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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
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.

Ask a homework question - tutors are online