Answer Key 969 BC 97 BC. C. E. B. D 5. E 6. B 7. D 8. C 9. D. A. B. E. C. D 5. B 6. B 7. B 8. E 9. C. A. B. E. D. C 5. A 6. C 7. C 8. D 9. C. D. C. B. A. D 5. A 6. B 7. D 8. A 9. D. E. D. B. E. E 5. E. A. D. A. C 5. B 6. D 7. D 8. B 9. A. A. E. D. D. A 5. C 6. A 7. C 8. D 9. D. E. B. C. C. A 5. B 6. D 7. E 8. C 9. A. B. E. C. A. C 5. C 6. E 7. E 8. B 9. D. C. D. D. E. A 5. E AP Calculus Multiple-Choice Question Collection 5 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
985 BC 988 BC. D. A. B. D 5. D 6. E 7. A 8. C 9. B. A. A. A. B. C 5. C 6. C 7. B 8. C 9. D. C. B. A. C. D 5. C 6. E 7. E 8. E 9. D. B. D. E. C. A 5. B 6. E 7. A 8. C 9. A. A. C. E. E. A 5. D. A. D. B. E 5. C 6. C 7. A 8. A 9. D. D. A. B. B. A 5. E 6. A 7. D 8. E 9. B. E. D. E. E. D 5. D 6. C 7. B 8. E 9. B. C. C. E. E. C 5. A 6. E or D 7. D 8. C 9. C. E. B. A. A. A 5. B AP Calculus Multiple-Choice Question Collection 55 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
99 BC. A. C. E. B 5. D 6. A 7. A 8. B 9. D. E. E. E. C. B 5. D 6. A 7. A 8. B 9. B. E. A. B. D. C 5. D 6. B 7. C 8. A 9. E. C. A. B. A. E 5. A 6. E 7. B 8. C 9. C. C. C. E. A. E 5. D 997 BC. C. E. A. C 5. C 6. A 7. C 8. E 9. A. B. C. A. B. C 5. D 6. B 7. B 8. C 9. D. E. A. C. E. D 5. A 76. D 77. E 78. A 79. D 8. B 8. D 8. B 8. E 8. C 85. D 86. A 87. B 88. C 89. D 9. B AP Calculus Multiple-Choice Question Collection 57 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
998 BC. C. A. D. A 5. A 6. E 7. E 8. B 9. D. E. A. E. B. E 5. B 6. C 7. D 8. B 9. D. E. C. A. E. C 5. C 6. E 7. D 8. C 76. D 77. E 78. B 79. A 8. B 8. B 8. B 8. C 8. B 85. C 86. C 87. D 88. C 89. A 9. A 9. E 9. D AP Calculus Multiple-Choice Question Collection 59 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
969 Calculus BC Solutions. C For horizontal asymptotes consider the limit as ± : t y= is an asymptote For vertical asymptotes consider the limit as y ± : t = is an asymptote. E + y = ( + ) tan, y = + tan + ( )() ( )( ) + + y = + = ( + + ) ( + ) y changes sign at = only. The point of inflection is (, ). B y =, y =. By the Mean Value Theorem we have The point is (,). c c = =.. D 8 d d = + = ( ) = + 8 5. E Using implicit differentiation, 6+ y + y+ y y =. Therefore When =, + y+ y = = y + y+ = ( y+ ) y = Therefore + y = and so dy is not defined at d =. y 6 y =. + y f 8 at. 8 6. B This is the derivative of ( ) = = f = 6 = 7 7. D k k With f( ) = +, we need = f ( ) = and so k =. Since f ( ) < for k =, f does have a relative maimum at =. 8. C ( ) h( ) = f( ) f( ) g ( ) g( ) = f( ) g ( ) g ( ) f( ) = f( ) g ( ) AP Calculus Multiple-Choice Question Collection 66 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
9. D ( cos ) ( cos ) ( cos ) 969 Calculus BC Solutions A= + θ dθ= d d + θ θ= + θ θ. A d d d ( ) + + = = = tan = = + + + +. B Let L be the distance from, and,. L = ( ) + dl L = + d ( ) + dl + = = = = d L L L L ( ) + + ( ) dl dl < for all < and > for all >, so the minimum distance occurs at =. d d The nearest point is the origin. t F = e dt F = e.. E By the Fundamental Theorem of Calculus, if ( ) then ( ). C k cos cos ; sin sin d d k sin sin = = k k sin k+ = sin k; sin k = k = 6 dy dy d y = + u = = = = du d du. D and, ( ) 5. B The graphs do not need to intersect (eg. f ( ) = e and g( ) = e ). The graphs could intersect (e.g. f ( ) = and g ( ) = ). However, if they do intersect, they will intersect no more than once because f ( ) grows faster than g. ( ) AP Calculus Multiple-Choice Question Collection 67 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
