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

03 4 4 all options have the same value at x 0 we want

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Unformatted text preview: ce from ⎜ x , − ⎟ and ⎜ 0, − ⎟ . ⎜ ⎟ 2⎠ 2⎠ ⎝ ⎝ 2 ⎛ x2 1 ⎞ L = ( x − 0) + ⎜ − ⎟ ⎜ 2 2⎟ ⎝ ⎠ ⎛ x2 1 ⎞ dL 2L ⋅ = 2x + 2 ⎜ − ⎟ ( x) ⎜ 2 2⎟ dx ⎝ ⎠ 2 2 ⎛ x2 1 ⎞ 2x + 2 ⎜ − ⎟ ( x) 2 ⎜ 2 2⎟ 2 x + x3 − x x3 + x x x + 1 dL ⎝ ⎠ = = = = 2L 2L 2L 2L dx ( ) dL dL < 0 for all x < 0 and > 0 for all x > 0 , so the minimum distance occurs at x = 0 . dx dx The nearest point is the origin. 12. A 13. C 4 1 ⎛4⎞ = 2⎜ ⎟ ⇒ x − 1 = 4 x − 2; x = 2x −1 3 ⎝ x −1 ⎠ π ⎛ π⎞ ⎛ ⎞ cos x dx ; sin k − sin ⎜ − ⎟ = 3 ⎜ sin − sin k ⎟ 2 ⎝ 2⎠ ⎝ ⎠ π sin k + 1 = 3 − 3sin k ; 4sin k = 2 ⇒ k = 6 k π2 ∫ −π 2 cos x dx = 3∫ k 14. E y = x5 − 1 has an inverse x = y 5 − 1 ⇒ y = 5 x + 1 15. B The graphs do not need to intersect (eg. f ( x) = −e − x and g ( x) = e − x ) . The graphs could intersect (e.g. f ( x) = 2 x and g ( x) = x ). However, if they do intersect, they will intersect no more than once because f ( x) grows faster than g ( x) . 16. B y′ > 0 ⇒ y is increasing; y′′ < 0 ⇒ the graph is concave down . Only B meets these conditions. 17. B y′ = 20 x3 − 5 x 4 , y′′ = 60 x 2 − 20 x3 = 20 x 2 ( 3 − x ) . The only sign change in y′′ is at x = 3 . The only point of inflection is (3,162). AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 161 1969 Calculus AB Solutions 18. E There is no derivative at the vertex which is located at x = 3 . 19. C dv 1 − ln t dv = 2 > 0 for 0 < t < e and < 0 for t > e , thus v has its maximum at t = e . dt dt t 20. A y (0) = 0 and y′(0) = y= 1 2 x2 1− 4 x =0 1 = 4 − x2 x =0 = 1 . The tangent line is 2 1 x ⇒ x − 2y = 0 . 2 21. B f ′ ( x ) = 2 x − 2e−2 x , f ′ ( 0 ) = −2 , so f is decreasing 22. E ln e2 x = 2 x ⇒ 2 2x e 0 ( 2 1 2x e 2 23. C ∫ 24. C y = ln sin x , y′ = 25. A 26. C 27. C 2m 1 ∫m x dx = ) d d ln e2 x = ( 2 x ) = 2 dx dx dx = ln x 0 = ( ) 14 e −1 2 cos x = cot x sin x 2m m = ln ( 2m ) − ln ( m ) = ln 2 so the area is independent of m. 11 1 ( x − 1)2 = ∫0 0 0 02 2 Alternatively, the graph of the region is a right triangle with vertices at (0,0), (0,1), and (1,0). 1 The area is . 2 1 x 2 − 2 x + 1 dx = ∫ 1 1 x − 1 dx = ∫ − ( x − 1) dx = − sin x ∫ tan x dx = ∫ cos x dx = − ln cos x + C = ln sec x + C AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 162 1969 Calculus AB Solutions 28. C 3 cos x + 3sin x can be thought of as the expansion of sin ( x + y ) . Since 3 and 3 are too large for values of sin y and cos y , multiply and divide by the result of the Pythagorean Theorem used on those values, i.e. 2 3 . Then ⎛3 ⎞ ⎛1 ⎞ 3 3 3 cos x + 3sin x = 2 3 ⎜ ⎜ 2 3 cos x + 2 3 sin x ⎟ = 2 3 ⎜ 2 cos x + 2 sin x ⎟ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = 2 3 ( sin y cos x + cos y sin x ) = 2 3 sin ( y + x ) ⎛1⎞ where y = sin −1 ⎜ ⎟ . The amplitude is 2 3 . ⎝2⎠ Alternatively, the function f ( x) is periodic with period 2π . f ′( x) = − 3 sin x + 3cos x = 0 π 4π ⎛π⎞ when tan x = 3 . The solutions over one period are x = , . Then f ⎜ ⎟ = 2 3 and 33 ⎝3⎠ ⎛ 4π ⎞ f ⎜ ⎟ = −2 3 . So the amplitude is 2 3 . ⎝3⎠ π2 cos x dx = ln ( sin x ) sin x π2 1 = ln 2 2 29. A ∫π 4 30. E Because f is continuous for all x, the Intermediate Value Theorem implies that the graph of f must intersect the x-axis. The graph must also intersect the y-axis since f is defined for all x, in particular, at x = 0. π4 = ln1 − ln 31. C dy = − y ⇒ y = ce − x and 1 = ce−1 ⇒ c = e ; y = e ⋅ e− x = e1− x dx 32. B If a < 0 then lim y = ∞ and lim y = −∞ which would mean that there is at least one root. x→−∞ x→∞ If a > 0 then lim y = −∞ and lim y = ∞ which would mean that there is at least one root. x→−∞ x→∞ In both cases the equation has at least one root. 33. A 12 3 2 1⎛ 3 4 1 3⎞ ∫ −1 3t − t dt = 3 ⎜ 4 t − 3 t ⎟ 3 ⎝ ⎠ 34. D y′ = − 1 x 2 1 ⎛⎛ 8 ⎞ ⎛ 3 1 ⎞ ⎞ 11 = ⎜ ⎜12 − ⎟ − ⎜ + ⎟ ⎟ = −1 3 ⎝ ⎝ 3 ⎠ ⎝ 4 3 ⎠⎠ 4 2 , so the desired curve satisfies y′ = x 2 ⇒ y = AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 13 x +C 3 163 1969 Calculus AB Solutions 35. A a ( t ) = 24t 2 , v(t ) = 8t 3 + C and v(0) = 0 ⇒ C = 0. The particle is always moving to the right, so distance = 36. B ∫0 8t 3dt = 2t 4 y = 4 + sin x , y (0) = 2, y′(0) = L( x ) = 2 + 37. D 2 2 0 = 32 . cos 0 1 = . The linear approximation to y is 2 4 + sin 0 4 1 1 x . L(1.2) = 2 + (1.2) = 2.03 4 4 All options have the same value at x = 0 . We want the one that has the same first and...
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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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