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

A f x x sin x 5 a lim e x 0 y 0 is a horizontal

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Unformatted text preview: they do intersect, they will intersect no more than once because f ( x) grows faster than g ( x) . AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 167 1969 Calculus BC Solutions 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) . 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 f ( x) = ∫ 23. D 2 2 dy − xe− x = ⇒ 2 y dy = −2 xe − x dx ⇒ y 2 = e − x + C dx y x 1 0 3 dt , f ( −1) = ∫ t +2 f ( −1) < 0 so E is false. −1 1 0 3 t +2 dt = − ∫ 0 1 −1 3 t +2 dt < 0 2 2 2 4 = 1 + C ⇒ C = 3 ; y 2 = e− x + 3 ⇒ y = e− x + 3 24. C 25. A y = ln sin x , y′ = 2m 1 ∫m x dx = ln x cos x = cot x sin x 2m m = ln ( 2m ) − ln ( m ) = ln 2 so the area is independent of m. AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 168 1969 Calculus BC Solutions 26. C 1 1 1 2 x − 1) = ( ∫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 27. C ∫ tan x dx = ∫ cos x dx = − ln 28. D Use L’Hôpital’s Rule: lim 29. C Make the subsitution x = 2sin θ ⇒ dx = 2 cos θ d θ. x →0 ∫0 (4 − x 1 30. D 2 ) −3 2 dx = ∫ π 6 0 Substitute − x for x in cos x + C = ln sec x + C e2 x − 1 2e 2 x = lim =2 tan x x→0 sec 2 x π 61 1π 1 3 3 d θ = ∫ 6 sec 2 θ d θ = tan θ = ⋅ = 3 0 4 4 43 12 8cos θ 0 2 cos θ ∞ xn ∑ n ! = e x to get n =0 ∞ ∑ n =0 ( −1)n x n n! = e− x 31. C dy = − y ⇒ y = ce− x and 1 = ce−1 ⇒ c = e ; y = e ⋅ e− x = e1− x dx 32. B 1 + 2 x + 3x + 4 x + + nx + ∞ =∑ 1 n =1 n p where p = − x . This is a p-series and is convergent if p > 1 ⇒ − x > 1 ⇒ x < −1 . 2 1 ⎛⎛ 8 ⎞ ⎛ 3 1 ⎞ ⎞ 11 = ⎜ ⎜12 − ⎟ − ⎜ + ⎟ ⎟ = −1 3 ⎝ ⎝ 3⎠ ⎝ 4 3⎠⎠ 4 33. A 12 3 2 1⎛ 3 4 1 3⎞ ∫ −1 3t − t dt = 3 ⎜ 4 t − 3 t ⎟ 3 ⎝ ⎠ 34. D y′ = − 35. A a ( t ) = 24t 2 , v(t ) = 8t 3 + C and v(0) = 0 ⇒ C = 0. The particle is always moving to the 1 , so the desired curve satisfies y′ = x 2 ⇒ y = x3 + C 3 x2 1 right, so distance = 2 ∫0 8t 3dt = 2t 4 2 0 = 32 . AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 169 1969 Calculus BC Solutions 36. B y = 4 + sin x , y (0) = 2, y′(0) = L( x ) = 2 + 37. D 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 This item uses the formal definition of a limit and is no longer part of the AP Course Description. Need to have (1 − 3 x) − (−5) < ε whenever 0 < x − 2 < δ. (1 − 3 x) − (−5) = 6 − 3 x = 3 x − 2 < ε if x−2 < ε/3. Thus we can use any δ < ε / 3 . Of the five choices, the largest satisfying this condition is δ = ε/4. 38. A 1 Note f (1) = . Take the natural logarithm of each side of the equation and then 2 differentiate. ( ) ln f ( x) = (2 − 3x) ln x 2 + 1 ; f ′( x) 2x = (2 − 3 x) ⋅ 2 − 3ln x 2 + 1 f ( x) x +1 ( ) ( ) 2 1 1 1 ⎛ ⎞ f ′(1) = f (1) ⎜ (−1) ⋅ − 3ln(2) ⎟ ⇒ f ′(1) = ( −1 − 3ln 2 ) = − ln e + ln 23 = − ln 8e 2 2 2 2 ⎝ ⎠ ( ) () dy dy du dv 1 ⎞⎛ 1 ⎞ 2 ⎛ = ⋅ ⋅ = sec 2 u ⎜ 1 + 2 ⎟ ⎜ ⎟ = (1)( 2 ) e−1 = dx du dv dx e ⎝ v ⎠⎝ x ⎠ 39. D x = e ⇒ v = 1, u = 0, y = 0; 40. E One solution technique is to evaluate each integral and note that the value is Another technique is to use the substitution u = 1 − x ; 1 ∫ 0 (1 − x ) Integrals do not depend on the variable that is used and so 41. D 3 ∫ −1 f ( x ) dx = ∫ 2 −1 ( ) 3 1⎞ ⎛ 8 − x 2 dx + ∫ x 2 dx = ⎜ 8 x − x3 ⎟ 2 3⎠ ⎝ 2 −1 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. + 13 x 3 0 1 1 n 1n ∫0 u 1 for each. n +1 0 dx = ∫ u n ( −du ) = ∫ u n du . du is the same as 3 2 = 27 1 1n ∫0 x dx . 3 170 1969 Calculus BC Solutions 42. B Use the technique of antiderivatives by parts to evaluate u = x2 du = 2 x dx ∫x 2 cos x dx dv = cos x dx v = sin x f ( x) − ∫ 2 x sin x dx = ∫ x 2 cos x dx = x 2 sin x − ∫ 2 x sin x dx + C f ( x) = x 2 sin x + C b 2 ( b ⎛ dy ⎞ 1 + ⎜ ⎟ dx = ∫ 1 + sec 2 x a ⎝ dx ⎠ ) 2 L=∫ 44. E y′′ − y′ − 2 y = 0, y′(0) = −2, y (0) = 2 ; the characteristic equation is r 2 − r − 2 = 0 . a dx = ∫ b 43. E a 1 + sec 4 x dx The solutions are r = −1, r = 2 so the general solution to the...
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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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