969 Calculus BC Solutions 6. B y > y is increasing; y < the graph is concave down. Only B meets these conditions. 7. B y 5, y 6 ( ) = = =. The only sign change in y is at =. The only point of inflection is (,6). 8. E There is no derivative at the verte which is located at =. 9. C. A. B dv lnt dv = > for < t < e and < for t > e, thus v has its maimum at t = e. dt t dt y() = and y () = = =. The tangent line is = = y = y =. f ( ) = e, f () =, so f is decreasing. E ( ) ( ) f = dt, f = dt = dt < t + t + t + f < so E is false. ( ). D. C dy e = ydy= e d y = e + C d y = + C C = ; cos y = ln sin, y = = cot sin y = e + y = e + m m = ln = ln ln = ln so the area is independent of m. m m 5. A d ( m) ( m) AP Calculus Multiple-Choice Question Collection 68 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
+ = = = = 6. C d d ( ) d ( ) 969 Calculus BC Solutions Alternatively, the graph of the region is a right triangle with vertices at (,), (,), and (,). The area is. 7. C sin tan d = d = ln cos + C = ln sec + C cos 8. D Use L Hôpital s Rule: e e lim = lim tan = sec 9. C Make the subsitution = sin θ d = cos θdθ. ( ) 6 6 θ d d 6 sec d tan cos = θ= θ θ= θ = = 8cos θ. D Substitute for in. C n = e to get n= n! n= ( ) n n! n = e dy = y y = ce and = ce c = e; y = e e = e d. B. A + + + + + n + = where p p =. This is a p-series and is n= n convergent if p> > <. t t dt = t t = + = 8. D y =, so the desired curve satisfies y = y = + C a t = t, v( t) = 8 t + C and v() = C =. The particle is always moving to the 5. A () right, so distance = 8tdt= t =. AP Calculus Multiple-Choice Question Collection 69 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
969 Calculus BC Solutions 6. B cos y = + sin, y() =, y () = =. The linear approimation to y is + sin L ( ) = +. L (.) = + (.) =. 7. D This item uses the formal definition of a limit and is no longer part of the AP Course Description. Need to have ( ) ( 5) <ε whenever < <δ. ( ) ( 5) = 6 = <ε if <ε /. Thus we can use any δ<ε /. Of the five choices, the largest satisfying this condition is δ=ε /. 8. A Note f () =. Take the natural logarithm of each side of the equation and then differentiate. f ( ) ( ) ( ) ln f( ) = ( )ln + ; = ( ) ln + f( ) + f () f() = ( ) ln() f () = ln = ln e+ ln = ln 8e ( ) ( ) dy dy du dv 9. D e v, u, y ; ( sec u) ()( )( e ) = = = = = = d du dv d + = = v e. E One solution technique is to evaluate each integral and note that the value is n + for each. n n n. Another technique is to use the substitution ; ( ) ( ) Integrals do not depend on the variable that is used and so f d d d. D ( ) ( ) u = d= u du = u du = 8 + = 8 + = 7 n n u du is the same as d. AP Calculus Multiple-Choice Question Collection 7 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
