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

2 2 ap calculus multiple choice question collection

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Unformatted text preview: . D ∫ 1 + x2 dx = 5 ∫ 1 + x 2 dx = 5 tan 33. A f (− x) = − f ( x) ⇒ f ′(− x) ⋅ (−1) = − f ′( x) ⇒ f ′(− x) = − f ′( x) thus f ′(− x0 ) = − f ′( x0 ) . 5 1 −1 ( x) + C AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 174 1973 Calculus AB Solutions 32 34. C 35. C 12 12 ∫ 0 x dx = 2 ⋅ 3 x 2 2 Washers: ∑πr Volume = π∫ 36. A 37. A 38. B 39. B 40. E 41. D π 4 0 2 0 3 1 2 = ⋅ 22 = 2 3 3 ∆x where r = y = sec x . sec 2 x dx = π tan x y = e nx , y′ = ne nx , y′′ = n 2 enx , π4 0 = π(tan π − tan 0) = π 4 , y ( n) = n n e nx dy = 4 y , y (0) = 4 . This is exponential growth. The general solution is y = Ce 4 x . Since dx y (0) = 4 , C = 4 and so the solution is y = 4e4 x . Let z = x − c . Then 5 = ∫ 2 1 f ( x − c ) dx = ∫ 2−c 1−c f ( z ) dz 12 x ) on 2 2 ⎛1 ⎞ the curve to the point (4,1) . The distance L satisfies the equation L2 = ( x − 4 ) + ⎜ x 2 − 1⎟ . ⎝2 ⎠ Determine where L is a maximum by examining critical points. Differentiating with respect dL dL ⎛1 ⎞ to x, 2 L ⋅ changes sign from positive to negative at = 2( x − 4) + 2 ⎜ x 2 − 1⎟ x = x3 − 8 . dx dx ⎝2 ⎠ x = 2 only. The point on the curve has coordinates (2, 2) . Use the distance formula to determine the distance, L, from any point ( x, y ) = ( x , 2 sec ( xy ) ⋅ ( xy′ + y ) = 1, xy′ sec2 ( xy ) + y sec2 ( xy ) = 1 , y′ = 1 ∫ −1 0 1 −1 0 f ( x) dx = ∫ ( x + 1) dx + ∫ cos(π x) dx = = 1 ( x + 1) 2 2 0 1 − y sec2 ( xy ) cos 2 ( xy ) − y = x x sec2 ( xy ) 1 + sin(π x) −1 π 1 0 11 1 + ( sin π − sin 0 ) = 2π 2 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 175 1973 Calculus AB Solutions 42. D 2 2 ⎞ 127 1 1 1⎛ ⎛4⎞ ⎛5⎞ ∆x = ; T = ⋅ ⎜12 + 2 ⎜ ⎟ + 2 ⎜ ⎟ + 22 ⎟ = ⎟ 54 3 2 3⎜ ⎝3⎠ ⎝3⎠ ⎝ ⎠ 43. E Solve x x = −1 and = 2; x = −2, 4 2 2 3 44. B 45. C 1− 1 Use the linearization of f ( x) = x at x = 16 . f ′( x) = x 4 , f ′(16) = 4 32 1 h L( x) = 2 + ( x − 16); f (16 + h) ≈ L(16 + h) = 2 + 32 32 4 This uses the definition of continuity of f at x = x0 . AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 176 1973 Calculus BC Solutions 1. A f ′( x) 3 1 = ex 1 ( x + 1) 2 2. D ∫0 3. A f ′( x) = 1 − 1 ⎛1⎞ d⎜ ⎟ 1 ex x ⎛ 1⎞ ⋅ ⎝ ⎠ = ex ⎜− 2 ⎟ = − 2 dx x ⎝ x⎠ 3 3 2⎛ 2 2 ⎞ 2 14 = ⎜ 4 − 1 ⎟ = (8 − 1) = ⎟3 0 3⎜ 3 ⎝ ⎠ 33 2 dx = ( x + 1) 2 3 1 x = 2 ( x + 1)( x − 1) x2 . f ′ ( x ) > 0 for x < −1 and for x > 1 . f is increasing for x ≤ −1 and for x ≥ 1 . 4. C The slopes will be negative reciprocals at the point of intersection. 3 x 2 = 3 ⇒ x = ±1 and x ≥ 0, thus x = 1 and the y values must be the same at x = 1 . 1 4 − + b =1⇒ b = 3 3 2 x 0 2 dx = ∫ −1 dx + ∫ dx = −1 + 2 = 1 0 −1 x 5. B ∫ −1 6. D f ′( x) = (1)( x + 1) − ( x − 1)(1) ( x + 1)2 , f ′(1) = 21 = 42 dy dx at (1, 0) ⇒ y′ = 2 = 2 2 2 1 x +y 2x + 2 y ⋅ 7. D dy = dx 8. B 4 y = sin x , y′ = cos x , y′′ = − sin x , y′′′ = − cos x , y ( ) = sin x 9. A y′ = 2 cos 3x ⋅ d d (cos 3x) = 2 cos 3x ⋅ (− sin 3x) ⋅ (3x) = 2 cos 3 x ⋅ (− sin 3 x) ⋅ 3 dx dx y′ = −6 sin 3x cos 3x AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 177 1973 Calculus BC Solutions 10. A L=∫ =∫ b 0 b 0 1 + ( y′ ) dx = ∫ 2 2 ⎛ sec x tan x ⎞ 1+ ⎜ ⎟ dx ⎝ sec x ⎠ b 0 1 + ( tan x ) dx = ∫ b 2 b sec2 x dx = ∫ sec x dx 0 0 11. E ⎛1 dy = ⎜ x ⋅ 1 + x 2 ⎜2 ⎝ 12. D k 1 xn = ∫ x n −1 dx = 1 n n 13. D v(t ) = 8t − 3t 2 + C and v(1) = 25 ⇒ C = 20 so v ( t ) = 8t − 3t 2 + 20 . ( ) − 1 2 ( ( 2x) + 1+ 1 22 x ) ⎞ ⎟ dx ; dy = (0 + 1)(2) = 2 ⎟ ⎠ 1 1 kn 1 kn 2 ⇒= −; = ⇒ k = 2n 1 nn nnn k ( 4 s (4) − s (2) = ∫ v(t ) dt = 4t 2 − t 3 + 20t 2 14. A dy = dx dy dt dx dt 2 ) 2 = 32 4 2et et = = 2t t 1 x e2 12 x 2e 2 0 15. C Area = ∫ 16. A sin t = t − 17. C t t dN = 3000e 5 , N = 7500e 5 + C and N (0) = 7500 ⇒ C = 0 dt 0 dx = t3 t5 t7 +−+ 3! 5! 7! 2 N 18. D 2 t = 7500e 5 = 2(e − 1) ⇒ t2 t4 t6 sin t = 1− + − + t 3! 5! 7! 2 , N (5) = 7500e2 D could be false, consider g ( x) = 1 − x on [0,1] . A is true by the Extreme Value Theorem, B is true because g is a function, C is true by the Intermediate Value Theorem, and E is true because g is continuous. AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 178 1973 Calculus BC Solutions 19. D 20. E I is a convergent p-series, p = 2 > 1 II is the Harmonic series and is known to be divergent, III is convergent by the Alternating Ser...
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