign from positive to negative is at x = −3 .

14. C 4. f 1 ; fx

( ) lnx x′ = + x ⋅ ′( )x changes sign from negative to positive only at x = e−1.

f e( − −1 1e

) = −e = −1 .

15. B The graph of y x2= − 4 is a parabola that changes from positive to negative at x = −2 andfrom negative to positive at x = 2 . Since g is always negative, f ′ changes sign opposite to

the way y x2= − 4 does. Thus f has a relative minimum at x = −2 and a relative maximum at x = 2 .

16. B Make a sketch. x < −2 one zero, − x< <2 5 no zeros, x > 5 one zero for a total of 2 zeros

17. E Students should know what the graph looks like without a calculator and choose option E.Or = −5y x − 2( ) ; =y 5( −x 2− −1 2) ;′ ′ =y −10′ −x 2( ) . <y−3 0 for ′′ x > 2 .

18. C y x= −− −x2 3 y =; 2− x−3 + x3 −4′ ′y; = x6 −′ 4 −1 x2 −5 = x6 −5 x( − 2) . The only domain value atwhich there is a sign change in y′′ is x = 2 . Inflection point at x = 2 .

19. D. y x2= +10 ;x′ ′y = 2x′ +10; y cha′′ nges sign at x = 5−

20. C Points of inflection occur where f ′′ changes sign. This is only at x = 0 and x = −1 . There isno sign change at x = 2.

21. B7. y x( ) 3x2= + 2ax + b ,′ ′y (x) 6= x′ + 2a , y (1) 0= ⇒′′ a = −3(1) = −6 so , 6− =1y a+ + − 4⇒ 6b − =1−3+ − 4⇒b = 0 b

22. C Inflection point will occur when f ′′ changes sign. f ′( ) 15x x= − 204 3 .x′′(f x) 60x= − 603 2x 60= x (2 x 1).− The only sign change is at x = 1.

23. D 1/ 3) = ⋅2 13

′(f xx

. This does not exist at x = 0 . D is false, all others are true.

24. C Consider the cases: I. fa iflse f x( ) = 1

II. This is true by the Mean Value Theorem

III. false if the graph of f is a parabola with vertex at2

x = +a b .

Only II must be true.

1988 AB4 Solution

(a) 0y =

(b) f ′( ) 2= −x xe +− −x x( )e = 2 −e (x 1− )x

There is a critical point at x = 1 where f (x) has a relative maximum since ′f x( ) > 0 for x <1 and ′(f x) < 0 for x >1.

(c) ′′f x( ) 2e= −( 1) (+ 2− −− ex x )(1 x− ) 2= e (x−x 2− )

The graph of f is concave down when:

−xe x2 ( 2− <) 02 0

2x

x− <

<

(d)

25.

1993 AB1 Solution

(a) ( )( )( )

( )

2 10 3

3 1 3

3

′f x = 3x x

= −x x

′f x x

− +

−

= =0 at 1 and x 3=

0

ign of f s ′

in f is increasing on −∞is increas , 1 g on [3,∞)

(b) ′′f x( ) = x6 1−

sign of f ′′ − +

3

++ −

13

3

53

The graph is concave down on 3

−∞, 5

(c) From (a), f has its relative minimum at x = 3, so

( )3 3= ( ) 5( )3 +3( )39 11

20

23f k

k

− += − +k =

=

26.

1993 AB4/BC3 Solution

(a) ( )

In [ ] 0 when = = 3x π ;2

f 1 cos ;x+2 sin x

, 2π π , cos x

′ x =

( )( )( )

ln 2 = 0.693

ln 2

ln 1 = 0

232

πππ

f x( )x

absolute maximum value is ln 2

absolute minimum value is 0

(b) f ( ) ( (x)

+2 s( in x)

( )

2

2

sin x 2) sin+ c) os−x x cos+2 s( in

10 when sin2

7 ,1166

′′f x x

x = π π

−′′ x =

2sin x 1 ;−= −

= = −

x

up

sign of f ′′

concavity

− −+

πdown down

76π 211

6π π

66x = a7π nd 11π since concavity changes as indicated at these points

27.