. B Use the technique of antiderivatives by parts to evaluate u = du = d dv = cos d v= sin f ( ) sind= cosd= sin sind+ C f( ) = sin+ C b b b sec sec a a a dy. E ( ) L= + d= + d= + d d 969 Calculus BC Solutions cos d. E y y y =, y () =, y() = ; the characteristic equation is r r =. The solutions are r =, r = so the general solution to the differential equation is with y = ce + c e y = ce + c e. Using the initial conditions we have the system: = c + c and = c + c c =, c =. The solution is f ( ) = e f() = e. 5. E The ratio test shows that the series is convergent for any value of that makes + <. The solutions to + = are the endpoints of the interval of convergence. Test = and = in the series. The resulting series are The interval is. k= ( ) k and k which are both convergent. k k= AP Calculus Multiple-Choice Question Collection 7 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
97 Calculus BC Solutions. A ( ) d f e e = = d = e. D. A ( + ) = ( + ) = = (8 ) = d ( + )( ) f ( ). = = f ( ) for and for > < >. f is increasing for and for.. C The slopes will be negative reciprocals at the point of intersection. = =± and, thus = and the y values must be the same at =. + b= b= 5. B 6. D 7. D 8. B d = d + d = + = ()( + ) ( )() f ( ) =, f () = = ( + ) dy + y dy = d at (, ) y = = d + y ( y = sin, y = cos, y = sin, y = cos, y ) = sin 9. A cos d d y = (cos ) = cos ( sin ) ( ) = cos ( sin ) d d y = 6sincos AP Calculus Multiple-Choice Question Collection 77 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
97 Calculus BC Solutions b sec tan b L = + y d = d + sec. A ( ) b b b ( ) = + tan d = sec d = sec d dy = + + + d ; dy = ( + )() =. E ( ) ( ) ( ). D n k d k n k = n = = ; n n n n n n k n = k = n n. D vt ( ) = 8t t + C and v() = 5 C= so vt () = 8t t +. ( ) s() s() = v( t) dt = t t + t =. A 5. C dy t t dy e e = dt = = d d t t dt Area = e d= e = ( e ) 6. A 5 7 6 t t t sin t t t t sin t = t + + = + +! 5! 7! t! 5! 7! 7. C dn dt t t 5 5 = e, N = 75 e + C and N() = 75 C = t 5 N = 75 e, N(5) = 75e 8. D D could be false, consider g ( ) = on [,]. A is true by the Etreme Value Theorem, B is true because g is a function, C is true by the Intermediate Value Theorem, and E is true because g is continuous. AP Calculus Multiple-Choice Question Collection 78 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
97 Calculus BC Solutions 9. D I is a convergent p-series, p = > II is the Harmonic series and is known to be divergent, III is convergent by the Alternating Series Test.. E ( ) ( ) ( ) ( ) d= d = + C = + C. B ( ) e d e (( ) d) ( e ) ( e e ) + + + e + = + = = =. C ( t) = t+ ( t) = ( t+ ) + C and () = C = ( t) = ( t+ ) + 5 5 5 5 () =, y() = ln ;, ln. C ln( + h) ln lim f () where f( ) ln h h = = = = ; f ( ) f ( ). A This item uses the formal definition of a limit and is no longer part of the AP Course ε Description. f ( ) 7 = ( + ) 7 = 6 = <ε whenever <. ε ε ε ε Any δ< will be sufficient and <, thus the answer is. tan sec tan = = = 5. B d ( ) d ( ) 6. D For in the interval (, ), Therefore g ( ) = = ( ) and so ( )( ) ( )( ) y y =, = = < ( ) ( ) y = ln g( ) = ln( ( )). AP Calculus Multiple-Choice Question Collection 79 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
