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

A 1 23 b d 3 x x 1 x 2 dx 24 d 16 2 2 x

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Unformatted text preview: ies Test. ∫ ( 1 2 +2 x ) 1 2 3 2 ( ) ( 1 x 4 − x dx = − ∫ 4 − x 2 2 2 ) 12 ( −2 x dx ) = − ⋅ 4 − x 2 23 1 1 x2 + 2 x 1 x2 + 2 x ∫ 0 e ( (2 x + 2) dx ) = 2 e 2 ( 1 + C = − 4 − x2 3 1 ) 3 2 +C 13 0 e3 − 1 e −e = 2 2 ( ) 21. B ∫0 22. C 1 1 1 1 x′(t ) = t + 1 ⇒ x(t ) = (t + 1) 2 + C and x(0) = 1 ⇒ C = ⇒ x(t ) = (t + 1) 2 + 2 2 2 2 ( x + 1) e x dx = 5 5 x(1) = , y (1) = ln ; 2 2 0 = 5⎞ ⎛5 ⎜ , ln ⎟ 2⎠ ⎝2 23. C ln(2 + h) − ln 2 = f ′(2) where f ( x) = ln x ; h →0 h 24. A This item uses the formal definition of a limit and is no longer part of the AP Course lim Description. Any δ < π 4 0 f ′( x) = 1 1 ⇒ f ′ ( 2) = x 2 f ( x ) − 7 = ( 3 x + 1) − 7 = 3x − 6 = 3 x − 2 < ε whenever ε εε ε will be sufficient and < , thus the answer is . 3 43 4 tan x dx = ∫ π 4 0 ε x−2 < . 3 (sec2 x − 1) dx = ( tan x − x ) 0π 4 = 1 − π 4 25. B ∫ 26. D For x in the interval (–1, 1), g ( x) = x 2 − 1 = −( x 2 − 1) and so y = ln g ( x) = ln(−( x 2 − 1)) . 2 Therefore y ′= 2x x2 − 1 , ( x2 − 1) ( 2) − ( 2 x )( 2 x ) = −2 x2 − 2 < 0 y′′ = 2 2 ( x2 − 1) ( x2 − 1) AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 179 1973 Calculus BC Solutions Alternative graphical solution: Consider the graphs of g ( x ) = x 2 − 1 and ln g ( x ) . concave down g ( x) = x 2 − 1 27. E f ′( x) = x 2 − 8 x + 12 = ( x − 2)( x − 6); the candidates are: x = 0, 2, 6,9 x 0 f ( x) 28. C 2 −5 17 3 6 9 the maximum is at x = 9 −5 22 1 π x = sin 2 y ⇒ dx = 2sin y cos y dy ; when x = 0, y = 0 and when x = , y = 2 4 ∫ 29. A ln x 2 − 1 1 2 0 π π x sin y dx = ∫ 4 ⋅ 2sin y cos y dy = ∫ 4 2sin 2 y dy 0 0 2 1− x 1 − sin y Let z = y′ . Then z = e when x = 0. Thus y′′ = 2 y′ ⇒ z′ = 2 z . Solve this differential equation. z = Ce 2 x ; e = Ce0 ⇒ C = e ⇒ y′ = z = e 2 x +1 . Solve this differential equation. ( ) 1 1 1 1 1 1 1 1 y = e2 x +1 + K ; e = e1 + K ⇒ K = e; y = e2 x +1 + e , y (1) = e3 + e = e e2 + 1 2 2 2 2 2 2 2 2 Alternative Solution: y′′ = 2 y′ ⇒ y′ = Ce 2 x = e ⋅ e 2 x . Therefore y′(1) = e3 . 