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

# A d 1 d 1 1 ln 1 x dx ln1 x 1 x

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Unformatted text preview: ly the solution students were expected to produce while solving this problem on the 1973 multiple-choice exam. However, the problem itself is not well-defined. A solution to an initial value problem should be a function that is differentiable on an interval containing the initial point. In this problem that would be the domain x < 0 since the solution requires the choice of the branch of the logarithm function with x < 0 . Thus one cannot ask about the value of the function at x = e −1. 45. E F ′ ( x ) = xg ′ ( x ) with x ≥ 0 and g ′ ( x ) < 0 ⇒ F ′ ( x ) < 0 ⇒ F is not increasing. AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 182 1985 Calculus AB Solutions 2 −3 x dx 1 1 = − x −2 2 2 1⎛1 ⎞ 3 = − ⎜ − 1⎟ = . 1 2⎝4 ⎠ 8 1. D ∫ 2. E f ′( x) = 4(2 x + 1)3 ⋅ 2, f ′′(1) = 4 ⋅ 3(2 x + 1) 2 ⋅ 22 , f ′′′(1) = 4 ⋅ 3 ⋅ 2(2 x + 1)1 ⋅ 23 , f ( 4) (1) = 4!⋅ 24 = 384 3. A y = 3(4 + x 2 ) −1 so y′ = −3(4 + x 2 ) −2 (2 x) = −6 x (4 + x 2 ) 2 ( 4 + x ) (0) − 3(2 x) = −6 x Or using the quotient rule directly gives y′ = (4 + x ) (4 + x ) 2 2 1 2 2 2 1 4. C ∫ cos(2 x) dx = 2 ∫ cos(2 x) (2 dx) = 2 sin(2 x) + C 5. D lim 6. C f ′( x) = 1 ⇒ f ′(5) = 1 7. E ∫1 8. B 1 1 ⎛ x⎞ y = ln ⎜ ⎟ = ln x − ln 2, y′ = , y′(4) = 4 x ⎝2⎠ 9. D Since e − x is even, n→∞ 4 4n 2 2 n + 10000n 4 =4 10000 1+ n = lim n→∞ 1 4 dt = ln t 1 = ln 4 − ln1 = ln 4 t 2 2 −1) 0 ∫ −1 e − x2 ( dx = 1 1 − x2 1 ∫ −1 e dx = 2 k 2 ) 2 d ( x 2 − 1) = 2 x ⋅10( x −1) ⋅ ln(10) dx 10. D y′ = 10( x 11. B v(t ) = 2t + 4 ⇒ a(t ) = 2 ∴ a(4) = 2 12. C f ( g ( x) ) = ln g ( x) 2 = ln x 2 + 4 ⇒ g ( x) = x 2 + 4 ⋅ ln(10) ⋅ ( ) ( ) AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 183 1985 Calculus AB Solutions 13. A 2 x + x ⋅ y′ + y + 3 y 2 ⋅ y′ = 0 ⇒ y′ = − 4 ∫0 2x + y x + 3y2 v ( t ) dt = ∫ 4 3⎞ 5⎞ ⎛1 ⎛3 2 + 5t 2 ⎟ dt = ⎜ 2t 2 + 2t 2 ⎟ ⎜ 3t ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 4 14. D Since v(t ) ≥ 0, distance = 15. C x2 − 4 > 0 ⇒ x > 2 16. B f ′( x) = 3 x 2 − 6 x = 3 x( x − 2) changes sign from positive to negative only at x = 0. 17. C 0 0 = 80 Use the technique of antiderivatives by parts: u=x dv = e− x dx du = dx v = −e − x ( − xe − x + ∫ e − x dx = − xe− x − e− x ) 0 = 1 − 2e − 1 1 18. C y = cos 2 x − sin 2 x = cos 2 x , y′ = −2sin 2 x 19. B Quick solution: lines through the origin have this property. Or, f ( x 1 ) + f ( x 2 ) = 2 x 1 + 2 x 2 = 2( x 1 + x 2 ) = f ( x 1 + x 2 ) 20. A dy 1 d − sin x = ⋅ ( cos x ) = 2 dx 1 + cos x dx 1 + cos 2 x 21. B x > 1 ⇒ x 2 > 1 ⇒ f ( x) < 0 for all x in the domain. lim f ( x) = 0 . lim f ( x) = −∞ . The only x →∞ x →1 option that is consistent with these statements is (B). 2 2 ( x + 1)( x − 1) 2 x2 − 1 1 dx = ∫ dx = ∫ ( x − 1) dx = ( x − 1) 2 1 1 x +1 x +1 2 22. A ∫1 23. B d −3 x − x −1 + x 2 dx 24. D 16 = ∫ ( 2 −2 ) x =−1 ( x 7 + k ) dx = ∫ ( = − 3 x −4 + x − 2 + 2 x 2 −2 x 7 dx + ∫ 2 −2 ) x =−1 2 1 = 1 2 = − 3 + 1 − 2 = −4 k dx = 0 + ( 2 − (−2) ) k = 4k ⇒ k = 4 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 184 1985 Calculus AB Solutions 25. E 26. E f ′(e) = lim h →0 f (e + h ) − f (e ) ee + h − ee = lim h →0 h h I: Replace y with (− y ) : (− y ) 2 = x 2 + 9 ⇒ y 2 = x 2 + 9 , no change, so yes. II: Replace x with (− x) : y 2 = (− x) 2 + 9 ⇒ y 2 = x 2 + 9 , no change, so yes. III: Since there is symmetry with respect to both axes there is origin symmetry. 27. D The graph is a V with vertex at x = 1 . The integral gives the sum of the areas of the two triangles that the V forms with the horizontal axis for x from 0 to 3. These triangles have areas of 1/2 and 2 respectively. 28. C Let x(t ) = −5t 2 be the position at time t. Average velocity = 29. D The tangent function is not defined at x = π 2 so it cannot be continuous for all real numbers. Option E is the only one that includes item III. In fact, the functions in I and II are a power and an exponential function that are known to be continuous for all real numbers x. 30. B ∫ tan(2 x) dx = − 2 ∫ 31. C 1 dV 1 ⎛ dr dh ⎞ 1 ⎛ ⎛1⎞ ⎛ 1 ⎞⎞ V = π r 2h , = π ⎜ 2rh + r 2 ⎟ = π ⎜ 2(6)(9) ⎜ ⎟ + 62 ⎜ ⎟ ⎟ = 24π 3 dt 3 ⎝ dt dt ⎠ 3 ⎝ ⎝2⎠ ⎝ 2 ⎠⎠ 32. D ∫0 33. B f ′ changes sign from positive to negative at x = –1 and therefore f changes from increasing to decreasing at x = –1. 1 π3 x(3) − x(0) −45 − 0 = = −15 3−0 3 −2sin(2 x) 1 dx = − ln cos(2 x) + C cos(2 x) 2 1 sin(3x) dx = − cos(3x) 3 π3 = 0 − 1 2 ( cos π − cos 0 ) = 3 3 Or f ′ changes sign from...
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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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