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Unformatted text preview: ch x = 3 from the left is 6 and the derivative as you approach x = 3
from the right is also 6. These two facts imply that f is differentiable at x = 3. The function
is clearly continuous and differentiable at all other values of x.
28. C The graph is a V with vertex at x = 3 . The
integral gives the sum of the areas of the two
triangles that the V forms with the horizontal
axis for x from 1 to 4. These triangles have
areas of 2 and 0.5 respectively. 29. B This limit gives the derivative of the function f ( x) = tan(3x) . f ′( x) = 3sec 2 (3 x) AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 196 1988 Calculus AB Solutions
30. A Shells (which is no longer part of the AB Course
Description) ∑ 2πrh∆x , where r = x, h = e2 x
1
Volume = 2π ∫ xe 2 x dx
0 31. C 32. A 33. A Let y = f ( x) and solve for x.
x
y
x
; xy + y = x ; x( y − 1) = − y ; x =
⇒ f −1 ( x) =
y=
x +1
1− y
1− x
⎛ x⎞
The period for sin ⎜ ⎟ is
⎝2⎠ 2π
= 4π .
1
2 Check the critical points and the endpoints.
f ′( x) = 3 x 2 − 6 x = 3 x( x − 2) so the critical points are 0 and 2.
x
−2 0 2 4
f ( x ) −8 12 8 28 Absolute maximum is at x = 4.
34. D 35. B The interval is x = a to x = c. The height of a rectangular slice is the top curve, f ( x) , minus
the bottom curve, g ( x) . The area of the rectangular slice is therefore ( f ( x) − g ( x))∆x . Set
up a Riemann sum and take the limit as ∆x goes to 0 to get a definite integral.
π⎞
⎛
⎛
⎛π⎞
⎛ π ⎞⎞
4 cos ⎜ x + ⎟ = 4 ⎜ cos x ⋅ cos ⎜ ⎟ − sin x ⋅ sin ⎜ ⎟ ⎟
3⎠
⎝
⎝3⎠
⎝ 3 ⎠⎠
⎝
⎛
1
3⎞
= 4 ⎜ cos x ⋅ − sin x ⋅
⎟ = 2 cos x − 2 3 sin x
⎜
2
2⎟
⎝
⎠ 36. C 3x − x 2 = x ( 3 − x ) > 0 for 0 < x < 3
Average value = ( ) 13
1⎛ 3 2 1 3⎞
2
∫ 0 3x − x dx = 3 ⎜ 2 x − 3 x ⎟
3
⎝
⎠ 3
0 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. = 3
2 197 1988 Calculus AB Solutions
37. D Since e x > 0 for all x, the zeros of f ( x) are the zeros of sin x , so x = 0 , π , 2π . 38. E ∫ ⎜ x ∫1
⎝ ⎛1 ( ln x )2
2
10 x du ⎞
1
⎟ dx = ∫ ln x dx =
u⎠
x ⎛ dx ⎞
⎟ .This is ∫ u du with u = ln x , so the value is
x⎠ ∫ ln x ⎜
⎝ +C f ( x) dx = − ∫ 3 3 40. B x 2 + y 2 = z 2 , take the derivative of both sides with respect to t. 2 x ⋅ 10 Divide by 2 and substitute: 4 ⋅
41. A 42. C 43. B 1 f ( x) dx − ∫ 3 f ( x) dx = 4 − (−7) = 11
dx
dy
dz
+ 2y ⋅
= 2z ⋅
dt
dt
dt dx
1 dx
dx
+ 3⋅
= 5 ⋅1 ⇒
=1
3 dt
dt
dt The statement makes no claim as to the behavior of f at x = 3 , only the value of f for input
arbitrarily close to x = 3 . None of the statements are true.
x
x 1
x
= lim
= lim
= 1.
1 x→∞
1
x→∞ x + 1 x→∞ x
1+
+
xx
x
None of the other functions have a limit of 1 as x → ∞
lim The crosssections are disks with radius r = y where y = Volume = π ∫
44. C f ( x) dx = ∫ 10 ∫3 f ( x) dx ; ∫1 10 39. E 3 −3 y 2 dx = 2π ∫ 3
0 ( ) 1
9 − x2 .
3 1
2π ⎛
1 3⎞
9 − x 2 dx =
⎜ 9x − x ⎟
9
9⎝
3⎠ 3
0 = 4π For I: p (− x) = f ( g (− x) ) = f ( − g ( x) ) = − f ( g ( x) ) = − p( x) ⇒ p is odd.
For II: r (− x) = f (− x) + g (− x) = − f ( x) − g ( x) = − ( f ( x) + g ( x) ) = − r ( x) ⇒ r is odd.
For III: s (− x) = f (− x) ⋅ g (− x) = ( − f ( x) ) ⋅ ( − g ( x) ) = f ( x) ⋅ g ( x) = s( x) ⇒ s is not odd. AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 198 1988 Calculus AB Solutions
45. D ( Volume = π r 2 h = 16π ⇒ h = 16r −2 . A = 2π rh + 2π r 2 = 2π 16 r −1 + r 2 ( ) ( ) ) dA
dA
dA
= 2π −16 r −2 + 2r = 4π r −2 −8 + r 3 ;
< 0 for 0 < r < 2 and
> 0 for r > 2
dr
dr
dr The minimum surface area of the can is when r = 2 ⇒ h = 4 . AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 199 1988 Calculus BC Solutions ∫0 (
1 ) x − x 2 dx = 12 13
x− x
2
3 1 111
− = only.
236 = 1. A 2. D ∫0 3. B 1
11
1
f ( x) = ln x = ln x ; f ′( x) = ⋅ ⇒ f ′′( x) = − 2
2
2x
2x 4. E ⎛ uv ⎞′ (uv′ + u ′v) w − uvw uv′w + u ′vw − uvw′
=
⎜ ⎟=
w2
w2
⎝w⎠ 5. C 1 ( x x2 + 2 ) 2 dx = 0 ( 11 2
x +2
2 ∫0 ) 2 ( 11
( 2 x dx ) = ⋅ x 2 + 2
23 ) 31
0 = ( ) 1 3 3 19
3 −2 =
6
6 lim f ( x) = f (a) for all values of a except 2. x →a lim f ( x) = lim− ( x − 2) = 0 ≠ −1 = f (2) x → 2− x →2 2 y ⋅ y′ − 2 x ⋅ y′ − 2 y = 0 ⇒ y′ = y
y−x 6. C 7. A ∫2 8. A f ′( x) = e x , f ′(2) = e2 , ln e 2 = 2 9. D ∞ dx
x2 = lim ∫ L L →∞ 2 ⎛ 1⎞ L
⎛1 1⎞ 1
= lim ⎜ − ⎟ = lim ⎜ − ⎟ =
2
L→∞ ⎝ x ⎠ 2 L→∞ ⎝ 2 L ⎠ 2...
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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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