Unformatted text preview: e of the normal line is 8.
2
8 6. A y 7. C The slope at x = 3 is 2. The equation of the tangent line is y − 5 = 2( x − 3) . 8. E Points of inflection occur where f ′ changes from increasing to decreasing, or from
decreasing to increasing. There are six such points. 9. A f increases for 0 ≤ x ≤ 6 and decreases for 6 ≤ x ≤ 8 . By comparing areas it is clear that f
increases more than it decreases, so the absolute minimum must occur at the left endpoint,
x = 0. 10. B y = xy + x 2 + 1; y′ = xy′ + y + 2 x; at x = −1, y = 1; y′ = − y′ + 1 − 2 ⇒ y′ = −
∞ 1
x(1 + x 2 ) −2 dx = lim − (1 + x 2 ) −1
L→∞ 2 L 1
2 1
1
1
−
=
L→∞ 4 2(1 + L2 )
4 11. C ∫1 12. A f ′ changes from positive to negative once and from negative to positive twice. Thus one
relative maximum and two relative minimums. 13. B a (t ) = 2t − 7 and v(0) = 6; so v(t ) = t 2 − 7t + 6 = (t − 1)(t − 6). Movement is right then left
with the particle changing direction at t = 1, , therefore it will be farthest to the right at t = 1. 1 = lim AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 222 1997 Calculus BC Solutions: Part A
14. C 3
3
3
so the sum will be S = 2 = 2.4
Geometric Series. r = < 1 ⇒ convergence. a =
3
8
2
1−
8 15. D x = cos3 t , y = sin 3 t for 0 ≤ t ≤ L=∫ π2 0 2 2 π 2 ⎛ dx ⎞
π
⎛ dy ⎞
. L=∫
⎜ ⎟ + ⎜ ⎟ dt
0
2
⎝ dt ⎠ ⎝ dt ⎠ (−3cos 2 t sin t )2 + (3sin 2 t cos t )2 dt = ∫ π2 0 9 cos 4 t sin 2 t + 9sin 4 t cos 2 t dt 16. B eh − 1 1
e h − e0 1
eh − 1 1
= lim
= f ′(0), where f ( x) = e x and f ′(0) = 1. lim
=
h →0 2 h
h →0 2 h
2 h →0 h
2
2 17. B f ( x) = ln(3 − x); f ′( x) = lim 1
1
2
, f ′′( x) = −
, f ′′′( x) =
;
x −3
( x − 3) 2
( x − 3)3 1
1
f (2) = 0, f ′(2) = −1, f ′′(2) = −1, f ′′′(2) = −2; a0 = 0, a1 = −1, a2 = − , a3 = −
2
3
2
3
( x − 2)
( x − 2)
f ( x) ≈ −( x − 2) −
−
2
3
2
dy 4t 3 + 4t − 8 4t 3 + 4t − 8
=
=
. Vertical tangents at t = 0,
2
dx
t (3t − 2)
3
3t − 2t 18. C x = t 3 − t 2 − 1, y = t 4 + 2t 2 − 8t ; 19. D ∫−4 f ( x)dx − 2∫−1 f ( x)dx = ( A1 − A2 ) − 2(− A2 ) = A1 + A2 4 ∞ ∑ 4 ( x − 2) n 20. E . The endpoints of the interval of convergence are when ( x − 2) = ±3; x = −1, 5 .
n ⋅ 3n
n =1
Check endpoints: x = −1 gives the alternating harmonic series which converges. x = 5 gives
the harmonic series which diverges. Therefore the interval is −1 ≤ x < 5 . 21. A Area = 2 ⋅ 22. C g ′( x) = f ( x). The only critical value of g on (a, d ) is at x = c . Since g ′ changes from
positive to negative at x = c , the absolute maximum for g occurs at this relative maximum. π2
1 π2
2
2
2
∫ 0 ((2 cos θ) − cos θ) d θ = ∫ 0 3cos θ d θ
2 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 223 1997 Calculus BC Solutions: Part A
23. E 24. D 25. A x = 5sin θ; dx
dθ
4 dx
⎛4⎞
= 5cos θ ⋅ ; When x = 3, cos θ = ;
= 5 ⎜ ⎟ (3) = 12
dt
dt
5 dt
⎝5⎠ ( x 2 )3
1
1
17
+ ⋅⋅⋅ = x 2 − x6 + ⋅⋅⋅ ⇒ f ( x) = x3 −
x + ⋅⋅⋅ The
3!
6
3
42
1
coefficieint of x7 is − .
42
f ′( x) = sin( x 2 ) = x 2 − This is the limit of a right Riemann sum of the function f ( x) = x on the interval [a, b] , so
n lim ∑ n→∞ i =1 xi ∆x = ∫ b
a 3b 2
x dx = x 2
3 a 3 3 2
= (b 2 − a 2 )
3 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 224 1997 Calculus BC Solutions: Part B
76. D 5
Sequence I → ; sequence II → ∞; sequence III → 1 . Therefore I and III only.
2 77. E Use shells (which is no longer part of the AP Course Description.) ∑ 2πrh∆x where r = x and
h = 4 x − x2 − x
Volume =
3 3 0 0 2π ∫ x(4 x − x 2 − x) dx = 2π ∫ (3x 2 − x3 ) dx 78. A ln(e + h) − 1
ln(e + h) − ln e
= lim
= f ′(e) where f ( x) = ln x
h →0
h →0
h
h 79. D Count the number of places where the graph of y (t ) has a horizontal tangent line. Six places. 80 Find the first turning point on the graph of y = f ′( x) . Occurs at x = 0.93 . B 81. D
82. B 83. E lim f assumes every value between –1 and 3 on the interval (−3, 6) . Thus f (c) = 1 at least once.
13 2 12
x − x ≥ x − 2 . Using the calculator, the greatest x value on
3
2
the interval [0, 4] that satisfies this inequality is found to occur at x = 1.3887 .
x ∫0 x (t 2 − 2t ) dt ≥ ∫ t dt ;
2 dy
= (1 + ln x) dx; ln y = x + x ln x − x + k = x ln x + k ;
y y = ek e x ln x ⇒ y = Ce x ln x . Since y = 1 when x = 1, C = 1 . Hence y = e x ln x . AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 225 1997 Calculus BC Solutions: Part B
84. C
∫x 2 sin x dx ;...
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