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Unformatted text preview: ce from ⎜ x , − ⎟ and ⎜ 0, − ⎟ .
⎜
⎟
2⎠
2⎠
⎝
⎝
2 ⎛ x2 1 ⎞
L = ( x − 0) + ⎜ − ⎟
⎜ 2 2⎟
⎝
⎠
⎛ x2 1 ⎞
dL
2L ⋅
= 2x + 2 ⎜ − ⎟ ( x)
⎜ 2 2⎟
dx
⎝
⎠
2 2 ⎛ x2 1 ⎞
2x + 2 ⎜ − ⎟ ( x)
2
⎜ 2 2⎟
2 x + x3 − x x3 + x x x + 1
dL
⎝
⎠
=
=
=
=
2L
2L
2L
2L
dx ( ) dL
dL
< 0 for all x < 0 and
> 0 for all x > 0 , so the minimum distance occurs at x = 0 .
dx
dx
The nearest point is the origin.
12. A 13. C 4
1
⎛4⎞
= 2⎜
⎟ ⇒ x − 1 = 4 x − 2; x =
2x −1
3
⎝ x −1 ⎠
π
⎛ π⎞
⎛
⎞
cos x dx ; sin k − sin ⎜ − ⎟ = 3 ⎜ sin − sin k ⎟
2
⎝ 2⎠
⎝
⎠
π
sin k + 1 = 3 − 3sin k ; 4sin k = 2 ⇒ k =
6
k π2 ∫ −π 2 cos x dx = 3∫ k 14. E y = x5 − 1 has an inverse x = y 5 − 1 ⇒ y = 5 x + 1 15. B The graphs do not need to intersect (eg. f ( x) = −e − x and g ( x) = e − x ) . The graphs could
intersect (e.g. f ( x) = 2 x and g ( x) = x ). However, if they do intersect, they will intersect no
more than once because f ( x) grows faster than g ( x) . 16. B y′ > 0 ⇒ y is increasing; y′′ < 0 ⇒ the graph is concave down . Only B meets these
conditions. 17. B y′ = 20 x3 − 5 x 4 , y′′ = 60 x 2 − 20 x3 = 20 x 2 ( 3 − x ) . The only sign change in y′′ is at x = 3 .
The only point of inflection is (3,162). AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 161 1969 Calculus AB Solutions
18. E There is no derivative at the vertex which is located at x = 3 . 19. C dv 1 − ln t
dv
= 2 > 0 for 0 < t < e and
< 0 for t > e , thus v has its maximum at t = e .
dt
dt
t 20. A y (0) = 0 and y′(0) = y= 1 2
x2
1−
4 x =0 1 = 4 − x2 x =0 = 1
. The tangent line is
2 1
x ⇒ x − 2y = 0 .
2 21. B f ′ ( x ) = 2 x − 2e−2 x , f ′ ( 0 ) = −2 , so f is decreasing 22. E ln e2 x = 2 x ⇒
2 2x
e
0 ( 2 1 2x
e
2 23. C ∫ 24. C y = ln sin x , y′ = 25. A 26. C 27. C 2m 1 ∫m x dx = ) d
d
ln e2 x = ( 2 x ) = 2
dx
dx dx = ln x 0 = ( ) 14
e −1
2 cos x
= cot x
sin x 2m m = ln ( 2m ) − ln ( m ) = ln 2 so the area is independent of m. 11
1
( x − 1)2 =
∫0
0
0
02
2
Alternatively, the graph of the region is a right triangle with vertices at (0,0), (0,1), and (1,0).
1
The area is .
2
1 x 2 − 2 x + 1 dx = ∫ 1 1 x − 1 dx = ∫ − ( x − 1) dx = − sin x ∫ tan x dx = ∫ cos x dx = − ln cos x + C = ln sec x + C AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 162 1969 Calculus AB Solutions
28. C 3 cos x + 3sin x can be thought of as the expansion of sin ( x + y ) . Since 3 and 3 are too large for values of sin y and cos y , multiply and divide by the result of the Pythagorean
Theorem used on those values, i.e. 2 3 . Then
⎛3
⎞
⎛1
⎞
3
3
3 cos x + 3sin x = 2 3 ⎜
⎜ 2 3 cos x + 2 3 sin x ⎟ = 2 3 ⎜ 2 cos x + 2 sin x ⎟
⎟
⎜
⎟
⎝
⎠
⎝
⎠
= 2 3 ( sin y cos x + cos y sin x ) = 2 3 sin ( y + x )
⎛1⎞
where y = sin −1 ⎜ ⎟ . The amplitude is 2 3 .
⎝2⎠
Alternatively, the function f ( x) is periodic with period 2π . f ′( x) = − 3 sin x + 3cos x = 0
π 4π
⎛π⎞
when tan x = 3 . The solutions over one period are x = ,
. Then f ⎜ ⎟ = 2 3 and
33
⎝3⎠
⎛ 4π ⎞
f ⎜ ⎟ = −2 3 . So the amplitude is 2 3 .
⎝3⎠
π2 cos x
dx = ln ( sin x )
sin x π2 1
= ln 2
2 29. A ∫π 4 30. E Because f is continuous for all x, the Intermediate Value Theorem implies that the graph of f
must intersect the xaxis. The graph must also intersect the yaxis since f is defined for all x,
in particular, at x = 0. π4 = ln1 − ln 31. C dy
= − y ⇒ y = ce − x and 1 = ce−1 ⇒ c = e ; y = e ⋅ e− x = e1− x
dx 32. B If a < 0 then lim y = ∞ and lim y = −∞ which would mean that there is at least one root.
x→−∞ x→∞ If a > 0 then lim y = −∞ and lim y = ∞ which would mean that there is at least one root.
x→−∞ x→∞ In both cases the equation has at least one root.
33. A 12 3 2
1⎛ 3 4 1 3⎞
∫ −1 3t − t dt = 3 ⎜ 4 t − 3 t ⎟
3
⎝
⎠ 34. D y′ = − 1
x 2 1 ⎛⎛
8 ⎞ ⎛ 3 1 ⎞ ⎞ 11
= ⎜ ⎜12 − ⎟ − ⎜ + ⎟ ⎟ =
−1 3 ⎝ ⎝
3 ⎠ ⎝ 4 3 ⎠⎠ 4
2 , so the desired curve satisfies y′ = x 2 ⇒ y = AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 13
x +C
3 163 1969 Calculus AB Solutions
35. A a ( t ) = 24t 2 , v(t ) = 8t 3 + C and v(0) = 0 ⇒ C = 0. The particle is always moving to the
right, so distance = 36. B ∫0 8t 3dt = 2t 4 y = 4 + sin x , y (0) = 2, y′(0) =
L( x ) = 2 + 37. D 2 2
0 = 32 . cos 0
1
= . The linear approximation to y is
2 4 + sin 0 4 1
1
x . L(1.2) = 2 + (1.2) = 2.03
4
4 All options have the same value at x = 0 . We want the one that has the same first and...
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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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