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

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Unformatted text preview: iption) gives xn+1 = xn − x2 = 1 − f ( xn ) x 3 + x −1 = xn − n 2 n f ′( xn ) 3xn + 1 1+1−1 3 = 3 +1 4 3 ⎛3⎞ 3 + −1 3 4⎟ 4 59 ⎝⎠ x3 = − = = 0.686 2 4 86 ⎛3⎞ 3⎜ ⎟ +1 ⎝4⎠ AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 211 1993 Calculus BC Solutions 1 1. A ⎛1 2 1 3⎞ 2 ∫ 0 ( x − x ) dx = ⎜ 2 x − 3 x ⎟ ⎝ ⎠ 2. C lim 3. E Q′( x) = p ( x) ⇒ degree of Q is n + 1 4. B If x = 2 then y = 5 . x 5. D r = 2 sec θ ; r cos θ = 2 ⇒ x = 2 . This is a vertical line through the point (2, 0) . 1 2 x2 + 1 − 1 x2 x →0 0 = 111 −= 236 =2 dy dx dx dx 6 +y = 0; 2(3) + 5 ⋅ = 0 ⇒ =− dt dt dt dt 5 ⎛ dy ⎞ d⎜ ⎟ 2 2 3 dy dx dy dy 3t dx = 2t , = 3t 2 thus = = t; = ⎝ ⎠= dt dt dx 2t 2 dx 2 dx 6. A 7. A ∫ 8. B f ( x) = ln e 2 x = 2 x, f ′( x) = 2 9. D f ′( x) = ⎛ dy ⎞ d⎜ ⎟ ⎝ dx ⎠ 3 dt = 2 = 3 dx 2t 4t dt 10. E 11. E 1 3 x4 x e dx 0 = 1 1 x4 1 x4 3 ∫ 0 e (4 x dx) = 4 e 4 1 0 = 1 ( e − 1) 4 21 . This does not exist at x = 0 . D is false, all others are true. ⋅ 3 x1 / 3 I. ln x is continuous for x > 0 II. e x is continuous for all x III. ln(e x − 1) is continuous for x > 0 . ∞ ∫4 −2 x ( 3 dx = lim 9 − x 2 3 b→∞ 2 9 − x2 ) 2 b 3 . This limit diverges. Another way to see this without 4 doing the integration is to observe that the denominator behaves like x 2 3 which has a smaller degree than the degree of the numerator. This would imply that the integral will diverge. AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 212 1993 Calculus BC Solutions 12. E v(t ) = 2 cos 2t + 3sin 3t , a (t ) = −4sin 2t + 9 cos 3t , a(π) = −9 . 13. C x dy 1 = x 2 dx, ln y = x3 + C1 , y = Ce 3 . Only C is of this form. 3 y 1 14. B 15. D 3 The only place that f ′( x) changes sign from positive to negative is at x = −3 . f ′( x) = e tan 2 x ⋅ ( d tan 2 x dx ) = 2 tan x ⋅ sec2 x ⋅ etan x 2 16. A I. Compare with p-series, p = 2 6 II. Geometric series with r = 7 III. Alternating harmonic series 17. A Using implicit differentiation, y + xy′ y + y′ = 1 . When x = 1, = 1 ⇒ y′ = 0 . xy y x ex xe x − e x e ( x − 1) . y′(1) = 0 Alternatively, xy = e , y = , y′ = = x x2 x2 x 18. B 19. B f ′( x) ⋅ e f ( x ) = 2 x ⇒ f ′( x) = 2x e f ( x) = 2x 1 + x2 Use cylindrical shells which is no longer part of the AP Course Description. Each shell is of the form 2π rh∆x where r = x and h = kx − x 2 . Solve the equation 10 = 2π∫ k=4 20. E k 0 k ⎛ kx3 x 4 ⎞ k4 x (kx − x 2 ) dx = 2π ⎜ − ⎟ = 2π ⋅ . ⎜3 4⎟ 12 ⎝ ⎠0 60 ≈ 2.0905 . π 1 1 v(t ) = − e−2t + 3 and x(t ) = e−2t + 3t + 4 2 4 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 213 1993 Calculus BC Solutions 21. A Use logarithms. 1 1 y′ 2x 2 ln y = ln x 2 + 8 − ln ( 2 x + 1) ; = ; at (0, 2), y′ = −1 . − 2 3 4 y 3 x + 8 4 ( 2 x + 1) ( ) ( ) 22. B f ′( x) = x 2e x + 2 xe x = xe x ( x + 2); f ′( x) < 0 for − 2 < x < 0 23. D L=∫ 24. C This is L’Hôpital’s Rule. 25. D dy dy dt −te−t + e−t 1 − t At t = 3, slope = = = = 2t dx dx et e dt 26. B 27. C 4 0 2 2 4 ⎛ dx ⎞ ⎛ dy ⎞ ⎜ ⎟ + ⎜ ⎟ dt = ∫ 0 ⎝ dt ⎠ ⎝ dt ⎠ 2e 2 x 1 + (e 2 x ) 2 = 4t 2 + 1 dt =− t =3 2 e6 = −0.005 2e 2 x 1 + e4 x This is a geometric series with r = convergent for −2 < x < 4 . x −1 . Convergence for −1 < r < 1 . Thus the series is 3 28. A ⎛ 2t + 2 ⎞ ⎛6 ⎞ ⎛3 ⎞ v=⎜ 2 , 4t ⎟ , v(2) = ⎜ ,8 ⎟ = ⎜ ,8 ⎟ ⎝8 ⎠ ⎝4 ⎠ ⎝ t + 2t ⎠ 29. E Use the technique of antiderivatives by parts: u = x and dv = sec 2 x dx ∫ x sec 30. C 2 x dx = x tan x − ∫ tan x dx = x tan x + ln cos x + C Each slice is a disk with radius r = sec x and width ∆x . Volume = π∫ π3 0 sec 2 x dx = π tan x π3 0 =π 3 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 214 1993 Calculus BC Solutions 100 31. A 1⎛ 5+ n ⎞ sn = ⎜ ⎟ 5⎝ 4+n⎠ 32. B Only II is true. To see that neither I nor III must be true, let f ( x) = 1 and let g ( x) = x 2 − 1 1 , lim sn = ⋅1 = n→∞ 5 5 128 15 on the interval [0, 5]. 33. A The value of this integral is 2. Option A is also 2 and none of the others have a value of 2. Visualizing the graphs of y = sin x and y = cos x is a useful approach to the problem. 34. E Let y = PR and x=RQ . dx dy 3 ⎛ dy ⎞ dy 3 x 2 + y 2 = 402 , 2 x + 2 y = 0, x ⋅ ⎜ − ⎟ + y =0 ⇒ y = x. dt dt dt 4 ⎝ dt ⎠ 4 9 25 2 Substitute into x 2 + y 2 = 402 . x 2 + x 2 = 402 , x = 402 , x = 32 16 16 35. A 36. E F (b) − F (a) 0 = = 0 . This means that there is b−a a number in the interval (a, b) for which F ′ is zero. However, F ′( x) =...
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