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Unformatted text preview: iption) gives
xn+1 = xn − x2 = 1 − f ( xn )
x 3 + x −1
= xn − n 2 n
f ′( xn )
3xn + 1 1+1−1 3
=
3 +1
4
3 ⎛3⎞ 3
+ −1
3 4⎟ 4
59
⎝⎠
x3 = −
=
= 0.686
2
4
86
⎛3⎞
3⎜ ⎟ +1
⎝4⎠ AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 211 1993 Calculus BC Solutions
1 1. A ⎛1 2 1 3⎞
2
∫ 0 ( x − x ) dx = ⎜ 2 x − 3 x ⎟
⎝
⎠ 2. C lim 3. E Q′( x) = p ( x) ⇒ degree of Q is n + 1 4. B If x = 2 then y = 5 . x 5. D r = 2 sec θ ; r cos θ = 2 ⇒ x = 2 . This is a vertical line through the point (2, 0) . 1 2 x2 + 1 − 1
x2 x →0 0 = 111
−=
236 =2 dy
dx
dx
dx
6
+y
= 0; 2(3) + 5 ⋅ = 0 ⇒
=−
dt
dt
dt
dt
5 ⎛ dy ⎞
d⎜ ⎟
2
2
3 dy
dx
dy
dy 3t
dx
= 2t ,
= 3t 2 thus
=
= t;
= ⎝ ⎠=
dt
dt
dx 2t 2 dx 2
dx 6. A 7. A ∫ 8. B f ( x) = ln e 2 x = 2 x, f ′( x) = 2 9. D f ′( x) = ⎛ dy ⎞
d⎜ ⎟
⎝ dx ⎠ 3
dt = 2 = 3
dx
2t 4t
dt 10. E 11. E 1 3 x4
x e dx
0 = 1 1 x4
1 x4
3
∫ 0 e (4 x dx) = 4 e
4 1
0 = 1
( e − 1)
4 21
. This does not exist at x = 0 . D is false, all others are true.
⋅
3 x1 / 3 I. ln x is continuous for x > 0
II. e x is continuous for all x
III. ln(e x − 1) is continuous for x > 0 .
∞ ∫4 −2 x ( 3
dx = lim 9 − x 2
3
b→∞ 2
9 − x2 ) 2 b
3 . This limit diverges. Another way to see this without
4 doing the integration is to observe that the denominator behaves like x 2 3 which has a
smaller degree than the degree of the numerator. This would imply that the integral will
diverge. AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 212 1993 Calculus BC Solutions
12. E v(t ) = 2 cos 2t + 3sin 3t , a (t ) = −4sin 2t + 9 cos 3t , a(π) = −9 . 13. C x
dy
1
= x 2 dx, ln y = x3 + C1 , y = Ce 3 . Only C is of this form.
3
y 1 14. B 15. D 3 The only place that f ′( x) changes sign from positive to negative is at x = −3 . f ′( x) = e tan 2 x ⋅ ( d tan 2 x
dx ) = 2 tan x ⋅ sec2 x ⋅ etan x
2 16. A I. Compare with pseries, p = 2
6
II. Geometric series with r =
7
III. Alternating harmonic series 17. A Using implicit differentiation, y + xy′
y + y′
= 1 . When x = 1,
= 1 ⇒ y′ = 0 .
xy
y x
ex
xe x − e x e ( x − 1)
. y′(1) = 0
Alternatively, xy = e , y = , y′ =
=
x
x2
x2
x 18. B
19. B f ′( x) ⋅ e f ( x ) = 2 x ⇒ f ′( x) = 2x
e f ( x) = 2x
1 + x2 Use cylindrical shells which is no longer part of the AP
Course Description. Each shell is of the form 2π rh∆x
where r = x and h = kx − x 2 . Solve the equation 10 = 2π∫
k=4 20. E k
0 k ⎛ kx3 x 4 ⎞
k4
x (kx − x 2 ) dx = 2π ⎜
− ⎟ = 2π ⋅ .
⎜3
4⎟
12
⎝
⎠0 60
≈ 2.0905 .
π 1
1
v(t ) = − e−2t + 3 and x(t ) = e−2t + 3t + 4
2
4 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 213 1993 Calculus BC Solutions
21. A Use logarithms.
1
1
y′
2x
2
ln y = ln x 2 + 8 − ln ( 2 x + 1) ; =
; at (0, 2), y′ = −1 .
−
2
3
4
y 3 x + 8 4 ( 2 x + 1) ( ) ( ) 22. B f ′( x) = x 2e x + 2 xe x = xe x ( x + 2); f ′( x) < 0 for − 2 < x < 0 23. D L=∫ 24. C This is L’Hôpital’s Rule. 25. D dy
dy dt −te−t + e−t 1 − t
At t = 3, slope =
=
=
= 2t
dx dx
et
e
dt 26. B 27. C 4
0 2 2 4
⎛ dx ⎞ ⎛ dy ⎞
⎜ ⎟ + ⎜ ⎟ dt = ∫ 0
⎝ dt ⎠ ⎝ dt ⎠ 2e 2 x
1 + (e 2 x ) 2 = 4t 2 + 1 dt =−
t =3 2
e6 = −0.005 2e 2 x
1 + e4 x This is a geometric series with r =
convergent for −2 < x < 4 . x −1
. Convergence for −1 < r < 1 . Thus the series is
3 28. A ⎛ 2t + 2
⎞
⎛6 ⎞ ⎛3 ⎞
v=⎜ 2
, 4t ⎟ , v(2) = ⎜ ,8 ⎟ = ⎜ ,8 ⎟
⎝8 ⎠ ⎝4 ⎠
⎝ t + 2t
⎠ 29. E Use the technique of antiderivatives by parts: u = x and dv = sec 2 x dx ∫ x sec
30. C 2 x dx = x tan x − ∫ tan x dx = x tan x + ln cos x + C Each slice is a disk with radius r = sec x and width ∆x .
Volume = π∫ π3 0 sec 2 x dx = π tan x π3
0 =π 3 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 214 1993 Calculus BC Solutions
100 31. A 1⎛ 5+ n ⎞
sn = ⎜
⎟
5⎝ 4+n⎠ 32. B Only II is true. To see that neither I nor III must be true, let f ( x) = 1 and let g ( x) = x 2 − 1
1
, lim sn = ⋅1 =
n→∞
5
5 128
15 on the interval [0, 5].
33. A The value of this integral is 2. Option A is also 2 and none of the others have a value of 2.
Visualizing the graphs of y = sin x and y = cos x is a useful approach to the problem. 34. E Let y = PR and x=RQ .
dx
dy
3 ⎛ dy ⎞
dy
3
x 2 + y 2 = 402 , 2 x + 2 y
= 0, x ⋅ ⎜ − ⎟ + y
=0 ⇒ y = x.
dt
dt
dt
4 ⎝ dt ⎠
4
9
25 2
Substitute into x 2 + y 2 = 402 . x 2 + x 2 = 402 ,
x = 402 , x = 32
16
16 35. A 36. E F (b) − F (a) 0
= = 0 . This means that there is
b−a
a
number in the interval (a, b) for which F ′ is zero. However, F ′( x) =...
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 Calculus

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