1998 Calculus AB Scoring Guidelines

2. Let f be the function given by f(x) = 2xe2x.(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 = bxebx, where b is a nonzero constant. Show

that the absolute minimum value of bxebx is the same for all nonzero values of b.

(a) limx→−∞

2xe2x = 0

limx→∞

2xe2x =∞ or DNE

2

{1: 0 as x→ −∞

1: ∞ or DNE as x→∞

(b) f ′(x) = 2e2x + 2x · 2 · e2x = 2e2x(1 + 2x) = 0

if x = −1/2

f(−1/2) = −1/e or −0.368 or −0.367

−1/e is an absolute minimum value because:

(i) f ′(x) < 0 for all x < −1/2 andf ′(x) > 0 for all x > −1/2

–or–

(ii)− +f ′(x)

−1/2

and x = −1/2 is the only criticalnumber

3

1: solves f ′(x) = 0

1: evaluates f at student’s critical point0/1 if not local minimum fromstudent’s derivative

1: justifies absolute minimum value0/1 for a local argument0/1 without explicit symbolicderivative

Note: 0/3 if no absolute minimum based onstudent’s derivative

(c) Range of f = [−1/e,∞)or [−0.367,∞)or [−0.368,∞)

1: answer

Note: must include the left–hand endpoint;exclude the right–hand “endpoint”

(d) y′ = bebx + b2xebx = bebx(1 + bx) = 0

if x = −1/b

At x = −1/b, y = −1/e

y has an absolute minimum value of −1/e forall nonzero b

3

1: sets y′ = bebx(1 + bx) = 0

1: solves student’s y′ = 0

1: evaluates y at a critical numberand gets a value independent of b

Note: 0/3 if only considering specificvalues of b

28.

1988 BC1 Solution

(a) 3)0 for 3,1.

e 2xx xe = ex′f x( ) = (x′(f x x

2 − +) = = −

(x + 3)(x 1− )

Therefore f is increasing for x < −3 and x >1.

(b) ′′f x( ) (x= −3+ 2x)e + (2x + 2)x xe = e (2 2xx + 4x 1− )4 20

2′′f x( ) 0 for = =x − ±

The points of inflection occur at x = − ±2 5 .

(c) (f x) has a relative minimum at x =1; (1) = −2f e . f x( ) has a relative maximum at x = −3 and limit (f x) = 0

x→−∞

So, f has an absolute minimum at x =1, = −2y e .

29.

30. B2

2dy 0> ⇒ y is increasing; d y 0< ⇒ graph is concave downdx dx

. This is only on <b x < .c

32. A From the graph it is clear that f is not continuous at x = a . All others are true.

31. D From the graph f (1) = 0 . Since f ′(1) represents the slope of the graph at x =1, f ′(1) > 0 .Also, since f ′′(1) represents the concavity of the graph at x =1, f ′′(1) <0 .

33. A The graph of the derivative would have to change from positive to negative. This is only truefor the graph of f ′ .

34. C Look for concavity changes, there are 3.

35. E Graphs A and B contradict f ′′ < 0 . Graph C contradicts f ′(0) does not exist. Graph Dcontradicts continuity on the interval −[ 2,3] . Graph E meets all given conditions.

1985 AB6 Solution

(a) f has a relative maximum at x = −2 because: f ′ changes from positive to negative at x = −2

or f changes from increasing to decreasing at x = −2

or ( 2)− = 0 andf f ′ ′ ( 2− )′ < 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 (0) 0 an df f′ ′ (0= >) 0′

(b) f is concave up on (−1,1) and (2,3) because: f ′ is increasing on those intervals

or f ′′ > 0 on those intervals

(c)

36.

37. D Want c so that 02 2 4 432 2

′(f c

π πf sin π π⎛ ⎞⎜ ⎟ ⎜ ⎟− ⎛f3 ⎞ ⎛ ⎞

⎜ ⎟ sin3 ⎛⎜− ⎞

⎟) ⎝ ⎠ ⎝= =⎠ ⎝ ⎠ ⎝ ⎠ =

π π π π−.

