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

14 a t3 t5 3 5 let t 2 x 1 n1 2 x 2n1 2n

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Unformatted text preview: 1 + x dx = 2 (1 + x )3 2 3 3 0 = ( ) 2 32 32 2 14 4 −1 = ( 8 − 1) = 3 3 3 ( 3x ) + ( 3x ) + u 2 u3 + + , then e3 x = 1 + 3x + Since e = 1 + u + 2! 3! 2! 3! 3 3 9 The coefficient we want is = 3! 2 u 2 3 Graphs A and B contradict f ′′ < 0 . Graph C contradicts f ′(0) does not exist. Graph D contradicts continuity on the interval [−2,3] . Graph E meets all given conditions. 3 3 dy dy = 3x 2 y ⇒ = 3 x 2 dx ⇒ ln y = x3 + K ; y = Ce x and y (0) = 8 so, y = 8e x dx y AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 192 1985 Calculus BC Solutions 45. D The expression is a Riemann sum with ∆x = The evaluation points are: 123 ,,, nnn , 1 and f ( x) = x 2 . n 3n n Thus the right Riemann sum is for x = 0 to x = 3 . The limit is equal to AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 32 ∫0 x dx . 193 1988 Calculus AB Solutions 1. C dy d d = x 2 ⋅ (e x ) + e x ⋅ ( x 2 ) = x 2e x + 2 xe x = xe x ( x + 2) dx dx dx 2. D x 2 − 4 ≥ 0 and x ≠ 3 ⇒ x ≥ 2 and x ≠ 3 3. A Distance = ∫ 4. E Students should know what the graph looks like without a calculator and choose option E. 2 0 2 v ( t ) dt = ∫ et dt = et 0 −1 Or y = −5 ( x − 2 ) ; y ′ = 5 ( x − 2 ) 5. A ∫ sec 2 −2 2 0 = e 2 − e0 = e 2 − 1 −3 ; y ′′ = −10 ( x − 2 ) . y ′′ < 0 for x > 2 . x dx = ∫ d ( tan x ) = tan x + C x⋅ ⎛1⎞ d d (ln x) − ln x ⋅ ( x) x ⋅ ⎜ ⎟ − ln x ⋅ (1) 1 − ln x x dx dx = ⎝ ⎠2 = x2 x x2 D dy = dx 7. D ∫ x(3x 8. B dy d2y > 0 ⇒ y is increasing; < 0 ⇒ graph is concave down . This is only on b < x < c . dx dx 2 9. E 1 + ( 2 x ⋅ y ′ + 2 y ) − 2 y ⋅ y ′ = 0; y ′ = 6. 2 + 5) − 1 2 1 1 1 − 1 1 1 dx = ∫ (3x 2 + 5) 2 ( 6 x dx ) = ⋅ 2(3x 2 + 5) 2 + C = (3x 2 + 5) 2 + C 6 6 3 1+ 2y . This cannot be evaluated at (1,1) and so y ′ does 2 y − 2x not exist at (1,1) . 10. C 1⎞ ⎛ 18 = ⎜ kx 2 − x3 ⎟ 3⎠ ⎝ k 0 = 23 k ⇒ k 3 = 27, so k = 3 3 11. A f ′( x) = x ⋅ 3(1 − 2 x) 2 (−2) + (1 − 2 x)3 ; f ′(1) = −7 . Only option A has a slope of –7. 12. B ⎛π⎞ ⎛π⎞ 1 f ′ ⎜ ⎟ = cos ⎜ ⎟ = ⎝3⎠ ⎝3⎠ 2 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 194 1988 Calculus AB Solutions 13. A By the Fundamental Theorem of Calculus π 2 0 −1 2 ∫0 π 12 2 14. D ∫ (1 + sin θ ) 15. B f ( x) = 2 x = 2 ⋅ x ; f ′( x) = 2 ⋅ 16. C c ( cos θ d θ ) = 2 (1 + sin θ ) 1 2x 0 f ′( x) dx = f ( x) =2 ( c 0 = f (c) − f (0) ) 2 −1 ; f ′(2) = 2 ⋅ 1 22 ( = 1 2 ) At rest when 0 = v(t ) = x′(t ) = 3t 2 − 6t − 9 = 3 t 2 − 2t − 3 = 3(t − 3)(t + 1) t = −1, 3 and t ≥ 0 ⇒ t = 3 1 ( 3x − 2 ) dx = 2 11 11 2 3 ∫ 0 ( 3x − 2 ) ( 3 dx ) = 3 ⋅ 3 ( 3x − 2 ) 3 1 1 (1 − ( −8) ) = 1 9 17. D ∫0 18. E ⎛ ⎛ 1 ⎛ x⎞ 1⎞ ⎛ x⎞ ⎛ x ⎞ ⎛ 1 ⎞⎞ ⎛ x⎞ y ′ = 2 ⋅ ⎜ − sin ⎜ ⎟ ⋅ ⎟ = − sin ⎜ ⎟ ; y ′′ = − ⎜ cos ⎜ ⎟ ⋅ ⎜ ⎟ ⎟ = − cos ⎜ ⎟ 2 ⎝ 2⎠ 2⎠ ⎝ 2⎠ ⎝ 2 ⎠ ⎝ 2 ⎠⎠ ⎝2⎠ ⎝ ⎝ 3 x ( ) 2 = 1 ( ln10 − ln 5) = 1 ln 2 2 2 1 3 2 x dx 1 = ln x 2 + 1 2 ∫ 2 x2 + 1 2 19. B ∫ 2 x2 + 1 20. C 0 = 3 Consider the cases: I. false if f ( x ) = 1 dx = II. This is true by the Mean Value Theorem III. false if the graph of f is a parabola with vertex at x = a+b . 2 Only II must be true. 21. C x = x 2 − 3x + 3 at x = 1 and at x = 3. Area = 22. C ∫1 ( ( 3 )) 4 ⎛ ⎞ ( − x2 + 4 x − 3) dx = ⎜ − 1 x3 + 2 x2 − 3x ⎟ 1 = 3 1 ⎝3 ⎠ x − x 2 − 3 x + 3 dx = ∫ 2 = ln x − ln 3 3 1 = ln x + ln x ⇒ ln x = 1 ⇒ x = e x AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 195 1988 Calculus AB Solutions 23. B By L’Hôpital’s rule (which is no longer part of the AB Course Description), f ( x) f ′( x) f ′(0) cos 0 1 = lim = = = =1 lim x→0 g ( x ) x→0 g ′( x ) g ′(0) 1 1 Alternatively, f ′( x) = cos x and f (0) = 0 ⇒ f ( x) = sin x . Also g ′( x) = 1 and f ( x) sin x g (0) = 0 ⇒ g ( x) = x . Hence lim = lim = 1. x→0 g ( x ) x →0 x 24. C Let y = x ln x and take the ln of each side. ln y = ln x ln x = ln x ⋅ ln x . Take the derivative of y′ 1 1 each side with respect to x. = 2 ln x ⋅ ⇒ y ′ = 2 ln x ⋅ ⋅ x ln x y x x 25. B Use the Fundamental Theorem of Calculus. f ′( x) = 26. E Use the technique of antiderivatives by parts: Let u = x and dv = cos x dx . ∫ 27. E π 2 0 ( x cos x dx = x sin x − ∫ sin x dx ) π 2 0 1 x = ( x sin x + cos x ) π 2 0 = π −1 2 The function is continuous at x = 3 since lim− f ( x) = lim+ f ( x) = 9 = f (3) . Also, the x→3 x→3 derivative as you approa...
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