97 Calculus BC Solutions Alternative graphical solution: Consider the graphs of g( ) and ln g( ) =. concave down g ( ) = ln 7. E f ( ) = 8+ = ( )( 6); the candidates are: =,,6,9 f ( ) 6 9 5 7 5 the maimum is at = 9 8. C = sin y d = sin y cos y dy; when =, y = and when =, y = sin y = = d sin y cos y dy sin y dy sin y 9. A Let z = y. Then z = e when =. Thus y = y z = z. Solve this differential equation. z = Ce ; e= Ce C = e y = z = e +. Solve this differential equation. ( ) + + y = e + K; e= e + K K = e; y = e + e, y() = e + e= e e + Alternative Solution: y = y y = Ce = e e. Therefore and so y () y () = y ( ) d = y ( ) d = y() y() y () y () + y() e + e y() = =. y () = e. AP Calculus Multiple-Choice Question Collection 8 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
. B d d ( ) ( ). E f ( ) 97 Calculus BC Solutions = = ln + = ln+ ln+ = ln d ( ln ) = d = = ln ln ln ln. C Take the log of each side of the equation and differentiate. ( ) ln y = ln = ln ln = ln y d ln = ln ( ln) = ln y = ln y d. A f ( ) = f( ) f ( ) ( ) = f ( ) f ( ) = f ( ) thus f ( ) = f ( ).. C d= = = 5. C Washers: r where r = y = sec. d Volume sec tan tan tan = = = = 6. E L + L + d = lim d = lim ln + L L + + L ( L L ) = lim ln + ln =. Divergent 7. E cos sin sin sin lim = lim = lim = = 8. B Let z c c =. 5 = ( ) = ( ) f c d f z dz c 9. D h ( ) = f ( g( ) ) g ( ) ; ( ) h () = f g() g () = f () g () = ( )( ) = AP Calculus Multiple-Choice Question Collection 8 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
. C Area ( cos ) d ( cos cos ) = θ θ= θ+ θ dθ 97 Calculus BC Solutions ; cos θ = ( + cos θ). D Area = cos ( cos ) d sin sin θ+ + θ θ= θ θ+ θ = f ( d ) = ( + ) d+ cos( d ) ( ) sin( ) ( sin sin ) = + + = + =. D 5 7 = ; T = + + + = 5. E Use the technique of antiderivatives by part: u = sin dv = d d du = v = sin d = sin d. A Multiply both sides of = f ( ) f( ) by. Then f ( ) f( ) d f( ) = = d. f( ) Thus we have = ln + C f( ) = ( ln + C) = ( ln ) since f ( ) =. Therefore ( ) f ( e ) = e ln e = e ( ) = e This was most likely the solution students were epected to produce while solving this problem on the 97 multiple-choice eam. 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 < since the solution requires the choice of the branch of the logarithm function with <. Thus one cannot ask about the value of the function at = e. 5. E ( ) ( ) ( ) ( ) F = g with and g < F < F is not increasing. AP Calculus Multiple-Choice Question Collection 8 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
985 Calculus BC Solutions. D. A. B. D ( + ) d= ( + ) = (6 + ) ( + ) = 7 f ( ) = 5 5 = 5 ( ) = 5 ( )( + ), changes sign from positive to negative only at =. So f has a relative maimum at = only. ( + ) d = = ln + = ln 8 ln + + + d ( ) d d dy d ( t) = t = t and = ; y( t) = t t = t 6 t and = t t dt dt dt dt d d y a() t =, = (, t t) a() = (,) dt dt 5. D Area = ( top curve bottom curve ), < ; Area = ( ( ) ( )) d f g d a 6. E ( ) tan sec f( ), f ( ), f = = tan = = tan 7. A du u d du = sin d sin C a = + a u 5 5 8. C f( ) f() lim = f () so the derivative of f at = is. 9. B Take the derivative of each side of the equation with respect to. yy + y + y + y =, substitute the point (,) ()() y + + ()() y + ()() = y =. A Take the derivative of the general term with respect to : ( ) d d d = = = d. A ln ( ln( ) ) n= n + n AP Calculus Multiple-Choice Question Collection 88 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