1 cos 02 ⎜ ⎟2

′f c( ) = c⎛ ⎞ = c⇒ = π⎝ ⎠

38. B f (0) 0, (3)= = 0,f f ′( )x 3= 2x 6 ;− x by the Mean Value Theorem,

3f c( ) = f′ (3) − (0)f 0 fo= r c (0,3)∈ .

So, 0 3 2= −c c6 = 3c c −( )2 . The only value in the open interval is 2.

39. D ′(f x) 3x2= − 4x , f (0) 0= and f (2) 0= . By the Mean Value Theorem,(2)

=−0 f f (0) f ′ c c( 2=) c3 − 4

2 0=

−for c∈( 0, 2 ) . So, c = 4

3 .

1989 AB1 Solution

(a) ( )( )(x )(

3 67

211, 2, 3

f x x= − x

x xx = =x x

+

= − − += −

3)

2

(b) ( )( ) ( )

( )

23 7

1 4, 1 1

12 4 1or4 8or

= − +x4 8

f

ff

xy

x + =y

y

′ x x= −′ − = − − =

− = − +

(c) ( )− (f )

( )2

3 1 12 013 2

3 7

133

f

c ′f c

c

= − =−

− = =

=

6

6

40.

41. B3. f ′ changes sign from positive to negative at x = –1 and therefore f changes from increasingto decreasing at x = –1.

O fr ′ changes sign from positive to negative at x = −1 and from negative to positive at x =1. Therefore f has a local maximum at x = −1 and a local minimum at x =1.

42. A The graph shows that f is increasing on an interval ( ,a c) and decreasing on the interval( ,c b) , where a c< < b . This means the graph of the derivative of f is positive on the interval ( ,a c) and negative on the interval ( ,c b) , 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.

43. 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 ex in I and f ( ) = sinx x in III.

44. E The graph of h has 2 turning points and one point of inflection. The graph of h′ will have 2x-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 representsthe 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 = π r h =16π⇒ h =16r2 2. A 2= π rh + 2π r 2= π(162 r− − + r )1 2

dA= π2 1− 6 +r r( )2 = π4− −2 2r (−8+ )r ;3

drdA

< 0dr

for < r <0 2 and dAdr

> 0 for r > 2

The minimum surface area of the can is when =r h⇒ =2 4 .

47. B5. (2A x)(2 )= =y 4xy and 249

= −4y x .

So 2189

A x= −1 x .

1122

12

22

22

1 1 1 28 1 129 9 9

−8 1⎛1 −(9 2x )9 9

A x x ⎟ ⎜− ⎟⎞x

x

−

−

⎞⎞ ⎛ ⎞ ⎛⎛⎛⎜′ = − ⎟ + ⎜x −⎝⎝⎜⎜ ⎟

⎠⎟⎠ ⎝ ⎠ ⎝ ⎠

⎞= ⎟⎜⎝ ⎠

A′ = 0 at 3, 32

x = ± . The maximum area occurs when 32

x = and y = 2. The value of

the largest area is 34 4 2 122

A x= =y ⋅ ⋅ =

48. = π6E = πv r ah nd +h 2π r = 30 ⇒ =v 2π(152 2r − π r ) for 0 < r < 15π

3 ; dv r 1( 0dr

− πr . The)

maximum volume is when r = 10π

,because dv 0 on 0,10⎛ ⎞ and dv> < 0 on 10⎛ 15 .⎞

dr π⎜ ⎟⎝ ⎠ dr ⎜

⎝ π π ⎟⎠

49. D (f x) 2= −x 2,′ ′f (2) = 2, and f (2) = 3 , so an equation for the tangent line is = 2y x −1. The

difference between the function and the tangent line is represented by x −( 2 2) . Solve

x( 2 2)− < 0.5 . This inequality is satisfied for all x such that 2 0.5− < x 2< + 0.5 . This is the same as 1.293 x< < 2.707 . Thus the largest value in the list that satisfies the inequality is 2.7.