1 1 0 0 y′(1) − y′(0) = ∫ y′′( x)dx = ∫ 2 y′( x)dx = 2 y (1) − 2 y (0) and so y (1) = y′(1) − y′(0) + 2 y (0) e3 + e . = 2 2 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 180 1973 Calculus BC Solutions 2 x−4 2 ⎛1 4⎞ ⎞ ⎛ dx = ∫ ⎜ − 4 x −2 ⎟ dx = ⎜ ln x + ⎟ 1 ⎝x x⎠ ⎠ ⎝ 30. B ∫1 31. E 2 = ( ln 2 + 2 ) − ( ln1 + 4 ) = ln 2 − 2 1 d ( ln x ) 1 =x= f ′ ( x ) = dx ln x ln x x ln x 32. C x2 1 Take the log of each side of the equation and differentiate. ln y = ln x ln x = ln x ⋅ ln x = ( ln x ) 2 y′ d 2 ⎛2 ⎞ = 2 ln x ⋅ ( ln x ) = ln x ⇒ y′ = xln x ⎜ ln x ⎟ y dx x ⎝x ⎠ 33. A f (− x) = − f ( x) ⇒ f ′(− x) ⋅ (−1) = − f ′( x) ⇒ f ′(− x) = − f ′( x) thus f ′(− x0 ) = − f ′( x0 ) . 34. C 12 12 ∫ 0 x dx = 2 ⋅ 3 x 2 2 32 35. C Washers: ∑ π r 2∆x Volume = π ∫ 36. E x +1 1 ∫0 π 4 0 x2 + 2 x − 3 lim sec x dx = π tan x dx = x →0 1 − cos 2 2 x x2 where r = y = sec x . 2 = 37. E 0 3 1 2 = ⋅ 22 = 2 3 3 π 4 0 π ⎛ ⎞ = π ⎜ tan − tan 0 ⎟ = π 4 ⎝ ⎠ L 1 2x + 2 1 lim− ∫ dx = lim− ln x 2 + 2 x − 3 2 L→1 0 x 2 + 2 x − 3 2 L→1 ( L 0 ) 1 lim− ln L2 + 2 L − 3 − ln − 3 = −∞ . Divergent 2 L→1 = lim sin 2 2 x x →0 2 x2 sin 2 x sin 2 x ⋅ ⋅ 4 = 1 ⋅1 ⋅ 4 = 4 x →0 2 x 2x = lim f ( x − c ) dx = ∫ 2−c f ( z ) dz 38. B Let z = x − c . 5 = ∫ 39. D h′( x) = f ′ ( g ( x) ) ⋅ g ′( x) ; h′(1) = f ′ ( g (1) ) ⋅ g ′(1) = f ′(2) ⋅ g ′(1) = (−4)(−3) = 12 1 1−c AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 181 1973 Calculus BC Solutions 40. C Area = ( ) π 1 2π 1 2 2 2 ∫ 0 (1 − cos θ ) d θ = ∫ 0 1 − 2 cos θ + cos θ d θ ; cos θ = 2 (1 + cos 2θ ) 2 π π⎛ 1 1 3 ⎞ ⎛3 ⎞ Area = ∫ ⎜ 1 − 2 cos θ + (1 + cos 2θ ) ⎟ d θ = ⎜ θ − 2sin θ + sin 2θ ⎟ = π 0⎝ 2 4 ⎠ ⎝2 ⎠0 2 41. D 1 0 = 42. D 43. E 1 ( x + 1) 2 2 0 1 + sin(π x) −1 π 1 0 = 11 1 + ( sin π − sin 0 ) = 2π 2 2 2 ⎞ 127 1 1 1⎛ 2 ⎛4⎞ ⎛5⎞ ∆x = ; T = ⋅ ⎜1 + 2 ⎜ ⎟ + 2 ⎜ ⎟ + 22 ⎟ = ⎟ 54 3 2 3⎜ ⎝3⎠ ⎝3⎠ ⎝ ⎠ Use the technique of antiderivatives by part: u = sin −1 x dv = dx du = ∫ sin 44. A 1 ∫ −1 f ( x) dx = ∫ −1 ( x + 1) dx + ∫ 0 cos(π x) dx dx 1 − x2 −1 v=x x dx = x sin −1 x − ∫ x 1 − x2 dx Multiply both sides of x = x f ′( x) − f ( x) by 1 2 . Then 1 x f ′( x) − f ( x) d ⎛ f ( x) ⎞ = =⎜ ⎟. x dx ⎝ x ⎠ x2 x f ( x) = ln x + C ⇒ f ( x) = x ( ln x + C ) = x ( ln x − 1) since f (−1) = 1 . Thus we have x ( ) Therefore f (e−1 ) = e−1 ln e−1 − 1 = e−1 (−1 − 1) = −2e−1 This was most like...
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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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