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

The answer is a b 0 2 lim ln 15e x 1 x ln

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Unformatted text preview: positive to negative at x = −1 and from negative to positive at x = 1 . Therefore f has a local maximum at x = −1 and a local minimum at x = 1 . 34. A ⎛ 3 3 ∫ 0 ( ( x + 8) − ( x + 8) ) dx = ∫ 0 ( x − x ) dx = ⎜ 1 1 1 2 1 4⎞ x− x⎟ 4⎠ ⎝2 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 1 0 = 1 4 185 1985 Calculus AB Solutions 35. D The amplitude is 2 and the period is 2. y = A sin Bx where A = amplitude = 2 and B = 36. B 2π 2π = =π period 2 II is true since − 7 = 7 will be the maximum value of f ( x) . To see why I and III do not ⎧ 5 if ⎪ have to be true, consider the following: f ( x ) = ⎨− x if ⎪ −7 if ⎩ For f 37. D 38. C x ≤ −5 −5 < x < 7 x≥7 ( x ) , the maximum is 0 and the minimum is –7. x =1 x→0 sin x lim x csc x = lim x →0 To see why I and II do not have to be true consider f ( x) = sin x and g ( x) = 1 + e x . Then f ( x) ≤ g ( x) but neither f ′( x) ≤ g ′( x) nor f ′′( x) < g ′′( x) is true for all real values of x. III is true, since f ( x) ≤ g ( x) ⇒ g ( x) − f ( x) ≥ 0 ⇒ 39. E f ′( x) = 2 1 1 1 ∫ 0 ( g ( x) − f ( x) ) dx ≥ 0 ⇒ ∫ 0 f ( x) dx ≤ ∫ 0 g ( x) dx 11 1 1 ⋅ − 2 ln x = 2 (1 − ln x) < 0 for x > e . Hence f is decreasing. for x > e . xx x x 2 40. D f ( x) dx ≤ ∫ 4 dx = 8 ∫0 41. E Consider the function whose graph is the horizontal line y = 2 with a hole at x = a . For this function lim f ( x) = 2 and none of the given statements are true. 0 x →a 42. C This is a direct application of the Fundamental Theorem of Calculus: f ′( x) = 1 + x 2 43. B y′ = 3 x 2 + 6 x , y′′ = 6 x + 6 = 0 for x = −1. y′(−1) = −3 . Only option B has a slope of –3. 44. A 122 3 x x +1 2 ∫0 ( ) 1 2 ( )( 11 2 dx = ⋅ ∫ x3 + 1 230 1 2 )( 3 x 2 dx = ) 13 x +1 6 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 3 2 ⋅ 2 3 2 0 = 26 9 186 1985 Calculus AB Solutions 45. A Washers: ∑ π ( R 2 − r 2 ) ∆y Volume = π ∫ 4 0 ( where R = 2, r = x ) 4 1⎞ ⎛ 22 − x 2 dy = π ∫ (4 − y ) dy = π ⎜ 4 y − y 2 ⎟ 0 2⎠ ⎝ AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 4 0 = 8π 187 1985 Calculus BC Solutions 2 2 1. D ∫0 2. A f ′( x) = 15 x 4 − 15 x 2 = 15 x 2 ( x 2 − 1) = 15 x 2 ( x − 1)( x + 1) , changes sign from positive to negative only at x = −1 . So f has a relative maximum at x = −1 only. 3. B 4. D 5. D Area = ∫ 6. E π 1− ⋅ 2 x tan x − x sec x ⎛π⎞ 4 f ( x) = , f ′( x) = , f ′⎜ ⎟ = 2 tan x 1 tan x ⎝4⎠ 7. A ∫ 8. C lim 9. B (4 x3 + 2) dx = ( x 4 + 2 x) = (16 + 4) − (1 + 2) = 17 0 x +1 2 ∫1 2 x + 2x dx = 1 2 ( 2 x + 2 ) dx 1 = ln x 2 + 2 x 2 ∫1 x 2 + 2 x 2 2 1 = dx d 2x dy d2y = 2t and 2 = 2; y (t ) = t 4 − 2t 3 ⇒ = 4t 3 − 6t 2 and = 12t 2 − 12t 2 dt dt dt dt ⎛ d 2x d 2 y ⎞ a ( t ) = ⎜ 2 , 2 ⎟ = (2, 12t 2 − 12t ) ⇒ a(1) = (2, 0) ⎜ dt dt ⎟ ⎝ ⎠ x(t ) = t 2 − 1 ⇒ x2 x1 a ( top curve – bottom curve ) dx , x1 < x2 ; Area = ∫ −1 ( f ( x) − g ( x) ) dx () 2 10. A 1 ( ln 8 − ln 3) 2 x →2 ⎛u⎞ du = sin −1 ⎜ ⎟ ⇒ ∫ ⎝a⎠ a2 − u2 du 2 = 1− π 2 ⎛ x⎞ dx = sin −1 ⎜ ⎟ + C ⎝5⎠ 25 − x 2 dx f ( x) − f (2) = f ′(2) so the derivative of f at x = 2 is 0. x−2 Take the derivative of each side of the equation with respect to x. 2 xyy′ + y 2 + 2 xy′ + 2 y = 0 , substitute the point (1, 2) 4 (1)(4) y′ + 22 + (2)(1) y′ + (2)(2) = 0 ⇒ y = − 3 Take the derivative of the general term with respect to x: ∞ ∑ ( −1) n +1 2 n − 2 x n =1 11. A d ⎛ ⎛ 1 ⎞⎞ d 1 ⎛ −1 ⎞ ⎜ ln ⎜ 1 − x ⎟ ⎟ = dx ( − ln(1 − x) ) = − ⎜ 1 − x ⎟ = 1 − x dx ⎝ ⎝ ⎠⎠ ⎝ ⎠ AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 188 1985 Calculus BC Solutions 12. A Use partial fractions to rewrite 1⎛1 1 1 ( x − 1)( x + 2 ) 1⎞ 1 as ∫ ( x − 1)( x + 2 ) dx= 3 ∫ ⎜ x − 1 − x + 2 ⎟ dx = 3 ( ln ⎝ ⎠ 1⎛ 1 1⎞ − ⎜ ⎟ 3 ⎝ x −1 x + 2 ⎠ 1 x −1 +C x − 1 − ln x + 2 ) + C = ln 3 x+2 13. B f (0) = 0, f (3) = 0, f ′( x) = 3 x 2 − 6 x; by the Mean Value Theorem, f (3) − f (0) f ′(c) = = 0 for c ∈ (0,3) . 3 So, 0 = 3c 2 − 6c = 3c ( c − 2 ) . The only value in the open interval is 2. 14. C I. convergent: p-series with p = 2 > 1 II. divergent: Harmonic series which is known to diverge 1 III. convergent: Geometric with r = < 1 3 15. C x(t ) = 4 + ∫ (2 w − 4) dw = 4 + ( w2 − 4 w) t t 0 0 = 4 + t 2 − 4t = t 2 − 4t + 4 or, x(t ) = t 2 − 4t + C , x(0) = 4 ⇒ C = 4 so, x(t ) = t 2 − 4t + 4 16. C For f ( x ) = 1 x3 2 1− we have continuity at x = 0 , however, f ...
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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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