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Unformatted text preview: second
derivatives at x = 0 as y = cos 2 x : y′(0) = −2sin 2 x
For y = 1 − 2 x 2 , y′(0) = −4 x
x2 x =0 x =0 = 0 and y′′(0) = −4 cos 2 x 1 − x3
1 − x3
1
2
∫ e (−3x dx) = − 3 e + C = − 3e− x3 + C
3 ∫ ex 39. D x = e ⇒ v = 1, u = 0, y = 0; 40. E One solution technique is to evaluate each integral and note that the value is dx = − ( dy dy du dv
=
⋅ ⋅ = sec 2 u
dx du dv dx Another technique is to use the substitution u = 1 − x ; ⎛⎞
) ⎛1 + v12 ⎞ ⎜ 1 ⎟ = (1)( 2) ( e−1 ) = 2
⎜
⎟x
e
⎝
⎠⎝ ⎠ 1 ∫ 0 (1 − x ) Integrals do not depend on the variable that is used and so
3 = −4. = 0 and y′′(0) = −4 and no other option works. 38. C 3 x =0 f ( x ) dx = ∫ 2 ( ) 3
1⎞
⎛
8 − x 2 dx + ∫ x 2 dx = ⎜ 8 x − x3 ⎟
2
3⎠
⎝ 2 n 1n
u
0 ∫ 13
x
3 1
for each.
n +1 0 1 1 0 dx = ∫ u n ( − du ) = ∫ u n du . du is the same as
3 1n
x
0 ∫ 41. D ∫ −1 42. D y = x3 − 3 x 2 + k , y′ = 3 x 2 − 6 x = 3 x( x − 2) . So f has a relative maximum at (0, k ) and a
relative minimum at (2, k − 4) . There will be 3 distinct xintercepts if the maximum and
minimum are on the opposite sides of the xaxis. We want k − 4 < 0 < k ⇒ 0 < k < 4 . 43. D dx . ∫ sin ( 2 x + 3) dx = − 2 cos ( 2 x + 3) + C −1 −1 + 2 = 27 1 3 1 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 164 1969 Calculus AB Solutions
44. C Since cos 2 A = 2 cos 2 A − 1 , we have 3 − 2 cos 2 2π
=3
⎛ 2π ⎞
⎜⎟
⎝3⎠ expression has period 45. D πx
2π x
= 3 − (1 + cos
) and the latter
3
3 Let y = f ( x3 ) . We want y′′ where f ′( x) = g ( x) and f ′′( x) = g ′( x) = f ( x 2 ) y = f ( x3 )
y′ = f ′( x3 ) ⋅ 3x 2 ( ) y′′ = 3 x 2 f ′′( x3 ) ⋅ 3x 2 + f ′( x3 ) ⋅ 6 x
= 9 x 4 f ′′( x3 ) + 6 x f ′( x3 ) = 9 x 4 f (( x3 ) 2 ) + 6 x g ( x3 ) = 9 x 4 f ( x6 ) + 6 x g ( x3 ) AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 165 1969 Calculus BC Solutions
1. C For horizontal asymptotes consider the limit as x → ±∞ : t → 0 ⇒ y = 0 is an asymptote
For vertical asymptotes consider the limit as y → ±∞ : t → −1 ⇒ x = −1 is an asymptote 2. E y = ( x + 1) tan −1 x , y′ = x +1
1+ x 2 + tan −1 x (1 + x2 ) (1) − ( x + 1)( 2 x ) + 1 = 2 − 2 x
y′′ =
2
2
1 + x2
(1 + x2 )
(1 + x2 ) ( y′′ changes sign at x = 1 only . The point of inflection is 1, π
3. B y = x , y′ = 1
2x . By the Mean Value Theorem we have 2 ) 1
2c = 2
⇒ c = 1.
4 The point is (1,1).
8 dx
dx = 2 1 + x
1+ x 8 4. D ∫0 5. E Using implicit differentiation, 6 x + 2 xy′ + 2 y + 2 y ⋅ y′ = 0 . Therefore y′ = 0 = 2(3 − 1) = 4 −2 y − 6 x
.
2x + 2 y When x = 1, 3 + 2 y + y 2 = 2 ⇒ 0 = y 2 + 2 y + 1 = ( y + 1) 2 ⇒ y = −1
dy
is not defined at x = 1 .
Therefore 2 x + 2 y = 0 and so
dx
6. B 1
This is the derivative of f ( x ) = 8 x8 at x = .
2
7 1
⎛1⎞
⎛1⎞
f ′ ⎜ ⎟ = 64 ⎜ ⎟ =
2
⎝2⎠
⎝2⎠ 7. D 8. C k
k
, we need 0 = f ′(−2) = 1 − and so k = 4. Since f ′′(−2) < 0 for k = 4, f
4
x
does have a relative maximum at x = −2 .
With f ( x) = x + h′( x) = 2 f ( x) ⋅ f ′( x) − 2 g ( x) ⋅ g ′( x) = 2 f ( x) ⋅ ( − g ( x) ) − 2 g ( x) ⋅ f ( x) = −4 f ( x) ⋅ g ( x) AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 166 1969 Calculus BC Solutions
9. D A=
1 1 2π
2 ∫0 x2 ( 3 + cos θ
1 2 dθ = 2 ⋅ 1π
2 ∫0 ( 3 + cos θ ) 2 dθ = ∫ π 0 ( 3 + cos θ ) d θ 1 ⎛ x2 + 1
1⎞
dx = ∫ ⎜ 2
− 2 ⎟ dx = x − tan −1 x
0 ⎜ x +1 x +1⎟
x2 + 1
⎝
⎠ x2 + 1 − 1 ( 10. A ∫0 11. B ) 0 = 1− π = 4 − π
4
4
1 ⎛
x2 ⎞
1⎞
⎛
Let L be the distance from ⎜ x , − ⎟ and ⎜ 0, − ⎟ .
⎜
2⎟
2⎠
⎝
⎝
⎠ x2 + 1 dx = ∫ ) 0 2 ⎛ x2 1 ⎞
L = ( x − 0) + ⎜ − ⎟
⎜ 2 2⎟
⎝
⎠
2
⎛x
1⎞
dL
= 2x + 2 ⎜ − ⎟ ( x)
2L ⋅
⎜ 2 2⎟
dx
⎝
⎠
2 2 ⎛ x2 1 ⎞
2x + 2 ⎜ − ⎟ ( x)
2
⎜ 2 2⎟
dL
2 x + x3 − x x3 + x x x + 1
⎝
⎠
=
=
=
=
dx
2L
2L
2L
2L ( ) 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.
x 2 2 12. E By the Fundamental Theorem of Calculus, if F ( x ) = ∫ e−t dt then F ′ ( x ) = e− x . 13. C ∫ −π 2 cos x dx = 3∫ k 0 k π2 π
⎛ π⎞
⎛
⎞
cos x dx ; sin k − sin ⎜ − ⎟ = 3 ⎜ sin − sin k ⎟
2
⎝ 2⎠
⎝
⎠ sin k + 1 = 3 − 3sin k ; 4sin k = 2 ⇒ k = π 6 dy dy dx
⎛1⎞
=⋅
= ( 2x ) ⎜ ⎟ = x
du dx du
⎝2⎠ 14. D y = x 2 + 2 and u = 2 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...
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 Fall '10
 smith
 Calculus

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