This preview shows page 1. Sign up to view the full content.
Unformatted text preview: positive to negative at x = −1 and from negative to positive at
x = 1 . Therefore f has a local maximum at x = −1 and a local minimum at x = 1 .
34. A ⎛
3
3
∫ 0 ( ( x + 8) − ( x + 8) ) dx = ∫ 0 ( x − x ) dx = ⎜
1 1 1 2 1 4⎞
x− x⎟
4⎠
⎝2 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 1
0 = 1
4
185 1985 Calculus AB Solutions
35. D The amplitude is 2 and the period is 2.
y = A sin Bx where A = amplitude = 2 and B = 36. B 2π
2π
=
=π
period 2 II is true since − 7 = 7 will be the maximum value of f ( x) . To see why I and III do not
⎧ 5 if
⎪
have to be true, consider the following: f ( x ) = ⎨− x if
⎪ −7 if
⎩ For f
37. D
38. C x ≤ −5
−5 < x < 7
x≥7 ( x ) , the maximum is 0 and the minimum is –7.
x
=1
x→0 sin x lim x csc x = lim x →0 To see why I and II do not have to be true consider f ( x) = sin x and g ( x) = 1 + e x . Then
f ( x) ≤ g ( x) but neither f ′( x) ≤ g ′( x) nor f ′′( x) < g ′′( x) is true for all real values of x.
III is true, since
f ( x) ≤ g ( x) ⇒ g ( x) − f ( x) ≥ 0 ⇒ 39. E f ′( x) =
2 1 1 1 ∫ 0 ( g ( x) − f ( x) ) dx ≥ 0 ⇒ ∫ 0 f ( x) dx ≤ ∫ 0 g ( x) dx 11 1
1
⋅ − 2 ln x = 2 (1 − ln x) < 0 for x > e . Hence f is decreasing. for x > e .
xx x
x
2 40. D f ( x) dx ≤ ∫ 4 dx = 8 ∫0 41. E Consider the function whose graph is the horizontal line y = 2 with a hole at x = a .
For this function lim f ( x) = 2 and none of the given statements are true. 0 x →a 42. C This is a direct application of the Fundamental Theorem of Calculus: f ′( x) = 1 + x 2 43. B y′ = 3 x 2 + 6 x , y′′ = 6 x + 6 = 0 for x = −1. y′(−1) = −3 . Only option B has a slope of –3. 44. A 122 3
x x +1
2 ∫0 ( ) 1 2 ( )( 11 2
dx = ⋅ ∫ x3 + 1
230 1 2 )( 3 x 2 dx = ) 13
x +1
6 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 3 2 ⋅ 2
3 2
0 = 26
9 186 1985 Calculus AB Solutions
45. A Washers: ∑ π ( R 2 − r 2 ) ∆y Volume = π ∫ 4
0 ( where R = 2, r = x ) 4
1⎞
⎛
22 − x 2 dy = π ∫ (4 − y ) dy = π ⎜ 4 y − y 2 ⎟
0
2⎠
⎝ AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 4
0 = 8π 187 1985 Calculus BC Solutions
2 2 1. D ∫0 2. A f ′( x) = 15 x 4 − 15 x 2 = 15 x 2 ( x 2 − 1) = 15 x 2 ( x − 1)( x + 1) , changes sign from positive to
negative only at x = −1 . So f has a relative maximum at x = −1 only. 3. B 4. D 5. D Area = ∫ 6. E π
1− ⋅ 2
x
tan x − x sec x
⎛π⎞
4
f ( x) =
, f ′( x) =
, f ′⎜ ⎟ =
2
tan x
1
tan x
⎝4⎠ 7. A ∫ 8. C lim 9. B (4 x3 + 2) dx = ( x 4 + 2 x) = (16 + 4) − (1 + 2) = 17
0 x +1 2 ∫1 2 x + 2x dx = 1 2 ( 2 x + 2 ) dx 1
= ln x 2 + 2 x
2 ∫1 x 2 + 2 x
2 2
1 = dx
d 2x
dy
d2y
= 2t and 2 = 2; y (t ) = t 4 − 2t 3 ⇒
= 4t 3 − 6t 2 and
= 12t 2 − 12t
2
dt
dt
dt
dt
⎛ d 2x d 2 y ⎞
a ( t ) = ⎜ 2 , 2 ⎟ = (2, 12t 2 − 12t ) ⇒ a(1) = (2, 0)
⎜ dt
dt ⎟
⎝
⎠
x(t ) = t 2 − 1 ⇒ x2
x1 a ( top curve – bottom curve ) dx , x1 < x2 ; Area = ∫ −1 ( f ( x) − g ( x) ) dx () 2 10. A 1
( ln 8 − ln 3)
2 x →2 ⎛u⎞
du = sin −1 ⎜ ⎟ ⇒ ∫
⎝a⎠
a2 − u2
du 2 = 1− π
2 ⎛ x⎞
dx = sin −1 ⎜ ⎟ + C
⎝5⎠
25 − x 2
dx f ( x) − f (2)
= f ′(2) so the derivative of f at x = 2 is 0.
x−2 Take the derivative of each side of the equation with respect to x.
2 xyy′ + y 2 + 2 xy′ + 2 y = 0 , substitute the point (1, 2)
4
(1)(4) y′ + 22 + (2)(1) y′ + (2)(2) = 0 ⇒ y = −
3
Take the derivative of the general term with respect to x: ∞ ∑ ( −1) n +1 2 n − 2 x n =1 11. A d ⎛ ⎛ 1 ⎞⎞ d
1
⎛ −1 ⎞
⎜ ln ⎜ 1 − x ⎟ ⎟ = dx ( − ln(1 − x) ) = − ⎜ 1 − x ⎟ = 1 − x
dx ⎝ ⎝
⎠⎠
⎝
⎠ AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 188 1985 Calculus BC Solutions
12. A Use partial fractions to rewrite
1⎛1 1 1 ( x − 1)( x + 2 )
1⎞ 1 as ∫ ( x − 1)( x + 2 ) dx= 3 ∫ ⎜ x − 1 − x + 2 ⎟ dx = 3 ( ln
⎝
⎠ 1⎛ 1
1⎞
−
⎜
⎟
3 ⎝ x −1 x + 2 ⎠
1
x −1
+C
x − 1 − ln x + 2 ) + C = ln
3
x+2 13. B f (0) = 0, f (3) = 0, f ′( x) = 3 x 2 − 6 x; by the Mean Value Theorem,
f (3) − f (0)
f ′(c) =
= 0 for c ∈ (0,3) .
3
So, 0 = 3c 2 − 6c = 3c ( c − 2 ) . The only value in the open interval is 2. 14. C I. convergent: pseries with p = 2 > 1
II. divergent: Harmonic series which is known to diverge
1
III. convergent: Geometric with r = < 1
3 15. C x(t ) = 4 + ∫ (2 w − 4) dw = 4 + ( w2 − 4 w) t t 0 0 = 4 + t 2 − 4t = t 2 − 4t + 4 or, x(t ) = t 2 − 4t + C , x(0) = 4 ⇒ C = 4 so, x(t ) = t 2 − 4t + 4 16. C For f ( x ) = 1
x3 2 1−
we have continuity at x = 0 , however, f ...
View
Full
Document
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

Click to edit the document details