. A Use partial fractions to rewrite ( )( ) 985 Calculus BC Solutions as + + d= d = ( ln ln + ) + C = ln + C + + + ( )( ). B f () =, f() =, f ( ) = 6 ; by the Mean Value Theorem, f() f() f () c = = for c (,). So, c 6c c( c ) = =. The only value in the open interval is.. C I. convergent: p-series with p = > II. divergent: Harmonic series which is known to diverge III. convergent: Geometric with r = < 5. C t t () t = + (w ) dw= + ( w w) = + t t = t t+ or, () t = t t+ C, () = C = so, () t = t t+ 6. C For ( ) 7. B ( ) f = we have continuity at =, however, ( ) d ( ) d f = () ln( ) + = ln( ) + = ln( ) + f = is not defined at =. 8. C 9. D sin( + ) d = sin( )( d) cos( ) C + = + + f ( ) f ( ) f ( ) f ( ) g( ) = e, g ( ) = e f ( ), g ( ) = e f ( ) + f ( ) e f ( ) f ( ( )) ( ) f ( ) ( ) g ( ) = e f ( ) + f ( ) = h( ) e h( ) = f ( ) + f ( ). C Look for concavity changes, there are. AP Calculus Multiple-Choice Question Collection 89 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
985 Calculus BC Solutions. B Use the technique of antiderivatives by parts: u = f dv= d ( ) sin ( ) cos du = f d v = f ( )sin d = f ( )cos + f ( )cos d and we are given that f ( )sin d= f( )cos + cos d f ( ) = f( ) =. A A= r, A= 6 when r = 8. Take the derivative with respect to t. da = r dr ; 96= (8) dr dr = 6 dt dt dt dt. C + h 5 + 8 d F( + h) F() 5 lim = lim = F () where F ( ) = + 8 h h h h. F ()= Alternate solution by L Hôpital s Rule: + h 5 5 + 8 d ( + h) + 8 lim = lim = 9= h h h d 8. D d ( ) Area = sin ( θ) θ= cos θ θ= θ sin θ = 5. C At rest when vt () =. t t t vt () = e te = e ( ) t, vt () = at t = only. 6. E Apply the log function, simplify, and differentiate. ln y = ln ( sin ) = ln ( sin ) y cos = ln ( sin ) + y = y ln ( sin ) + cot = sin ln sin + cot y sin ( ) ( ) ( ( ) ) 7. E Each of the right-hand sides represent the area of a rectangle with base length ( b a). I. Area under the curve is less than the area of the rectangle with height f ( b ). II. Area under the curve is more than the area of the rectangle with height f ( a ). III. Area under the curve is the same as the area of the rectangle with height f () c, a< c< b. Note that this is the Mean Value Theorem for Integrals. 8. E e d= e ( e d) + e e u. This is of the form, edu u= e, so + e e d= e + C e AP Calculus Multiple-Choice Question Collection 9 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
985 Calculus BC Solutions 9. D Let = t. sin sin t lim = lim = t t. B At t =, dy dy = dt = t + = = d d t t= 8 dt. D The center is =, so only C, D, or E are possible. Check the endpoints. ( ) At = : converges by alternating series test. n n= n At = : which is the harmonic series and known to diverge. n n=. E y( ) = 6, y ( ) = + 6+ 7 =, the slope of the normal is for the normal is = y+ 6 = ( + ) + y = 5. and an equation. C This is the differential equation for eponential growth. t t t y = y() e = e ; e = ; t = ln t = ln = ln. A This topic is no longer part of the AP Course Description. ρ s where ρ= = y d ( ) Surface Area = y + dy = y + y dy = y + 9y dy dy 5. B Use shells (which is no longer part of the AP Course Description) rh where r = and h = y = 6 ( ) 6 Volume = 6 d AP Calculus Multiple-Choice Question Collection 9 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