50. B Let f ( ) 3x x x= + −1 . Then Newton’s method (which is no longer part of the AP CourseDescription) gives

3

1 2

2

3

3 2

x( ) 1x( ) 3 1

1 1 1 313 4

3 3 13 594 4 0.6864 8633 1

4

n xn + x −nn = −xn = −xn

n n

fxf x

x

x

+ ′ +

+ −1= − =+

⎛ ⎞ + −⎜ ⎟= − ⎝ ⎠ = =

⎛ ⎞ +⎜ ⎟⎝ ⎠

51. B 2 2x y+ 2 ke the derivative of both sides with respect to .t 2x ⋅ dx 2+ y ⋅ dy 2= z ⋅ dzdt dt dt

at,z=

1Divide by 2 and substitute: 4 ⋅ dx+ ⋅3 1

3dx

dt dt dt= ⋅5 ⇒1 dx

=

52. D Let x and y represent the horizontal and vertical sides of the triangle formed by the ladder, thewall, and the ground.

2 y+ =2 25; 2 2 0; 2(24) 2(7)( 3− ) = 0;8

x dx+ y dy dx dxx

dt dt dt dt= + =

7 .

53. E Let =y PR and x=RQ .2 + =y2 3 340 ,2 2 2 0, 0

44x dxx +

dyy x dy ⎞ + dyy = ⇒ =y xdt dt dt dt

= ⋅ ⎛−⎜ ⎟⎝ ⎠

.

+x x =92 2

16 16x2 =Substitu inte to x y2 2+ = 40 . 2 40 ,2 25 40 ,2 x 32=

54. D 328

x S S= ⇒ x = S+

439 3

dx dSdt dt

= = 3 ⋅ 4=

55. D The area of a triangle is given by 12

=A bh . Taking the derivative with respect to t of both

sides of the equation yields 12

dA db= ⋅h b+ dh

⋅⎛ ⎞⎟⎜

⎝dt dt dt ⎠. Substitute the given rates to get

2 2dA (3= −3h b)1 3

= (hdt

)−b . The area will be decreasing whenever dAdt

< 0 . This is true

whenever >b h .

56. D = πA r2 and =C 2π ;r dA= 2π r dr and dC

= 2π drdt dt dt dt

. For dAdt dt

=dC , r =1.

57. B = πA r2 ⇒dA

= 2πr drdt dt

. However, = 2πC r and drdt

= −0.1 . Thus dA= −0.1C

dt.

58. A = πA r2 A =, 6 π4 when r = 8 . Take the derivative with respect to t.

dA= π r ⋅ dr2 ; 9 π6 = π2 ( ⋅8 dr) ⇒

dr= 6

dt dt dt dt

59. C 2 2 2(6)(9)⎛ 2= π1 1 ⎛ ⎞⎞1, 2 243 3 3 2 2

r h dV r drh + r dhVdt dt dt

⎛ ⎞ ⎛ ⎞⎜ ⎟ + 61

⎜ ⎟ = π= π⎜ ⎟⎠= π

1⎜⎝ ⎝ ⎠⎝ ⎝ ⎠⎟⎠

1985 AB5/BC2 Solution

(a) 2 343

= πV r +h πr

(3)2 31443

π = π(3) h + 4π

h = 12

At this instant, the height is 12 centimeters.

(b) dV= π

dh+ πr r drh2 4+ π2 2 drr

dt dt dt dt

261π = π(3) dh (3)(12)(2) (3)2 2(2)dt

+ 2π + 4π

dh= 5

dt

At this instant, the height is increasing at the rate of 5 centimeters per minute.

60.

61.

(a) x y2 2= +100 2

x dx=2 2y dy

dt dt

At y = 50 , x = 50 5 and ⋅3 50 3=

5 m/s550 5

dxdt

=

Explicitly: 2 22 2

21002 100

50 (3)12500

3 5 m/s5

dx y dy =x ydt dty

+ ⇒ =+

=

=

(b) 100 502

50 150 m/s

yA y

dA dydt dt

= =

= 50= 3⋅ =

(c) tan100

θ =y

sec2 1 3100 100

d dydt dtθ

θ = =

3 cos2100

ddtθ= θ

100 At y = 50 , cos50 5

θ = and therefore 23 2 3

100 1255ddtθ ⎛ ⎞= ⎜ ⎟ =

⎝ ⎠ radians/sec.

1988 BC3

xy

A

B

C100

θ

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 3 meters per second and the observer is 100 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 .