AP® Calculus AB AP® Calculus BC Free-Response Questions and Solutions 1989 – 1997

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1989 BC1 Let f be a function such that f ′′(x) = 6x + 8 . (a) Find f (x) if the graph of f is tangent to the line 3 x − y = 2 at the point (0, −2) . (b) Find the average value of f (x) on the closed interval [−1,1] .

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1989 BC1 Solution (a) f ′ ( x ) = 3 x 2 + 8 x + C f ′ (0 ) = 3

C =3 f ( x ) = x3 + 4 x 2 + 3x + d d = −2 f ( x ) = x3 + 4 x 2 + 3x − 2 (b)

1 1 x 3 + 4 x 2 + 3 x − 2 ) dx ( ∫ 1 − 1 − ( −1)

= =

1 1 4 4 3 3 2  x + x + x − 2x  2 4 3 2 

1

−1

1  1 4 3  1 4 3   + + − 2  −  − + + 2   2  4 3 2  4 3 2 

=−

2 3

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1989 BC2

Let R be the region enclosed by the graph of y =

x2 , the line x = 1, and the x-axis. x2 +1

(a) Find the area of R . (b) Find the volume of the solid generated when R is rotated about the y-axis.

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1989 BC2 Solution 1

x2 (a) Area = ⌠ dx  2 ⌡0 x + 1 1 1 =⌠ dx  1− 2 ⌡0 x +1 1

= x − arctan x 0 = 1−

π 4 1

⌠  x2  (b) Volume = 2π  x  2  dx ⌡0  x + 1  1

x dx = 2π ⌠  x− 2 x +1 ⌡0 1

 x2 1  = 2π  − ln x 2 + 1   2 2 0 = π (1 − ln 2 ) or 1/ 2

⌠  y  Volume = π  1 −  dy ⌡0  1− y  = π ( 2 y + ln y − 1 )

1/ 2 0

= π (1 − ln 2 )

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1989 BC3 Consider the function f defined by f (x) = e cos x with domain [0, 2π ] . x

(a) Find the absolute maximum and minimum values of f (x). (b) Find the intervals on which f is increasing. (c) Find the x-coordinate of each point of inflection of the graph of f .

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1989 BC3 Solution (a) f ′ ( x ) = −e x sin x + e x cos x

= e x [cos x − sin x ]

f ′ ( x ) = 0 when sin x = cos x, x =

, 4 4

f (x)

x 0

1

π

2 π /4 e 2 2 5π / 4 e − 2 e 2π

4 5π 4 2π

Max: e 2π ; Min: − (b)

π 5π

f ′( x)

2 5π / 4 e 2

−

+ 0

π 4

+ 5π 4

2π

 π   5π  Increasing on  0,  ,  , 2π   4  4  (c) f ′′ ( x ) = e x [− sin x − cos x ] + e x [cos x − sin x ] = −2e x sin x f ′′ ( x ) = 0 when x = 0, π , 2π

Point of inflection at x = π

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1989 BC4 Consider the curve given by the parametric equations x = 2t 3 − 3t 2 and y = t 3 − 12t (a) In terms of t , find

dy . dx

(b) Write an equation for the line tangent to the curve at the point where t = −1. (c) Find the x- and y-coordinates for each critical point on the curve and identify each point as having a vertical or horizontal tangent.

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1989 BC4 Solution

(a)

dy = 3t 2 − 12 dt dx = 6t 2 − 6t dt (t + 2 )(t − 2 ) dy 3t 2 − 12 t2 − 4 = 2 = 2 = 2t (t − 1) dx 6t − 6t 2t − 2t

(b) x = −5, y = 11 dy 3 =− dx 4 3 y − 11 = − ( x + 5 ) 4 or

3 29 y = − x+ 4 4 4 y + 3 x = 29 (c)

t −2 0 1 2

( x, y )

type

( −28,16 ) (0, 0 ) ( −1, − 11) ( 4, −16 )

horizontal vertical vertical horizontal

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1989 BC5 At any time t ≥ 0 , the velocity of a particle traveling along the x-axis is given by the dx differential equation − 10x = 60e 4 t . dt (a) Find the general solution x(t) for the position of the particle. (b) If the position of the particle at time t = 0 is x = − 8 , find the particular solution x(t) for the position of the particle. (c) Use the particular solution from part (b) to find the time at which the particle is at rest.

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1989 BC5 Solution

(a) Integrating Factor: e

∫

− 10 dt

= e −10t

d xe −10t ) = 60e 4t e −10t ( dt xe −10t = −10e −6t + C x (t ) = −10e 4t + Ce10t or xh (t ) = Ce10t x p = Ae 4t 4 Ae 4t − 10 Ae 4t = 60e 4t A = −10 x (t ) = Ce10t − 10e 4t (b) −8 = C − 10; C = 2

x (t ) = 2e10t − 10e 4t (c)

dx = 20e10t − 40e 4t dt 20e10t − 40e 4t = 0 1 t = ln 2 6 or dx − 10 ( −10e 4t + 2e10t ) = 60e 4t dt 0 + 100e 4t − 20e10t = 60e 4t 1 t = ln 2 6

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1989 BC6 Let f be a function that is everywhere differentiable and that has the following properties. (i) f (x + h) =

f (x) + f (h) for all real numbers h and x . f (− x) + f (− h)

(ii) f (x) > 0 for all real numbers x . (iii) f ′(0) = −1. (a) Find the value of f (0) . (b) Show that f (− x) =

1 for all real numbers x . f (x)

(c) Using part (b), show that f (x + h) = f (x) f (h) for all real numbers h and x . (d) Use the definition of the derivative to find f ′( x) in terms of f ( x) .

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1989 BC6 Solution (a) Let x = h = 0 f (0 ) = f (0 + 0 ) =

f (0 ) + f (0 ) f (0 ) + f (0 )

=1

(b) Let h = 0

f ( x + 0) = f ( x ) =

f ( x ) + f (0 ) f ( − x ) + f ( −0 )

Use f (0 ) = 1 and solve for f ( x ) =

1 f (− x )

or Note that f (− x + 0) =

(c) f ( x + h ) =

=

f (− x) + f (0) is the reciprocal of f(x). f ( x) + f (0)

f ( x ) + f (h) 1 1 + f ( x ) f (h )

f ( x ) + f (h )

f (h ) + f ( x )

f ( x ) f (h)

= f ( x ) f (h)

f ( x + h) − f ( x) h →0 h f ( x ) f (h) − f ( x ) = lim h →0 h f (h) −1 = f ( x ) lim h→0 h = f ( x ) f ′ (0 ) = − f ( x )

(d) f ′ ( x ) = lim

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1990 BC1 A particle starts at time t = 0 and moves along the x-axis so that its position at any time 3 t ≥ 0 is given by x(t) = (t − 1) (2t − 3). (a)

Find the velocity of the particle at any time t ≥ 0 .

(b)

For what values of t is the velocity of the particle less than zero?

(c) Find the value of t when the particle is moving and the acceleration is zero.

1990 BC1 Solution (a) v (t ) = x′ (t ) = 3 (t − 1) ( 2t − 3) + 2 (t − 1) 2

3

= (t − 1) (8t − 11) 2

(b) v (t ) < 0 when (t − 1) (8t − 11) < 0 2

Therefore 8t − 11 < 0 and t ≠ 1 or t