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AP Calculus AB AP Calculus BC

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AP Calculus AB AP Calculus BC AP® Calculus AB AP® Calculus BC Free-Response Questions and Solutions 1989 – 1997 Copyright © 2003 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. AP Central is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered trademark jointly owned by the College Entrance Examination Board and the National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service. Other products and services may be trademarks of their respective owners. For the College Board’s online home for AP professionals, visit AP Central at apcentral.collegeboard.com. 1989 BC1 Let f be a function such that f ′′ (x) = 6x + 8. (a) Find f (x) if the graph of f is tangent to the line 3xy−=2 at the point (0,−2) . (b) Find the average value of f (x) on the closed interval [1− ,1]. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC1 Solution (a) fx′()=+38x2 x+C f ′()03= C = 3 f ()xx=+3243x+x+d d =−2 fx()=+x3243x+x−2 1 1 (b) ∫ ()xx32++43x−2dx 11−−()−1 1 114 3 =+xx432+x−2x 24 3 2 −1 1143 143 =++−22−−++ 2432 432 =−2 3 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC2 x2 Let R be the region enclosed by the graph of y = , the line x = 1, and the x-axis. x 2 +1 (a) Find the area of R . (b) Find the volume of the solid generated when R is rotated about the y-axis. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC2 Solution ⌠ 1 x2 (a) Area = dx 2 + ⌡0 x 1 ⌠ 1 1 =− 1 2 dx ⌡0 x +1 =−x arctan x 1 0 π =−1 4 1 ⌠ x2 (b) Volume = 2π xdx 2 + ⌡0 x 1 ⌠ 1 x =−2π xdx 2 + ⌡0 x 1 1 x2 1 =−2lπ nx2 +1 22 0 =−π ()1ln2 or 1/2 ⌠ y Volume =−π 1 dy − ⌡ 0 1 y 1/2 =+π ()2lyyn−1 0 =−π ()1ln2 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC3 Consider the function f defined by f (x) = ex cos x with domain [0,2π ] . (a) Find the absolute maximum and minimum values of f (x). (b) Find the intervals on which f is increasing. (c) Find the x-coordinate of each point of inflection of the graph of f . Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC3 Solution (a) fx′()=−exxsin x+ecos x =−exx []cos sin x ππ5 fx′()==0 when sin xcos x, x=, 44 () x fx 0 1 π 2 π e /4 4 2 π 5 2 5/π 4 − e 4 2 π 2π 2 e ππ2 Max: ee25;Min: − /4 2 (b) f ′()x + − + π 5π 2π 0 4 4 ππ5 Increasing on 0, , , 2π 44 (c) f ′′ ()xe=−xx[]sin x−cos x+e[cos x−sin x] =−2sexx in fx′′ ()==0 when x0,ππ, 2 Point of inflection at x = π Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC4 Consider the curve given by the parametric equations x =−2tt323 and y=t3−12t dy (a) In terms of t , find . dx (b) Write an equation for the line tangent to the curve at the point where t =−1. (c) Find the x- and y-coordinates for each critical point on the curve and identify each point as having a vertical or horizontal tangent. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC4 Solution dy (a) =−31t 2 2 dt dx =−66tt2 dt 22−−()tt+−22() dy ===31t 2t 4 dx 66t 22−−t 2t 2t 2t ()t −1 (b) xy=−5, =11 dy 3 =− dx 4 3 yx−=11 −()+5 4 or 329 yx=− + 44 43yx+=29 (c) tx(),ty ype −−2 ()28,16 horizontal 00(),0 vertical 11()−−,11vertical − 2 ()4, 16 horizontal Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC5 At any time t ≥ 0 , the velocity of a particle traveling along the x-axis is given by the dx differential equation −10x = 60e4 t . dt (a) Find the general solution x(t) for the position of the particle. (b) If the position of the particle at time t = 0 is x =−8, find the particular solution x(t) for the position of the particle. (c) Use the particular solution from part (b) to find the time at which the particle is at rest. