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Unformatted text preview: ; y′(4,3) = − ;
3
25
⎛ 4⎞ ⎛ 4⎞
x + y ⋅ y′ = 0 ⇒ 1+y ⋅ y′′ + y′ ⋅ y′ = 0; 1 + (3) y′′ + ⎜ − ⎟ ⋅ ⎜ − ⎟ = 0; y′′ = −
27
⎝ 3⎠ ⎝ 3⎠ ∫ π
4
0 ∫ 18. C 13
2
∫ −2 (5 − ( x + 1))dx = 2(4 x − 3 x )
2 The area of the region is given by π
4
0 e tan x
2 cos x
e tan x
2 cos x dx is of the form
dx = e tan x f ( x) = ln x 2 − 1 ; π
4
0 ∫e u du where u = tan x. . = e1 − e0 = e − 1 f ′( x) = 1 d2
2x
( x − 1) = 2
x − 1 dx
x −1
2 ⋅ 15
1
1
∫ −3 cos x dx = 8 (sin 5 − sin(−3)) = 8 (sin 5 + sin 3) ; Note: Since the sine is an odd function,
8
sin(−3) = − sin(3) .
x
is nonexistent since lim ln x = 0 and lim x ≠ 0 .
x→1 ln x
x→1
x→1 21. E lim 22. D f ( x) = ( x 2 − 3)e − x ; f ′( x) = e− x (− x 2 + 2 x + 3) = −e− x ( x − 3)( x + 1); f ′( x) > 0 for − 1 < x < 3 23. A 2 2 0 0 Disks where r = x . V = π ∫ x 2 dy = π ∫ y 4 dy = π5
y
5 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 2
0 = 32π
5 218 1997 Calculus AB Solutions: Part A
24. B Let [ 0,1] be divided into 50 subintervals. ∆x =
Using f ( x) = x , the right Riemann sum 1
1
2
3
; x1 = , x2 = , x3 = , ⋅⋅⋅, x50 = 1
50
50
50
50 50 ∑ f ( xi )∆x is an approximation for i =1 25. A 1 ∫0 x dx . Use the technique of antiderivatives by parts, which was removed from the AB Course
Description in 1998.
u=x
du = dx dv = sin 2 x dx
1
v = − cos 2 x
2 1 1 1 1 ∫ x sin(2 x) dx = − 2 x cos(2 x) + ∫ 2 cos(2 x) dx = − 2 x cos(2 x) + 4 sin(2 x) + C AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 219 1997 Calculus AB Solutions: Part B 76. E f ( x) = e2 x
2e2 x ⋅ 2 x − 2e2 x e 2 x (2 x − 1)
; f ′( x) =
=
2x
4x2
2 x2 77. D y = x3 + 6 x 2 + 7 x − 2 cos x . Look at the graph of y′′ = 6 x + 12 + 2 cos x in the window
[–3,–1] since that domain contains all the option values. y′′ changes sign at x = −1.89 . 78. D F (3) − F (0) = ∫ f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx = 2 + 2.3 = 4.3 3 1 3 0 0 1 ( Count squares for 1 ∫ 0 f ( x)dx ) 79. C The stem of the questions means f ′(2) = 5 . Thus f is differentiable at x = 2 and therefore
continuous at x = 2. We know nothing of the continuity of f ′ . I and II only. 80. A f ( x) = 2e4 x ; f ′( x) = 16 xe4 x ; We want 16 xe4 x = 3. Graph the derivative function and the
function y = 3, then find the intersection to get x = 0.168 . 81. A Let x be the distance of the train from the
dx
crossing. Then
= 60 .
dt
dS
dx
dS x dx
S 2 = x 2 + 702 ⇒ 2 S
= 2x ⇒
=
.
dt
dt
dt S dt
After 4 seconds, x = 240 and so S = 250 .
dS 240
=
(60) = 57.6
Therefore
dt 250 82. B P ( x) = 2 x 2 − 8 x; P′( x) = 4 x − 8; P′ changes from negative to positive at x = 2. P (2) = −8 83. C cos x = x at x = 0.739085. Store this in A. 84. C Cross sections are squares with sides of length y. 2 2 e e 1 2 1 Volume = ∫ y 2 dx = ∫ ln x dx = (x ln x − x)
85. C
86. A A ∫ 0 (cos x − x) dx = 0.400 e
1 = (e ln e − e) − (0 − 1) = 1 Look at the graph of f ′ and locate where the y changes from positive to negative. x = 0.91
f ( x) = x ; f ′( x) = 1
2x ; 1
2c = 2⋅ 1
21 ⇒ c= 1
4 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 220 1997 Calculus AB Solutions: Part B
87. B 1
a(t ) = t + sin t and v(0) = −2 ⇒ v(t ) = t 2 − cos t − 1; v(t ) = 0 at t = 1.48
2 88. E f ( x) = ∫ h( x)dx ⇒ f (a) = 0, therefore only (A) or (E) are possible. But f ′( x) = h( x) and x a therefore f is differentiable at x = b. This is true for the graph in option (E) but not in option
(A) where there appears to be a corner in the graph at x = b. Also, Since h is increasing at
first, the graph of f must start out concave up. This is also true in (E) but not (A).
89. B 90. D 11
T = ⋅ (3 + 2 ⋅ 3 + 2 ⋅ 5 + 2 ⋅ 8 + 13) = 12
22
1
F ( x) = sin 2 x
2
1
F ( x) = cos 2 x
2
1
F ( x) = − cos(2 x)
4 F ′( x) = sin x cos x Yes F ′( x) = − cos x sin x No 1
F ′( x) = sin(2 x) = sin x cos x
2 Yes AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 221 1997 Calculus BC Solutions: Part A
1 1
x 2 dx = 5 3 22 22
x+x
5
3 1 16
15 1. C ∫0 2. E x = e 2t , y = sin(2t ); 3. A f ( x) = 3 x5 − 4 x3 − 3 x; f ′( x) = 15 x 4 − 12 x 2 − 3 = 3(5 x 2 + 1)( x 2 − 1) = 3(5 x 2 + 1)( x + 1)( x − 1) ;
f ′ changes from positive to negative only at x = −1 . 4. C eln x = x 2 ; so xeln x = x3 and 5. C x ( x + 1) dx = ∫ 3
12
x
0 2 f ( x) + dy 2 cos(2t ) cos(2t )
=
=
dx
2e 2 t
e 2t d3
( x ) = 3x2
dx 2 3
= ( x − 1) 2 1
= (16 − x) 2 ; 0 = 1 1
3
1
31
+ e x −2 ; f ′( x) = ( x − 1) 2 + e x −2 ; f ′(2) = + = 2
2
2
2
22
1 −
1
1
y′ = − (16 − x) 2 ; y′(0) = − ; The slop...
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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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