1969, 1973, 1985, 1993, 1997, 1998 AP Multiple Choice Sections, AB and BC, Solutions (620)

2rhx where r x and h 4 x x2 x volume 3 3 0 0 2

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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 Multiple-Choice 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 Multiple-Choice 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 t-axis for 0 ≤ t ≤ 6 , and let A 2 be the area between the graph and the t-axis 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 Multiple-Choice 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 non-negative, 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 right-sided 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.

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