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

A 3 c a 4 d 1 1 1 3 1 1 3 f x x3 x

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Unformatted text preview: + 3 x −4 ; y′′ = 6 x −4 − 12 x −5 = 6 x −5 ( x − 2) . The only domain value at which there is a sign change in y′′ is x = 2 . Inflection point at x = 2 . 22. E ∫ 23. C A quick way to do this problem is to use the effect of the multiplicity of the zeros of f on the graph of y = f ( x) . There is point of inflection and a horizontal tangent at x = −2 . There is a horizontal tangent and turning point at x = 3 . There is a horizontal tangent on the interval (−2,3) . Thus, there must be 3 critical points. Also, f ′( x) = ( x − 3)3 ( x + 2) 4 (9 x − 7) . 1 2 x − 2x + 2 dx = ∫ ( 1 2 ( x − 2 x + 1) + 1 ) − 1 3 1 dx = ∫ ( x − 1) 2 +1 dx = tan −1 ( x − 1) + C 24. A 2 f ′( x) = x 2 − 2 x − 1 3 25. C dx (2 ) = 2 x ⋅ ln 2 dx 26. D v(t ) = 4sin t − t ; a (t ) = 4 cos t − 1 = 0 at t = cos −1 (1 4) = 1.31812; v(1.31812) = 2.55487 27. C f ′( x) = 3 x 2 + 12 > 0 . Thus f is increasing for all x. 28. B ∫1 500 (13x − 11x ) dx + ∫ 500 2 ( 2x − 2) , 2 4 f ′(0) = ⋅ (−1) ⋅ (−2) = 3 3 (11x − 13x ) dx = ∫ ⎛ 13x 11x ⎞ 2 = ∫ (13x − 11x ) dx = ⎜ − ⎜ ln13 ln11 ⎟ ⎟ 1 ⎝ ⎠ 2 1 = 500 1 (13x − 11x ) dx − ∫ 500 2 (13x − 11x ) dx 132 − 13 112 − 11 − = 14.946 ln13 ln11 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 208 1993 Calculus AB Solutions 29. C Use L’Hôpital’s Rule (which is no longer part of the AB Course Description). lim θ→0 1 − cos θ 2 2sin θ sin θ 1 1 = lim = θ→0 4sin θ cos θ θ→0 4 cos θ 4 = lim A way to do this without L’Hôpital’s rule is the following lim θ→0 30. C 1 − cos θ 2sin 2 θ = lim 1 − cos θ θ→0 2(1 − cos 2 θ) 1 − cos θ 1 1 = lim = θ→0 2(1 − cos θ)(1 + cos θ) θ→0 2(1 + cos θ) 4 = lim Each slice is a disk whose volume is given by π r 2 ∆x , where r = x . 3 3 0 0 Volume = π∫ ( x ) 2 dx = π ∫ x dx = 2 π2 x 2 3 0 = 9 π. 2 6 31. E f ( x) = e3ln( x ) = eln( x ) = x6 ; f ′( x) = 6 x5 32. A ∫ ∫0 33. B ⎛u⎞ = sin −1 ⎜ ⎟ + C , a > 0 ⎝a⎠ a2 − u 2 du 3 ⎛ x⎞ = sin −1 ⎜ ⎟ 2 ⎝2⎠ 4− x dx 3 0 ⎛ 3⎞ π −1 = sin −1 ⎜ ⎜ 2 ⎟ − sin (0) = 3 ⎟ ⎝ ⎠ 1 −1 = 2x + C ; y = . Substitute the point (1, −1) y 2x + C −1 1 1 ⇒ C = −1, so y = . When x = 2, y = − . to find the value of C. Then −1 = 2+C 1− 2x 3 Separate the variables. y −2 dy = 2dx ; − 34. D Let x and y represent the horizontal and vertical sides of the triangle formed by the ladder, the wall, and the ground. dx dy dx dx 7 x 2 + y 2 = 25; 2 x + 2 y = 0; 2(24) + 2(7)(−3) = 0; =. dt dt dt dt 8 35. E For there to be a vertical asymptote at x = −3 , the value of c must be 3. For y = 2 to be a horizontal asymptote, the value of a must be 2. Thus a + c = 5 . 36. D Rectangle approximation = e0 + e1 = 1 + e ( ) Trapezoid approximation. = 1 + 2e + e 4 / 2 . Difference = (e 4 − 1) / 2 = 26.799 . AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 209 1993 Calculus AB Solutions 37. C I and II both give the derivative at a. In III the denominator is fixed. This is not the derivative of f at x = a . This gives the slope of the secant line from ( a , f (a ) ) to ( a + h , f (a + h) ) . 38. A 1 f ′( x) = x 2 − sin x + C , f ( x) = x3 + cos x + Cx + K . Option A is the only one with this form. 3 39. D A = π r 2 and C =2π r ; 40. C The graph of y = f x > 0 , x and x dA dr dC dr dA dC = 2π r and = 2π . For = , r = 1. dt dt dt dt dt dt ( x ) is symmetric to the y-axis. This leaves only options C and E. For are the same, so the graphs of f ( x ) and f ( x ) must be the same. This is option C. 41. D Answer follows from the Fundamental Theorem of Calculus. t 42. B ⎛ 3.5 ⎞ 2 This is an example of exponential growth. We know from pre-calculus that w = 2 ⎜ ⎟ is ⎝2⎠ an exponential function that meets the two given conditions. When t = 3 , w = 4.630 . Using calculus the student may translate the statement “increasing at a rate proportional to its weight” to mean exponential growth and write the equation w = 2e kt . Using the given conditions, 3.5 = 2e 43. B t⋅ ln(1.75) ; w = 2e ; ln(1.75) = 2k ; k = 2 ln(1.75) 2 . When t = 3 , w = 4.630 . Use the technique of antiderivative by parts, which is no longer in the AB Course Description. The formula is ∫ u dv = uv − ∫ v du . Let u = f ( x) and dv = x dx. This leads to 1 ∫ x f ( x) dx = 2 x 44. C 2k 2 f ( x) − 1 x 2 f ′( x) dx . 2∫ 1 f ′( x) = ln x + x ⋅ ; f ′( x) changes sign from negative to positive only at x = e −1 . x 1 f (e−1 ) = −e−1 = − . e AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 210 1993 Calculus AB Solutions 45. B Let f ( x) = x3 + x − 1 . Then Newton’s method (which is no longer part of the AP Course Descr...
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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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