985 Calculus BC Solutions 6. E d d lim d lim = = + + = which does not eist. L L L L 7. A This topic is no longer part of the AP Course Description. y = yh + yp where dy the solution to the homogeneous equation y solution to the given differential equation. Substitute d + = and p ( ) p yh = ce is y = A + B e is a particular y into the differential equation to determine the values of A and B. The answer is A=, B=. 8. C ( ) ( + e ) ln 5 5 e ln( 5 ) lim ln( 5e ) lim lim + e + + 5e lim + 5e = lim e = e = e = e = e 9. A Square cross sections: y where y = e. ( V = e d= e = e 6 ). A u =, du = d; when =, u = and when =, u = u u d = du = du u u = = + = + = + = = =. C y, L ( y ) d d ( ) ( ) ( 8 ). E Since e u = + u+ u + u +, then!! 9 The coefficient we want is =! ( ) ( ) e = + + + +!!. E Graphs A and B contradict f <. Graph C contradicts f () does not eist. Graph D contradicts continuity on the interval [,]. Graph E meets all given conditions.. A dy dy = y = d ln y = + K; y = Ce and y() = 8 so, y= 8e d y AP Calculus Multiple-Choice Question Collection 9 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
985 Calculus BC Solutions 5. D The epression is a Riemann sum with = and f( ) =. n The evaluation points are:,,,, n n n n n Thus the right Riemann sum is for = to =. The limit is equal to d. AP Calculus Multiple-Choice Question Collection 9 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
988 Calculus BC Solutions. A ( ) d= = = only. 6. D ( ) d ( ) ( d) ( ) ( ) 9 + = + = + = = 6 6. B. E f( ) = ln = ln ; f ( ) = f ( ) = uv ( uv + u v) w uvw uv w + u vw uvw = w = w w 5. C lim f ( ) = f ( a ) for all values of a ecept. a lim f ( ) = lim ( ) = = f () y 6. C y y y y = y = y 7. A 8. A d L d = lim = lim = lim = L L L L f ( ) = e, f () = e, lne = L 9. D II does not work since the slope of f at = is not equal to f ( ) work. For eample, f ( ) = e in I and f ( ) = sin in III.. D This limit is the derivative of sin.. A The slope of the line is. Both I and III could, so the slope of the tangent line at = is 7 f () = 7. 7. B vt ( ) = t+ C and v() = C= and vt ( ) = t+. Distance = ( t+ ) dt = + t = AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
5 t t. B The Maclaurin series for sin t is t +. Let t =.! 5! 5 n n ( ) ( ) ( ) ( ) sin( ) = + + +! 5! (n )! 988 Calculus BC Solutions A Use the Fundamental Theorem of Calculus: d( ) 6 + ( ) = + d 5. E d d dy d y = t +, t, ; y ln(t ), ; dt = dt = = + dt = t + dt = (t + ) 6. A Use the technique of antiderivatives by parts u = dv= e d du = d v = e e e d= e e + C 7. D Use partial fractions: 8 d = d = ( ln ln + ) = ln ln 5 ln+ ln = ln + + ( )( ) 5 ( ) e e e e 8. E = =, T = () + + + = ( e + e + e + e ) 9. B Make a sketch. < one zero, < < 5 no zeros, > 5 one zero for a total of zeros. E This is the definition of a limit. = = =. D d ( ) ln ln ln ln. E Quick Solution: f must have a factor of f which makes E the only option. Or, f ( ) ln f( ) = ln( + ) = + ln( + ) f ( ) = f( ) ln( ) f( ) + + + + AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
988 Calculus BC Solutions. E r = when cos θ= θ=±. The region is for the interval from θ= to θ=. 6 6 6 6 6 Area = ( cos ) d θ θ. D f ( ) =, f() = and f() =. By the Mean Value Theorem, f() f() = = f ( c) = c c for c (, ). So, c =. 5. D Square cross-sections: where. Volume 6 6 6 y y = = d= = 5 5. 6. C This is not true if f is not an even function. 7. B 8. E y ( ) = + a+ b, y ( ) = 6+ a, y () = a= y() = 6 so, 6 = + a+ b 6 = + b b= d cos d sin sin d d = = = tan cos cos cos 9. B Disks: Volume = y where = y. 5 5 ( y ) dy = ( y ) = 8 AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
