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

3 3 0 0 volume x 2 dx x dx 2 2 x 2 3 0 9 2 6 31

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Unformatted text preview: x dx II does not work since the slope of f at x = 0 is not equal to f ′ ( 0 ) . Both I and III could work. For example, f ( x) = e x in I and f ( x) = sin x in III. 10. D This limit is the derivative of sin x . 11. A 1 The slope of the line is − , so the slope of the tangent line at x = 1 is 7 ⇒ f ′(1) = 7 . 7 12. B v(t ) = 3t + C and v(2) = 10 ⇒ C = 4 and v(t ) = 3t + 4. Distance = ∫ 2 ( 3t + 4 ) dt = 0 32 x + 4t 2 2 0 = 14 AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 200 1988 Calculus BC Solutions 13. B The Maclaurin series for sin t is t − (2 x)3 (2 x)5 sin(2 x) = 2 x − + − 3! 5! 14 A t3 t5 +− 3! 5! . Let t = 2 x . (−1) n−1 (2 x) 2n−1 + + (2n − 1)! Use the Fundamental Theorem of Calculus: 1 + ( x 2 )3 ⋅ d ( x2 ) = 2 x 1 + x6 dx dx d 2x dy 2 d2y 4 = 2t , = 2; y = ln(2t + 3) , = ; =− dt dt 2t + 3 dt 2 dt 2 (2t + 3) 2 15. E x = t 2 + 1, 16. A Use the technique of antiderivatives by parts u=x dv = e 2 x dx 1 2x e 2 1 2x 1 2x 1 1 xe − ∫ e dx = xe 2 x − e2 x + C 2 2 2 4 du = dx 17. D v= Use partial fractions: 3 ⎛1 1⎞ 8 ∫ 2 ( x − 1)( x + 1) dx = ∫ 2 ⎜ x − 1 − x + 2 ⎟ dx = ( ln x − 1 − ln x + 2 ) 2 = ln 2 − ln 5 − ln1 + ln 4 = ln 5 ⎝ ⎠ 3 3 3 4 − (−2) 1 ⎛ e4 e2 e0 e−2 ⎞ 1 4 −2 2 0 = 2, T = (2) ⎜ + 2 ⋅ + 2 ⋅ + ⎟ = e + 2e + 2e + e ⎜2 ⎟2 3 2⎝ 2 2 2⎠ ( ) 18. E ∆x = 19. B Make a sketch. x < −2 one zero, −2 < x < 5 no zeros, x > 5 one zero for a total of 2 zeros 20. E This is the definition of a limit. 21. D 3 1 31 1 1 1 ∫1 x dx = 2 ln x 1 = 2 ( ln 3 − ln1) = 2 ln 3 2 22. E Quick Solution: f ′ must have a factor of f which makes E the only option. Or, ln f ( x) = x ln( x 2 + 1) ⇒ ⎛ 2 x2 ⎞ f ′( x) 2x = x⋅ 2 + ln( x 2 + 1) ⇒ f ′( x) = f ( x) ⋅ ⎜ 2 + ln( x 2 + 1) ⎟ ⎜ x +1 ⎟ f ( x) x +1 ⎝ ⎠ AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 201 1988 Calculus BC Solutions 23. E r = 0 when cos 3θ = 0 ⇒ θ = ± π π π . The region is for the interval from θ = – to θ = . 6 6 6 π 1 2 Area = ∫ 6 ( 4 cos 3θ ) d θ π 2− 6 24. D f ′( x) = 3 x 2 − 4 x , f (0) = 0 and f (2) = 0. By the Mean Value Theorem, f (2) − f (0) 0= = f ′(c) = 3c 2 − 4c for c ∈ ( 0, 2 ) . So, c = 4 . 3 2−0 25. D Square cross-sections: 26. C This is not true if f is not an even function. 27. B y′( x) = 3 x 2 + 2ax + b , y′′( x) = 6 x + 2a , y′′(1) = 0 ⇒ a = −3 y (1) = −6 so, − 6 = 1 + a + b − 4 ⇒ − 6 = 1 − 3 + b − 4 ⇒ b = 0 28. E 29. B 1 ∑ y 2 ∆x where y = 4 x 2 . Volume = ∫ 16 x 4 dx = 0 16 4 x 5 1 0 = 16 . 5 d⎛ ⎛ π ⎞⎞ ⎛π⎞ ⎛ π ⎞ ⎛π⎞ d ⎛π⎞ ⎜ cos ⎜ x ⎟ ⎟ − sin ⎜ ⎟ ⋅ ⎜ ⎟ − sin ⎜ ⎟ ⋅ ⎜ − 2 ⎟ π dx ⎝ ⎛π⎞ ⎝ ⎠⎠ ⎝x⎠ ⎝ x ⎠ = ⎝ x ⎠ dx ⎝ x ⎠ = = tan ⎜ ⎟ 2 ⎛π⎞ ⎛π⎞ ⎛π⎞ ⎝x⎠ x cos ⎜ ⎟ cos ⎜ ⎟ cos ⎜ ⎟ ⎝x⎠ ⎝ x⎠ ⎝x⎠ Disks: ∑ πx2∆y where x 2 = y − 1 . 5 Volume = π ∫ ( y − 1) dy = 1 π ( y − 1) 2 2 5 1 = 8π AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 202 1988 Calculus BC Solutions 30. C This is an infinite geometric series with ratio 1 1 and first term n . 3 3 ⎛1⎞ ⎜ n⎟ 3 1 first ⎛⎞ Sum = = ⎝ 3 ⎠ = ⋅⎜ n ⎟ 1 − ratio 1 − 1 2 ⎝ 3 ⎠ 3 31. C This integral gives ( ) 1 of the area of the circle with center at the origin and radius = 2. 4 1 π ⋅ 22 = π 4 32. E No longer covered in the AP Course Description. The solution is of the form y = yh + y p where yh is the solution to y′ − y = 0 and the form of y p is Ax 2 + Bx + K . Hence yh = Ce x . Substitute y p into the original differential equation to determine the values of A, B, and K. Another technique is to substitute each of the options into the differential equation and pick the one that works. Only (A), (B), and (E) are viable options because of the form for yh . Both (A) and (B) fail, so the solution is (E). 2 2 L=∫ 34. C dy 3 dy dt = 4t + 4t = At t = 1, dx dx 3t 2 + 1 dt 35. A 0 () 2 ⎛ dy ⎞ 1 + ⎜ ⎟ dx = ∫ 1 + 3 x 2 0 ⎝ dx ⎠ 33. E t =1 = 2 dx = ∫ 2 0 1 + 9 x 4 dx 8 = 2; the point at t = 1 is (2,3) . y = 3 + 2 ( x − 2 ) = 2 x − 1 4 Quick solution: For large x the exponential function dominates any polynomial, so xk lim x = 0 . x→+∞ e Or, repeated use of L’Hôpital’s rule gives ⇒ lim x→+∞ xk e x AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. = lim k! x→+∞ e x =0 203 1988 Calculus BC Solutions 36. E ∑ π( R 2 − r 2 )∆x where R = 1, r = sin x π2 Volume = π ∫ (1 − sin 2 x) dx 0 Disks: Note that the expression in (E) can also be written as π∫ π2...
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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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