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

All rights reserved available at

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: differential equation is y = c1e − x + c2e 2 x with y′ = −c1e− x + 2c2 e2 x . Using the initial conditions we have the system: 2 = c1 + c2 and − 2 = −c1 + 2c2 ⇒ c2 = 0, c1 = 2 . The solution is f ( x) = 2e− x ⇒ f (1) = 2e −1 . 45. E The ratio test shows that the series is convergent for any value of x that makes x + 1 < 1 . The solutions to x + 1 = 1 are the endpoints of the interval of convergence. Test x = −2 and x = 0 in the series. The resulting series are The interval is −2 ≤ x ≤ 0 . ∞ ∑ k =1 ( −1)k k 2 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. and ∞ 1 ∑ k2 which are both convergent. k =1 171 1973 Calculus AB Solutions 14 32 x − x +C 4 2 1. E ∫ (x 2. E g ( x) = 5 ⇒ g ( f ( x) ) = 5 3. B y = ln x 2 ; y′ = 4. A f ( x) = x + sin x ; 5. A lim e x = 0 ⇒ y = 0 is a horizontal asymptote 6. D 7. B 3 − 3x) dx = 2x x 2 = 2 2 . At x = e 2 , y′ = 2 . x e f ′( x) = 1 − cos x x→−∞ f ′( x) = (1)( x + 1) − ( x − 1)(1) ( x + 1) 2 , f ′(1) = 21 = 42 Replace x with (− x) and see if the result is the opposite of the original. This is true for B. −(− x)5 + 3(− x) = x5 − 3 x = −(− x5 + 3 x) . 8. B 9. A 10. C 11. B Distance = ∫ 2 1 2 1 t 2 dx = ∫ t 2 dt = t 3 1 3 1 7 = (23 − 13 ) = 13 3 d d ( cos 3x ) = 2 cos 3x ⋅ ( − sin 3x ) ⋅ ( 3x ) = 2 cos 3x ⋅ ( − sin 3x ) ⋅ ( 3) dx dx y′ = −6 sin 3x cos 3x y′ = 2 cos 3x ⋅ x 4 x5 4 x3 − ; f ′( x) = − x 4 ; f ′′ ( x ) = 4 x 2 − 4 x3 = 4 x 2 (1 − x ) 3 5 3 f ′′ > 0 for x < 1 and f ′′ < 0 for x > 1 ⇒ f ′ has its maximum at x = 1 . f ( x) = Curve and line have the same slope when 3x 2 = ⎛1 3 tangency is ⎜ , ⎝2 8 12. C 2 3 1 ⇒ x = . Using the line, the point of 4 2 3 3 ⎛1⎞ 1 ⎞ ⎟ . Since the point is also on the curve, = ⎜ ⎟ + k ⇒ k = . 8 ⎝2⎠ 4 ⎠ Substitute the points into the equation and solve the resulting linear system. 3 = 16 + 4 A + 2 B − 5 and − 37 = −16 + 4 A − 2 B − 5 ; A = −3, B = 2 ⇒ A + B = −1 . AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 172 1973 Calculus AB Solutions 13. D v(t ) = 8t − 3t 2 + C and v(1) = 25 ⇒ C = 20 so v(t ) = 8t − 3t 2 + 20 . 4 4 2 2 s (4) − s (2) = ∫ v ( t ) dt = (4t 2 − t 3 + 20t ) 14. D f ( x) = x1 3 ( x − 2 ) = 32 23 2 1 1 ( x − 2 )−1 3 + ( x − 2 )2 3 ⋅ x −2 3 = x −2 3 ( x − 2 )−1 3 ( 3x − 2 ) 3 3 3 ′ is not defined at x = 0 and at x = 2 . f f ′ ( x ) = x1 3 ⋅ 2 x 2 e x2 2 2e 0 = 2 ( e − 1) 15. C Area = ∫ 16. C t t dN = 3000e 5 , N = 7500e 5 + C and N (0) = 7500 ⇒ C = 0 dt 0 dx = 2 N 17. C 2 t 5 = 7500e 2 , N ( 5 ) = 7500e2 Determine where the curves intersect. − x 2 + x + 6 = 4 ⇒ x 2 − x − 2 = 0 ( x − 2)( x + 1) = 0 ⇒ x = −1, x = 2 . Between these two x values the parabola lies above the line y = 4. 2 1 ⎛1 ⎞2 9 Area = ∫ (− x 2 + x + 6) − 4 dx = ⎜ − x3 + x 2 + 2 x ⎟ = −1 2 ⎝3 ⎠ −1 2 ( ) 18. D d 1 d 2 2 ⋅ ( 2x) = = ( arcsin 2 x ) = 2 2 dx 1 − 4 x2 1 − ( 2 x ) dx 1− ( 2x) 19. D If f is strictly increasing then it must be one to one and therefore have an inverse. 20. D By the Fundamental Theorem of Calculus, 1 2 b ∫a f ( x) dx = F (b) − F ( a) where F ′( x) = f ( x) . ( 1 1 x2 +2 x 1 x2 +2 x ∫ 0 e ( (2 x + 2) dx ) = 2 e 2 ) 1 13 0 e3 − 1 e −e = 2 2 ( ) 21. B x ∫ 0 ( x + 1) e 22. B f ( x) = 3 x5 − 20 x3 ; f ′( x) = 15 x 4 − 60 x 2 ; f ′′( x) = 60 x3 − 120 x = 60 x x 2 − 2 +2 x dx = 0 = ( ) The graph of f is concave up where f ′′ > 0 : f ′′ > 0 for x > 2 and for − 2 < x < 0 . AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 173 1973 Calculus AB Solutions 23. C ln ( 2 + h ) − ln 2 1 1 = f ′ ( 2 ) where f ( x ) = ln x ; f ′ ( x ) = ⇒ f ′ ( 2 ) = h →0 h x 2 24. B f ( x) = cos ( arctan x ) ; − lim π π < arctan x < and the cosine in this domain takes on all values in 2 2 the interval (0,1]. π 4 0 ∫ 26. E π4 0 π 4 dV dr dr = 4πr 2 ⋅ = S ⋅ = 100π ( 0.3) = 30π dt dt dt 27. E 28. C tan x dx = ∫ π 4 0 25. B 2 1 2x ∫0 2 1 − x2 (sec 2 x − 1) dx = (tan x − x) dx = − ∫ 1 0 1 12 1 − x2 2 1 − 2 1 − x2 2 ( ) ( −2 x dx ) = − 2 ( v ( t ) = 8 − 6t changes sign at t = = 2− 3 4 ⎛4⎞ ⎛4⎞ 5 . Distance = x(1) − x ⎜ ⎟ + x(2) − x ⎜ ⎟ = . 3 ⎝3⎠ ⎝3⎠ 3 2 1 30. B ) 0 Alternative Solution: Distance = ∫ 29. C = 1− v ( t ) dt = ∫ 2 1 8 − 6t dt = 5 3 3 11 1 3 is . −1 ≤ sin x ≤ 1 ⇒ − ≤ sin x − ≤ ; The maximum for sin x − 2 22 2 2 2 ∫1 x−4 x2 2 ⎛1 4⎞ ⎞ ⎛ dx = ∫ ⎜ − 4 x −2 ⎟ dx = ⎜ ln x + ⎟ 1 ⎝x x⎠ ⎠ ⎝ () a 2 1 = ( ln 2 + 2 ) − ( ln1 + 4 ) = ln 2 − 2 1 a 1 = ⇒ log a 2 = ⇒ 2 = a 4 ; a = 16 4 4 31. D log a 2 32...
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