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Unformatted text preview: + 3 x −4 ; y′′ = 6 x −4 − 12 x −5 = 6 x −5 ( x − 2) . The only domain value at
which there is a sign change in y′′ is x = 2 . Inflection point at x = 2 . 22. E ∫ 23. C A quick way to do this problem is to use the effect of the multiplicity of the zeros of f on the
graph 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)3 ( x + 2) 4 (9 x − 7) . 1
2 x − 2x + 2 dx = ∫ ( 1
2 ( x − 2 x + 1) + 1 ) − 1
3 1 dx = ∫ ( x − 1) 2 +1 dx = tan −1 ( x − 1) + C 24. A 2
f ′( x) = x 2 − 2 x − 1
3 25. C dx
(2 ) = 2 x ⋅ ln 2
dx 26. D v(t ) = 4sin t − t ; a (t ) = 4 cos t − 1 = 0 at t = cos −1 (1 4) = 1.31812; v(1.31812) = 2.55487 27. C f ′( x) = 3 x 2 + 12 > 0 . Thus f is increasing for all x. 28. B ∫1 500 (13x − 11x ) dx + ∫ 500
2 ( 2x − 2) , 2
4
f ′(0) = ⋅ (−1) ⋅ (−2) =
3
3 (11x − 13x ) dx = ∫ ⎛ 13x 11x ⎞
2
= ∫ (13x − 11x ) dx = ⎜
−
⎜ ln13 ln11 ⎟
⎟
1
⎝
⎠ 2
1 = 500
1 (13x − 11x ) dx − ∫ 500
2 (13x − 11x ) dx 132 − 13 112 − 11
−
= 14.946
ln13
ln11 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 208 1993 Calculus AB Solutions
29. C Use L’Hôpital’s Rule (which is no longer part of the AB Course Description).
lim θ→0 1 − cos θ
2 2sin θ sin θ
1
1
= lim
=
θ→0 4sin θ cos θ θ→0 4 cos θ
4 = lim A way to do this without L’Hôpital’s rule is the following
lim θ→0 30. C 1 − cos θ
2sin 2 θ = lim 1 − cos θ θ→0 2(1 − cos 2 θ) 1 − cos θ
1
1
= lim
=
θ→0 2(1 − cos θ)(1 + cos θ) θ→0 2(1 + cos θ)
4 = lim Each slice is a disk whose volume is given by π r 2 ∆x , where r = x .
3 3 0 0 Volume = π∫ ( x ) 2 dx = π ∫ x dx =
2 π2
x
2 3
0 = 9
π.
2 6 31. E f ( x) = e3ln( x ) = eln( x ) = x6 ; f ′( x) = 6 x5 32. A ∫
∫0 33. B ⎛u⎞
= sin −1 ⎜ ⎟ + C , a > 0
⎝a⎠
a2 − u 2
du 3 ⎛ x⎞
= sin −1 ⎜ ⎟
2
⎝2⎠
4− x
dx 3
0 ⎛ 3⎞
π
−1
= sin −1 ⎜
⎜ 2 ⎟ − sin (0) = 3
⎟
⎝
⎠ 1
−1
= 2x + C ; y =
. Substitute the point (1, −1)
y
2x + C
−1
1
1
⇒ C = −1, so y =
. When x = 2, y = − .
to find the value of C. Then −1 =
2+C
1− 2x
3
Separate the variables. y −2 dy = 2dx ; − 34. D Let x and y represent the horizontal and vertical sides of the triangle formed by the ladder, the
wall, and the ground.
dx
dy
dx
dx 7
x 2 + y 2 = 25; 2 x + 2 y
= 0; 2(24) + 2(7)(−3) = 0;
=.
dt
dt
dt
dt 8 35. E For there to be a vertical asymptote at x = −3 , the value of c must be 3. For y = 2 to be a
horizontal asymptote, the value of a must be 2. Thus a + c = 5 . 36. D Rectangle approximation = e0 + e1 = 1 + e ( ) Trapezoid approximation. = 1 + 2e + e 4 / 2 .
Difference = (e 4 − 1) / 2 = 26.799 . AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 209 1993 Calculus AB Solutions
37. C I and II both give the derivative at a. In III the denominator is fixed. This is not the derivative
of f at x = a . This gives the slope of the secant line from ( a , f (a ) ) to ( a + h , f (a + h) ) . 38. A 1
f ′( x) = x 2 − sin x + C , f ( x) = x3 + cos x + Cx + K . Option A is the only one with this form.
3 39. D A = π r 2 and C =2π r ; 40. C The graph of y = f
x > 0 , x and x dA
dr
dC
dr
dA dC
= 2π r
and
= 2π . For
=
, r = 1.
dt
dt
dt
dt
dt
dt ( x ) is symmetric to the yaxis. This leaves only options C and E. For
are the same, so the graphs of f ( x ) and f ( x ) must be the same. This is option C.
41. D Answer follows from the Fundamental Theorem of Calculus.
t 42. B ⎛ 3.5 ⎞ 2
This is an example of exponential growth. We know from precalculus that w = 2 ⎜
⎟ is
⎝2⎠
an exponential function that meets the two given conditions. When t = 3 , w = 4.630 . Using
calculus the student may translate the statement “increasing at a rate proportional to its
weight” to mean exponential growth and write the equation w = 2e kt . Using the given conditions, 3.5 = 2e
43. B t⋅
ln(1.75)
; w = 2e
; ln(1.75) = 2k ; k =
2 ln(1.75)
2
. When t = 3 , w = 4.630 . Use the technique of antiderivative by parts, which is no longer in the AB Course
Description. The formula is ∫ u dv = uv − ∫ v du . Let u = f ( x) and dv = x dx. This leads to
1 ∫ x f ( x) dx = 2 x
44. C 2k 2 f ( x) − 1
x 2 f ′( x) dx .
2∫ 1
f ′( x) = ln x + x ⋅ ; f ′( x) changes sign from negative to positive only at x = e −1 .
x
1
f (e−1 ) = −e−1 = − .
e AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 210 1993 Calculus AB Solutions
45. B Let f ( x) = x3 + x − 1 . Then Newton’s method (which is no longer part of the AP Course
Descr...
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This note was uploaded on 12/29/2010 for the course MATH 214 taught by Professor Smith during the Fall '10 term at Oregon Tech.
 Fall '10
 smith
 Calculus

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