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

# When x 1 1 y 0 xy y x ex xe x e x e x 1 y1

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Unformatted text preview: 0 cos 2 x dx = −π ∫ = π∫ 0 π2 π2 0 cos 2 ( π − x ) dx 2 sin 2 x dx and therefore option (D) is also a correct answer. 37. D x+S S = ⇒ x = 3S 8 2 dx dS 44 =3 = 3⋅ = dt dt 93 38. C Check x = −1, Check x = 1, ∞ ∑ ( −1)n n n =1 ∞ 1 ∑n which is convergent by alternating series test which is the harmonic series and known to diverge. n =1 39. C dy = sec2 x dx ⇒ ln y = tan x + k ⇒ y = Ce tan x . y (0) = 5 ⇒ y = 5e tan x y 40. E Since f and g are inverses their derivatives at the inverse points are reciprocals. Thus, 1 g ′(−2) ⋅ f ′(5) = 1 ⇒ g ′(−2) = = −2 1 − 2 41. B Take the interval [0,1] and divide it into n pieces of equal length and form the right Riemann Sum for the function f ( x) = x . The limit of this sum is what is given and its value is given by 1 ∫0 x dx AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 204 1988 Calculus BC Solutions 42. A Let 5 − x = u , dx = − du , substitute 4 ∫1 43. A 1 4 4 1 1 This is an example of exponential growth, B = B0 ⋅ 2 3B0 = B0 ⋅ 2 44. A 4 f (5 − x) dx = ∫ f (u ) (− du ) = ∫ f (u ) du = ∫ f ( x) dx = 6 t 3 ⇒3=2 t 3 t 3 . Find the value of t so B = 3B0 . t 3ln 3 ⇒ ln3= ln 2 ⇒ t = 3 ln 2 I. Converges by Alternate Series Test n 1⎛3⎞ ⎜ ⎟ ≠0 n→∞ n ⎝ 2 ⎠ II Diverges by the nth term test: lim III Diverges by Integral test: 45. B ∞ ∫2 1 dx = lim ln ( ln x ) L →∞ x ln x L 2 =∞ 4 A = (2 x)(2 y ) = 4 xy and y = 4 − x 2 . 9 1 So A = 8 x 1 − x 2 . 9 ⎛ ⎛ 1 ⎞ 12 1 ⎛ 1 ⎞− 12 ⎛ 2 ⎞ ⎞ A′ = 8 ⎜ ⎜1 − x 2 ⎟ + x ⎜ 1 − x 2 ⎟ ⎜ − x ⎟ ⎟ ⎜⎝ 9 ⎠ 2 ⎝ 9 ⎠ ⎝ 9 ⎠⎟ ⎝ ⎠ 8⎛ 1 ⎞ = ⎜1 − x 2 ⎟ 9⎝ 9 ⎠ − 12 (9 − 2 x 2 ) 3 3 . The maximum area occurs when x = and y = 2. The value of 2 2 3 the largest area is A = 4 xy = 4 ⋅ ⋅ 2 = 12 2 A′ = 0 at x = ±3, AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 205 1993 Calculus AB Solutions 1 1 1. C 2. B 3 3 3 f ′( x) = x 2 ; f ′(4) = ⋅ 4 2 = ⋅ 2 = 3 2 2 2 Summing pieces of the form: ( vertical) ⋅ (small width) , vertical = ( d − f ( x) ) , width = ∆x Area = b ∫ a ( d − f ( x) ) dx 3n3 − 5n 3− 5 n2 3. D Divide each term by n3 . lim 4. A 3x 2 + 3 ( y + x ⋅ y′ ) + 6 y 2 ⋅ y′ = 0; y′(3 x + 6 y 2 ) = −(3 x 2 + 3 y ) n→∞ n3 y′ = − 3x2 + 3 y 3x + 6 y 2 =− − 2n 2 + 1 = lim n→∞ 21 1− + 3 nn =3 x2 + y x + 2 y2 5. A x2 − 4 ( x + 2)( x − 2) lim = lim = lim ( x − 2) = −4. For continuity f (−2) must be –4. x→−2 x + 2 x→−2 x→−2 x+2 6. D Area = ∫ 7. B y′ = 8. E y′ = sec 2 x + csc2 x 9. E h( x ) = f 4 3 1 dx = ( ln x − 1 x −1 2 ⋅ (3 x − 2) − (2 x + 3) ⋅ 3 (3x − 2) 2 ( x )=3 x 2 ) 3 = ln 3 − ln 2 = ln 3 2 4 ; y′(1) = −13 . Tangent line: y − 5 = −13( x − 1) ⇒ 13x + y = 18 − 1 = 3x2 − 1 10. D f ′( x) = 2( x − 1) ⋅ sin x + ( x − 1) 2 cos x ; f ′(0) = (−2) ⋅ 0 + 1⋅1 = 1 11. C a (t ) = 6t − 2; v(t ) = 3t 2 − 2t + C and v(3) = 25 ⇒ 25 = 27 − 6 + C ; v(t ) = 3t 2 − 2t + 4 x(t ) = t 3 − t 2 + 4t + K ; Since x(1) = 10, K = 6; x(t ) = t 3 − t 2 + 4t + 6 . AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 206 1993 Calculus AB Solutions 12. B The only one that is true is II. The others can easily been seen as false by examples. For example, let f ( x) = 1 and g ( x) = 1 with a = 0 and b = 2. Then I gives 2 = 4 and III gives 2 = 2 , both false statements. 2π 2π = B 3 13. A period = 14. A Let u = x3 + 1. Then 15. D f ′( x) = ( x − 3) 2 + 2( x − 2)( x − 3) = ( x − 3)(3 x − 7); f ′( x) changes from positive to negative at 7 x= . 3 16. B 17. E 18. D 19. E 3x 2 ∫ 3 x +1 dx = ∫ u −1/ 2 du = 2u1/ 2 + C = 2 x3 + 1 + C sec x tan x = 2 tan x; y′(π 4) = 2 tan(π 4) = 2 . The slope of the normal line sec x 1 1 − =− y′(π 4) 2 y′ = 2 Expand the integrand. ∫ (x 2 + 1) 2 dx = ∫ ( x 4 + 2 x 2 + 1) dx = 15 23 x + x + x+C 5 3 ⎛ 3π ⎞ ⎛π⎞ ⎛ 3π ⎞ ⎛ π⎞ f ⎜ ⎟ − f ⎜ ⎟ sin ⎜ ⎟ − sin ⎜ ⎟ 2 ⎝2⎠ = ⎝4⎠ ⎝4⎠ = 0 . Want c so that f ′(c) = ⎝ ⎠ 3π π π π − 22 1 ⎛c⎞ f ′(c) = cos ⎜ ⎟ = 0 ⇒ c = π 2 ⎝2⎠ The only one that is true is E. A consideration of the graph of y = f ( x) , which is a standard cubic to the left of 0 and a line with slope 1 to the right of 0, shows the other options to be false. AP Calculus Multiple-Choice Question Collection Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 207 1993 Calculus AB Solutions 20. B Use Cylindrical Shells which is no part of the AP Course Description. The volume of each 7 1 3 shell is of the form (2π rh) ∆x with r = x and h = y. Volume = 2π ∫ x ( x + 1) dx . 0 21. C y = x −2 − x −3 ; y′ = −2 x −3...
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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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