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC5 Solution −∫ 10dt (a) Integrating Factor: ee= −10t d −− ()xe 10tt= 60e4 e 10t dt xee−−10tt=−10 6 +C x()te=−10 41tt+Ce0 or ()= 10t xth Ce = 4t xAp e 41Ae44tt−=0Ae 60e4t A =−10 xt()=−Ce10tt10e4 (b) −=81CC−0;=2 x()te=−2110tt0e4 dx (c) =20ee10tt−40 4 dt 20ee10tt−=40 4 0 1 t = ln 2 6 or dx −−10()10ee41tt+2 0=60e4t dt 0 +−100ee41tt20 0=60e4t 1 t = ln 2 6 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC6 Let f be a function that is everywhere differentiable and that has the following properties. f (x) + f (h) (i) f (x + h) = for all real numbers h and x . f (−x) + f (−h) (ii) f (x) > 0 for all real numbers x . (iii) f ′ (0) =−1. (a) Find the value of f (0) . 1 (b) Show that f (−x) = for all real numbers x . f (x) (c) Using part (b), show that f (x + h) = f (x) f (h) for all real numbers h and x . (d) Use the definition of the derivative to find f ′()xf in terms of (x). Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1989 BC6 Solution (a) Let xh==0 ff()00+ () ff()00=+( 0)= =1 ff()00+ () (b) Let h = 0 fx()+ f(0) fx()+=0 fx()= fx()−+f(−0) 1 Use ff()0 ==1 and solve for (x) f ()−x or f ()−+xf(0) Note that fx(0−+)= is the reciprocal of f(x). fx()+ f(0) fx()+ f()h (c) fx()+=h 11+ fx() f()h fx()+ f()h = f ()xf()h fh()+ f()x = fx()f()h f ()xh+−f(x) (d) fx′()= lim h→0 h f ()xf()h− f()x = lim h→0 h fh()−1 = fx()lim h→0 h ==f ()xf′(0) −f()x Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 1990 BC1 A particle starts at time t = 0 and moves along the x-axis so that its position at any time t ≥ 0 is given by x(t) = (t −1) 3 (2t − 3). (a) Find the velocity of the particle at any time t ≥ 0. (b) For what values of t is the velocity of the particle less than zero? (c) Find the value of t when the particle is moving and the acceleration is zero. 1990 BC1 Solution (a) vt()= x′()t =−31()tt23(2−3)+2()t−1 =−()tt182 (−11) (b) vt()<−0 when (t 1)2 (8t−11)<0 Therefore 8tt−<11 0 and ≠1 11 or tt<≠ and 1 8 11 Since tt≥≤0, answer is 0 <, except t=1 8 (c) at()= v′()t =−21()tt(8−11)+8()t−12 =−61()tt(4−5) 5 at()==0 when t 1, t= 4 5 but particle not moving at tt==1 so 4 1990 BC2 − Let R be the region in the xy-plane between the graphs of y = ex and y = e x from xx==0 to 2. (a) Find the volume of the solid generated when R is revolved about the x-axis. (b) Find the volume of the solid generated when R is revolved about the y-axis. 1990 BC2 Solution 2 (a) Ve=−π ∫ ()22xxe− dx 0 2 1122xx− =+π ee 220 11− 11 =+π ee44−+ 22 22 π − =+ee44−2 2 2 (b) Vx=−2π ∫ exxe− dx 0 2 =+π xx−−−xx+ 2 xe()e ∫ e e dx 0 2 =+2π xe()xxe−−−(ex−ex) 0 22−−22 =+22π ()ee−(e−e)−0−()1−1 − =+23π ee22 1990 BC3 Let f (x) = 12 − x 2 for x ≥ 0 and f (x) ≥ 0. (a) The line tangent to the graph of f at the point (,kf(k)) intercepts the x-axis at x = 4. What is the value of k ? (b) An isosceles triangle whose base is the interval from (0,0) to (,c 0) has its vertex on the graph of f . For what value of c does the triangle have maximum area? Justify your answer. 1990 BC3 Solution (a) fx()=−12 x2 ; f′()x=−2x slope of tangent line at ()kf,2()k =− k line through ()4,0 &()kf, (k) has slope fk()− 0 12 − k 2 = −− kk44 12 − k 2 so −=2kk⇒2 −8k+12 =0 k − 4 kk==2 or 6 but f()6 =−24 so 6 is not in the domain. k = 2 11cc2 (b) Ac=⋅f=c12 − 22 24 3 c =−6c on 0,4 3 8 dA 33c22c =−6;6− =0 when c=4.
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