AP Calculus AB AP Calculus BC

116 Pages·2013·1.23 MB·English

®
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 3x− 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)=3x2 +8x+C
f′(0)=3
C =3
f (x)= x3 +4x2 +3x+d
d =−2
f (x)= x3 +4x2 +3x−2
(b) 1 ∫1 (x3 +4x2 +3x−2)dx
1−(−1) −1
11 4 3  1
=  x4 + x3 + x2 −2x
24 3 2 
−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

x2
Let R be the region enclosed by the graph of y = , 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
⌡ x2 +1
0
⌠1 1
= 1− dx
⌡ x2 +1
0
= x−arctanx1
0
π
=1−
4
⌠1  x2 
(b) Volume=2π x dx
⌡  x2 +1
0
⌠1 x
=2π x− dx
⌡ x2 +1
0
 x2 1 1
=2π − ln x2 +1
 2 2 
0
=π(1−ln2)
or
⌠1/2 y 
Volume=π 1− dy
⌡  1− y
0
=π(2y+ln y−1)1/2
0
=π(1−ln2)
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1989 BC3

Consider the function f defined by f(x) =excosx with domain [0,2π].

(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)=−exsinx+excosx
[ ]
=ex cosx−sinx
f′(x)=0 when sinx=cosx, x=π,5π
4 4

( )
x f x

0 1

π
2
eπ/4
4 2
5π 2
− e5π/4
4 2
2π e2π
2
Max:e2π; Min: − e5π/4
2
(b) f′(x) + − +
π 5π 2π
0

4 4
 π 5π 
Increasing on 0, ,  ,2π
 4  4 
(c) f′′(x)=ex[−sinx−cosx]+ex[cosx−sinx]
=−2exsinx
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=2t3−3t2 and y =t3 −12t

dy
(a) In terms of t , find .
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

dy
(a) =3t2 −12
dt
dx
=6t2 −6t
dt
dy 3t2 −12 t2 −4 (t+2)(t−2)
= = =
dx 6t2 −6t 2t2 −2t 2t(t−1)
(b) x=−5, y =11
dy 3
=−

dx 4
y−11=−3(x+5)
4
or
3 29
y =− x+
4 4
4y+3x=29

( )
(c) t x,y type

−2 (−28,16) horizontal
( )
0 0,0 vertical
1 (−1,−11) vertical
2 (4,−16) 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 =60e4t.
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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