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

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

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Unformatted text preview: ch x = 3 from the left is 6 and the derivative as you approach x = 3 from the right is also 6. These two facts imply that f is differentiable at x = 3. The function is clearly continuous and differentiable at all other values of x. 28. C The graph is a V with vertex at x = 3 . The integral gives the sum of the areas of the two triangles that the V forms with the horizontal axis for x from 1 to 4. These triangles have areas of 2 and 0.5 respectively. 29. B This limit gives the derivative of the function f ( x) = tan(3x) . f ′( x) = 3sec 2 (3 x) AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 196 1988 Calculus AB Solutions 30. A Shells (which is no longer part of the AB Course Description) ∑ 2πrh∆x , where r = x, h = e2 x 1 Volume = 2π ∫ xe 2 x dx 0 31. C 32. A 33. A Let y = f ( x) and solve for x. x y x ; xy + y = x ; x( y − 1) = − y ; x = ⇒ f −1 ( x) = y= x +1 1− y 1− x ⎛ x⎞ The period for sin ⎜ ⎟ is ⎝2⎠ 2π = 4π . 1 2 Check the critical points and the endpoints. f ′( x) = 3 x 2 − 6 x = 3 x( x − 2) so the critical points are 0 and 2. x −2 0 2 4 f ( x ) −8 12 8 28 Absolute maximum is at x = 4. 34. D 35. B The interval is x = a to x = c. The height of a rectangular slice is the top curve, f ( x) , minus the bottom curve, g ( x) . The area of the rectangular slice is therefore ( f ( x) − g ( x))∆x . Set up a Riemann sum and take the limit as ∆x goes to 0 to get a definite integral. π⎞ ⎛ ⎛ ⎛π⎞ ⎛ π ⎞⎞ 4 cos ⎜ x + ⎟ = 4 ⎜ cos x ⋅ cos ⎜ ⎟ − sin x ⋅ sin ⎜ ⎟ ⎟ 3⎠ ⎝ ⎝3⎠ ⎝ 3 ⎠⎠ ⎝ ⎛ 1 3⎞ = 4 ⎜ cos x ⋅ − sin x ⋅ ⎟ = 2 cos x − 2 3 sin x ⎜ 2 2⎟ ⎝ ⎠ 36. C 3x − x 2 = x ( 3 − x ) > 0 for 0 < x < 3 Average value = ( ) 13 1⎛ 3 2 1 3⎞ 2 ∫ 0 3x − x dx = 3 ⎜ 2 x − 3 x ⎟ 3 ⎝ ⎠ 3 0 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. = 3 2 197 1988 Calculus AB Solutions 37. D Since e x > 0 for all x, the zeros of f ( x) are the zeros of sin x , so x = 0 , π , 2π . 38. E ∫ ⎜ x ∫1 ⎝ ⎛1 ( ln x )2 2 10 x du ⎞ 1 ⎟ dx = ∫ ln x dx = u⎠ x ⎛ dx ⎞ ⎟ .This is ∫ u du with u = ln x , so the value is x⎠ ∫ ln x ⎜ ⎝ +C f ( x) dx = − ∫ 3 3 40. B x 2 + y 2 = z 2 , take the derivative of both sides with respect to t. 2 x ⋅ 10 Divide by 2 and substitute: 4 ⋅ 41. A 42. C 43. B 1 f ( x) dx − ∫ 3 f ( x) dx = 4 − (−7) = 11 dx dy dz + 2y ⋅ = 2z ⋅ dt dt dt dx 1 dx dx + 3⋅ = 5 ⋅1 ⇒ =1 3 dt dt dt The statement makes no claim as to the behavior of f at x = 3 , only the value of f for input arbitrarily close to x = 3 . None of the statements are true. x x 1 x = lim = lim = 1. 1 x→∞ 1 x→∞ x + 1 x→∞ x 1+ + xx x None of the other functions have a limit of 1 as x → ∞ lim The cross-sections are disks with radius r = y where y = Volume = π ∫ 44. C f ( x) dx = ∫ 10 ∫3 f ( x) dx ; ∫1 10 39. E 3 −3 y 2 dx = 2π ∫ 3 0 ( ) 1 9 − x2 . 3 1 2π ⎛ 1 3⎞ 9 − x 2 dx = ⎜ 9x − x ⎟ 9 9⎝ 3⎠ 3 0 = 4π For I: p (− x) = f ( g (− x) ) = f ( − g ( x) ) = − f ( g ( x) ) = − p( x) ⇒ p is odd. For II: r (− x) = f (− x) + g (− x) = − f ( x) − g ( x) = − ( f ( x) + g ( x) ) = − r ( x) ⇒ r is odd. For III: s (− x) = f (− x) ⋅ g (− x) = ( − f ( x) ) ⋅ ( − g ( x) ) = f ( x) ⋅ g ( x) = s( x) ⇒ s is not odd. AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 198 1988 Calculus AB Solutions 45. D ( Volume = π r 2 h = 16π ⇒ h = 16r −2 . A = 2π rh + 2π r 2 = 2π 16 r −1 + r 2 ( ) ( ) ) dA dA dA = 2π −16 r −2 + 2r = 4π r −2 −8 + r 3 ; < 0 for 0 < r < 2 and > 0 for r > 2 dr dr dr The minimum surface area of the can is when r = 2 ⇒ h = 4 . AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 199 1988 Calculus BC Solutions ∫0 ( 1 ) x − x 2 dx = 12 13 x− x 2 3 1 111 − = only. 236 = 1. A 2. D ∫0 3. B 1 11 1 f ( x) = ln x = ln x ; f ′( x) = ⋅ ⇒ f ′′( x) = − 2 2 2x 2x 4. E ⎛ uv ⎞′ (uv′ + u ′v) w − uvw uv′w + u ′vw − uvw′ = ⎜ ⎟= w2 w2 ⎝w⎠ 5. C 1 ( x x2 + 2 ) 2 dx = 0 ( 11 2 x +2 2 ∫0 ) 2 ( 11 ( 2 x dx ) = ⋅ x 2 + 2 23 ) 31 0 = ( ) 1 3 3 19 3 −2 = 6 6 lim f ( x) = f (a) for all values of a except 2. x →a lim f ( x) = lim− ( x − 2) = 0 ≠ −1 = f (2) x → 2− x →2 2 y ⋅ y′ − 2 x ⋅ y′ − 2 y = 0 ⇒ y′ = y y−x 6. C 7. A ∫2 8. A f ′( x) = e x , f ′(2) = e2 , ln e 2 = 2 9. D ∞ dx x2 = lim ∫ L L →∞ 2 ⎛ 1⎞ L ⎛1 1⎞ 1 = lim ⎜ − ⎟ = lim ⎜ − ⎟ = 2 L→∞ ⎝ x ⎠ 2 L→∞ ⎝ 2 L ⎠ 2...
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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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