988 Calculus BC Solutions. C This is an infinite geometric series with ratio and first term n. Sum = first n = = ratio n. C This integral gives ( ) = of the area of the circle with center at the origin and radius =.. E No longer covered in the AP Course Description. The solution is of the form y = y + y where y is the solution to y y = and the form of y is A + B+ K. Hence h p h yh of A, B, and K. = Ce. Substitute y p into the original differential equation to determine the values 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 y h. Both (A) and (B) fail, so the solution is (E). 9 dy. E ( ) L= + d= + d= + d d p. C At t =, dy dy dt t + t 8 = = = = ; the point at t = is (,) d d t + t= dt. ( ) y = + = 5. A Quick solution: For large the eponential function dominates any polynomial, so k lim =. + e k! Or, repeated use of L Hôpital s rule gives lim = lim = + e + e k AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
988 Calculus BC Solutions 6. E Disks: Volume = ( R r ) where R =, r = sin ( sin ) d Note that the epression in (E) can also be written as cos d cos ( ) = d = sin d and therefore option (D) is also a correct answer. 7. D + S S = = S 8 d ds = = = dt dt 9 8. C Check Check n= ( ) n =, which is convergent by alternating series test n =, which is the harmonic series and known to diverge. n n= 9. C dy sec d ln y tan k y Ce. y() 5 y 5e y = = + = = = tan tan. E Since f and g are inverses their derivatives at the inverse points are reciprocals. Thus, g ( ) f (5) = g ( ) = =. B Take the interval [,] and divide it into n pieces of equal length and form the right Riemann Sum for the function f ( ) value is given by d =. The limit of this sum is what is given and its AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
988 Calculus BC Solutions. A Let 5 = u, d= du, substitute f (5 ) d = f ( u)( du) = f ( u) du = f ( ) d = 6. A This is an eample of eponential growth, B = B t. Find the value of t so B = B. t ln B = B = t = ln t t ln= ln. A I. Converges by Alternate Series Test II Diverges by the nth term test: n lim n n ln III Diverges by Integral test: d lim ln ( ln ) = = L L 5. B A= ( )( y) = y and y =. 9 So A= 8. 9 A = 8 + 9 9 9 8 = (9 ) 9 9 A = at =±, the largest area is. The maimum area occurs when A= y = = = and y =. The value of AP Calculus Multiple-Choice Question Collection 5 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
99 Calculus BC Solutions. A. C ( ) d= = = 6 + lim =. E Q ( ) = p( ) degree of Q is n+. B If = then y = 5. dy d d d 6 + y = ; () + 5 = = dt dt dt dt 5 5. D r = sec θ; rcosθ= =. This is a vertical line through the point (,). 6. A dy d dy d d = = = = = = = = dt dt d t d d d t t dt d dy dy t d y d t, t thus t; dt 7. A e d = e ( d) = e = ( e ) 8. B 9. D f( ) = lne =, f ( ) = f ( ) =. This does not eist at =. D is false, all others are true. /. E I. ln is continuous for > II. e is continuous for all III. ln( e ) is continuous for >.. E ( ) d = lim 9 9 b b. This limit diverges. Another way to see this without doing the integration is to observe that the denominator behaves like which has a smaller degree than the degree of the numerator. This would imply that the integral will diverge. AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
