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

A d 1 d 1 1 ln 1 x dx ln1 x 1 x

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ly the solution students were expected to produce while solving this problem on the 1973 multiple-choice exam. However, the problem itself is not well-defined. A solution to an initial value problem should be a function that is differentiable on an interval containing the initial point. In this problem that would be the domain x < 0 since the solution requires the choice of the branch of the logarithm function with x < 0 . Thus one cannot ask about the value of the function at x = e −1. 45. E F ′ ( x ) = xg ′ ( x ) with x ≥ 0 and g ′ ( x ) < 0 ⇒ F ′ ( x ) < 0 ⇒ F is not increasing. AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 182 1985 Calculus AB Solutions 2 −3 x dx 1 1 = − x −2 2 2 1⎛1 ⎞ 3 = − ⎜ − 1⎟ = . 1 2⎝4 ⎠ 8 1. D ∫ 2. E f ′( x) = 4(2 x + 1)3 ⋅ 2, f ′′(1) = 4 ⋅ 3(2 x + 1) 2 ⋅ 22 , f ′′′(1) = 4 ⋅ 3 ⋅ 2(2 x + 1)1 ⋅ 23 , f ( 4) (1) = 4!⋅ 24 = 384 3. A y = 3(4 + x 2 ) −1 so y′ = −3(4 + x 2 ) −2 (2 x) = −6 x (4 + x 2 ) 2 ( 4 + x ) (0) − 3(2 x) = −6 x Or using the quotient rule directly gives y′ = (4 + x ) (4 + x ) 2 2 1 2 2 2 1 4. C ∫ cos(2 x) dx = 2 ∫ cos(2 x) (2 dx) = 2 sin(2 x) + C 5. D lim 6. C f ′( x) = 1 ⇒ f ′(5) = 1 7. E ∫1 8. B 1 1 ⎛ x⎞ y = ln ⎜ ⎟ = ln x − ln 2, y′ = , y′(4) = 4 x ⎝2⎠ 9. D Since e − x is even, n→∞ 4 4n 2 2 n + 10000n 4 =4 10000 1+ n = lim n→∞ 1 4 dt = ln t 1 = ln 4 − ln1 = ln 4 t 2 2 −1) 0 ∫ −1 e − x2 ( dx = 1 1 − x2 1 ∫ −1 e dx = 2 k 2 ) 2 d ( x 2 − 1) = 2 x ⋅10( x −1) ⋅ ln(10) dx 10. D y′ = 10( x 11. B v(t ) = 2t + 4 ⇒ a(t ) = 2 ∴ a(4) = 2 12. C f ( g ( x) ) = ln g ( x) 2 = ln x 2 + 4 ⇒ g ( x) = x 2 + 4 ⋅ ln(10) ⋅ ( ) ( ) AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 183 1985 Calculus AB Solutions 13. A 2 x + x ⋅ y′ + y + 3 y 2 ⋅ y′ = 0 ⇒ y′ = − 4 ∫0 2x + y x + 3y2 v ( t ) dt = ∫ 4 3⎞ 5⎞ ⎛1 ⎛3 2 + 5t 2 ⎟ dt = ⎜ 2t 2 + 2t 2 ⎟ ⎜ 3t ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 4 14. D Since v(t ) ≥ 0, distance = 15. C x2 − 4 > 0 ⇒ x > 2 16. B f ′( x) = 3 x 2 − 6 x = 3 x( x − 2) changes sign from positive to negative only at x = 0. 17. C 0 0 = 80 Use the technique of antiderivatives by parts: u=x dv = e− x dx du = dx v = −e − x ( − xe − x + ∫ e − x dx = − xe− x − e− x ) 0 = 1 − 2e − 1 1 18. C y = cos 2 x − sin 2 x = cos 2 x , y′ = −2sin 2 x 19. B Quick solution: lines through the origin have this property. Or, f ( x 1 ) + f ( x 2 ) = 2 x 1 + 2 x 2 = 2( x 1 + x 2 ) = f ( x 1 + x 2 ) 20. A dy 1 d − sin x = ⋅ ( cos x ) = 2 dx 1 + cos x dx 1 + cos 2 x 21. B x > 1 ⇒ x 2 > 1 ⇒ f ( x) < 0 for all x in the domain. lim f ( x) = 0 . lim f ( x) = −∞ . The only x →∞ x →1 option that is consistent with these statements is (B). 2 2 ( x + 1)( x − 1) 2 x2 − 1 1 dx = ∫ dx = ∫ ( x − 1) dx = ( x − 1) 2 1 1 x +1 x +1 2 22. A ∫1 23. B d −3 x − x −1 + x 2 dx 24. D 16 = ∫ ( 2 −2 ) x =−1 ( x 7 + k ) dx = ∫ ( = − 3 x −4 + x − 2 + 2 x 2 −2 x 7 dx + ∫ 2 −2 ) x =−1 2 1 = 1 2 = − 3 + 1 − 2 = −4 k dx = 0 + ( 2 − (−2) ) k = 4k ⇒ k = 4 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 184 1985 Calculus AB Solutions 25. E 26. E f ′(e) = lim h →0 f (e + h ) − f (e ) ee + h − ee = lim h →0 h h I: Replace y with (− y ) : (− y ) 2 = x 2 + 9 ⇒ y 2 = x 2 + 9 , no change, so yes. II: Replace x with (− x) : y 2 = (− x) 2 + 9 ⇒ y 2 = x 2 + 9 , no change, so yes. III: Since there is symmetry with respect to both axes there is origin symmetry. 27. D The graph is a V with vertex at x = 1 . The integral gives the sum of the areas of the two triangles that the V forms with the horizontal axis for x from 0 to 3. These triangles have areas of 1/2 and 2 respectively. 28. C Let x(t ) = −5t 2 be the position at time t. Average velocity = 29. D The tangent function is not defined at x = π 2 so it cannot be continuous for all real numbers. Option E is the only one that includes item III. In fact, the functions in I and II are a power and an exponential function that are known to be continuous for all real numbers x. 30. B ∫ tan(2 x) dx = − 2 ∫ 31. C 1 dV 1 ⎛ dr dh ⎞ 1 ⎛ ⎛1⎞ ⎛ 1 ⎞⎞ V = π r 2h , = π ⎜ 2rh + r 2 ⎟ = π ⎜ 2(6)(9) ⎜ ⎟ + 62 ⎜ ⎟ ⎟ = 24π 3 dt 3 ⎝ dt dt ⎠ 3 ⎝ ⎝2⎠ ⎝ 2 ⎠⎠ 32. D ∫0 33. B f ′ changes sign from positive to negative at x = –1 and therefore f changes from increasing to decreasing at x = –1. 1 π3 x(3) − x(0) −45 − 0 = = −15 3−0 3 −2sin(2 x) 1 dx = − ln cos(2 x) + C cos(2 x) 2 1 sin(3x) dx = − cos(3x) 3 π3 = 0 − 1 2 ( cos π − cos 0 ) = 3 3 Or f ′ changes sign from...
View Full Document

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.

Ask a homework question - tutors are online