Unformatted text preview: f ( x) . So, f ( x) = 0 for
some number in the interval (a, b).
Apply the Mean Value Theorem to F. F ′(c) = ( ) 15 dv
;
= 6π r (10 − π r ) . The
π dr
10
dv
dv
⎛ 10 ⎞
⎛ 10 15 ⎞
because
maximum volume is when r =
> 0 on ⎜ 0, ⎟ and
< 0 on ⎜ , ⎟ .
π
dr
dr
⎝ π⎠
⎝π π⎠
v = π r 2 h and h + 2π r = 30 ⇒ v = 2π 15r 2 − π r 3 for 0 < r < e 1 e 1
1
3
dx = + ln e =
x
2
2 37. B ∫ 0 f ( x) dx = ∫ 0 x dx + ∫1 38. C dN
1
= kN ⇒ N = Ce kt . N (0) = 1000 ⇒ C = 1000 . N (7) = 1200 ⇒ k = ln(1.2) . Therefore
dt
7
12 ln(1.2) N (12) = 1000e 7 39. C 40. C Want ≈ 1367 . y (4) − y (1)
ln 4 − ln1 1
1
2
= ln 4 = ln 22 = ln 2 .
where y ( x) = ln x + C . This gives
4 −1
3
3
3
3 1
4
( 0 + 4 ln 2 + 0 ) = ln 2 . Note that
3
3
Simpson’s rule is no longer part of the BC Course Description.
The interval is [0, 2] , x0 = 0, x1 = 1, x2 = 2 . S = AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 215 1993 Calculus BC Solutions
41. C 42. E (
f ′( x) = ( 2 x − 3) e x 2 −3 x ) 2 3
3
and f ′ > 0 for x > .
2
2
3
Thus f has its absolute minimum at x = .
2 ( Suppose lim ln (1 + 2 x )
x →0 ; f ′ < 0 for x < csc x ) = A . The answer to the given question is e
( ) A . ln(1 + 2 x)
2
1
= lim
⋅
= 2.
x→0 sin x
x→0 1 + 2 x cos x Use L’Hôpital’s Rule: lim ln (1 + 2 x)csc x = lim
x →0 x3 x5
+−
3! 5! ⇒ sin x 2 = x 2 − ( x 2 )3 ( x 2 )5
+
−
3!
5! = x2 − x6 x10
+
−
3! 5! 43. A sin x = x − 44. E By the Intermediate Value Theorem there is a c satisfying a < c < b such that f (c) is equal
to the average value of f on the interval [a,b]. But the average value is also given by
b
1
f ( x) dx . Equating the two gives option E.
b − a ∫a
t Alternatively, let F (t ) = ∫ f ( x) dx . By the Mean Value Theorem, there is a c satisfying
a b
F (b) − F (a)
= F ′(c) . But F (b) − F (a ) = ∫ f ( x) dx , and F ′(c) = f (c) by
a
b−a
the Fundamental Theorem of Calculus. This again gives option E as the answer. This result is
called the Mean Value Theorem for Integrals. a < c < b such that 45. D This is an infinite geometric series with a first term of sin 2 x and a ratio of sin 2 x .
π
sin 2 x
The series converges to
= tan 2 x for x ≠ (2k + 1) , k an integer. The answer is
2
2
1 − sin x
2
therefore tan 1 = 2.426. AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 216 1997 Calculus AB Solutions: Part A
1. C 2 ∫1 (4 x3 − 6 x) dx = ( x 4 − 3 x 2 ) f ( x) = 1
x(2 x − 3) 2 ; 2
1 = (16 − 12) − (1 − 3) = 6 1
f ′( x) = (2 x − 3) 2 + x(2 x − 3) − 1
2 = (2 x − 3) − 1
2 (3 x − 3) = (3x − 3)
2x − 3 2. A 3. C ∫a 4. D 1
1
1
= −3 + 1 − 1 = −3
f ( x) = − x3 + x + ; f ′( x) = −3 x 2 + 1 − 2 ; f ′(−1) = −3(−1) 2 + 1 −
x
x
(−1) 2 5. E b b b a a ( f ( x) + 5) dx = ∫ f ( x)dx + 5∫ 1 dx = a + 2b + 5(b − a) = 7b − 4a y = 3 x 4 16 x3 + 24 x 2 + 48; y′ = 12 x3 − 48 x 2 + 48 x; y′′ = 36 x 2 − 96 x + 48 = 12(3x − 2)( x − 2)
2
2
y′′ < 0 for
< x < 2, therefore the graph is concave down for < x < 2
3
3
t t 6. C 12
e dt = e 2 + C
2∫ 7. D d
⎛d
⎞
⎛d
⎞
cos 2 ( x3 ) = 2 cos( x3 ) ⎜ (cos( x3 ) ⎟ = 2 cos( x3 )(− sin( x3 ) ⎜ ( x3 ) ⎟
dx
⎝ dx
⎠
⎝ dx
⎠
= 2 cos( x3 )(− sin( x3 )(3 x 2 ) 8. C The bug change direction when v changes sign. This happens at t = 6 . 9. B Let A1 be the area between the graph and taxis for 0 ≤ t ≤ 6 , and let A 2 be the area between
the graph and the taxis for 6 ≤ t ≤ 8 Then A1 = 12 and A 2 = 1 . The total distance is
A1 + A 2 = 13 . 10. E π⎞
⎛π⎞
⎛π⎞
⎛
y = cos(2 x); y′ = −2sin(2 x); y′ ⎜ ⎟ = −2 and y ⎜ ⎟ = 0; y = −2 ⎜ x − ⎟
4⎠
⎝4⎠
⎝4⎠
⎝ 11. E Since f ′ is positive for −2 < x < 2 and negative for x < −2 and for x > 2, we are looking
for a graph that is increasing for −2 < x < 2 and decreasing otherwise. Only option E. 12. B y= 12
1
1
⎛1 1⎞
x ; y′ = x; We want y′ =
⇒ x = . So the point is ⎜ , ⎟ .
2
2
2
⎝ 2 8⎠ AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 217 1997 Calculus AB Solutions: Part A
4 − x2 ; f is decreasing when f ′ < 0 . Since the numerator is nonnegative, this is
x−2
only when the denominator is negative. Only when x < 2 . 13. A f ′( x) = 14. C f ( x) ≈ L( x) = 2 + 5( x − 3); L( x) = 0 if 0 = 5 x − 13 ⇒ x = 2.6 15. B Statement B is true because lim− f ( x) = 2 = lim+ f ( x) . Also, lim f ( x) does not exist
x →a x→b x→a because the left and rightsided limits are not equal, so neither (A), (C), nor (D) are true.
16. D 17. A 19. D 20. E 2 ⎛ 8 ⎞ 32
= 2⎜8 − ⎟ =
0
⎝ 3⎠ 3 4
x 2 + y 2 = 25; 2 x + 2 y ⋅ y′ = 0; x + y ⋅ y′ = 0...
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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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