99 Calculus BC Solutions. E vt ( ) = cos t+ sin t, at ( ) = sin t+ 9cos t, a( ) = 9.. C dy y,ln, = d y = + C y = Ce. Only C is of this form.. B The only place that f ( ) changes sign from positive to negative is at =. 5. D ( tan ) d f ( ) = e = tan sec e d tan tan 6. A I. Compare with p-series, p = 6 II. Geometric series with r = 7 III. Alternating harmonic series 7. A y+ y y+ y Using implicit differentiation, =. When =, = y =. y y Alternatively, ( ) e e e e y = e, y =, y = =. y () = 8. B f ( ) f ( ) e = f ( ) = = e + f ( ) 9. B Use cylindrical shells which is no longer part of the AP Course Description. Each shell is of the form rh where r = and h= k. Solve the equation k k k k k d. = ( ) = = k = 6.95.. E t vt () = e + and t () t = e + t+ AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
99 Calculus BC Solutions. A Use logarithms. y ln y = ln ( + 8) ln ( + ); = ; at (, ), y =. y + 8 + ( ) ( ). B f ( ) = e + e = e ( + ); f ( ) < for < <. D d dy L= + dt = t + dt dt dt. C This is L Hôpital s Rule. 5. D At t =, dy t t dy slope dt te + e t = = = = = =.5 d d t t 6 e e t= e dt 6. B e e = + ( e ) + e 7. C This is a geometric series with convergent for < <. r =. Convergence for < r <. Thus the series is 8. A t + 6 v=, t, v(),8,8 = = t + t 8 9. E Use the technique of antiderivatives by parts: u = and dv= sec d sec d= tan tan d= tan + ln cos + C. C Each slice is a disk with radius r = sec and width. d Volume = sec = tan = AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
99 Calculus BC Solutions. A s n 5+ n =, lim sn = = 5 + n n 5 5. B Only II is true. To see that neither I nor III must be true, let f( ) = and let on the interval [, 5]. 8 g ( ) = 5. A The value of this integral is. Option A is also and none of the others have a value of. Visualizing the graphs of y = sin and y = cos is a useful approach to the problem.. E Let y = PR and = RQ. d dy dy dy + y =, + y =, + y = y=. dt dt dt dt 9 5 Substitute into + y =. + =, =, = 6 6 5. A Fb ( ) Fa ( ) Apply the Mean Value Theorem to F. F () c = = =. This means that there is b a a number in the interval ( ab, ) for which F is zero. However, F ( ) = f( ). So, f( ) = for some number in the interval ( ab, ). 5 dv = + = = for < r < ; = 6r( r). The dr dv dv 5 maimum volume is when r = because > on, and < on,. dr dr 6. E v r h and h r v ( 5r r ) 7. B 8. C e e f( ) d= d+ d= + lne= dn kn N Ce kt = =. N() = C =. dt N() = e 67. 7 ln(.) N(7) = k = ln(.). Therefore 7 9. C Want y() y() where y ( ) = ln + C. This gives ln ln = ln = ln = ln.. C The interval is [,], =, =, =. S = ( + ln + ) = ln. Note that Simpson s rule is no longer part of the BC Course Description. AP Calculus Multiple-Choice Question Collection 5 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
= ; f < for < and f > for >. Thus f has its absolute minimum at =.. C f ( ) ( ) e ( ) csc ( ). E Suppose lim ln ( ) 99 Calculus BC Solutions + = A. The answer to the given question is csc Use L Hôpital s Rule: ( ) A e. ln( + ) lim ln ( + ) = lim = lim =. sin + cos. A 5 5 6 ( ) ( ) sin = + sin = + = +! 5!! 5!! 5!. E By the Intermediate Value Theorem there is a c satisfying a< c< b such that f ( c ) is equal to the average value of f on the interval [a,b]. But the average value is also given by b ( ) b a f d. Equating the two gives option E. a t a Alternatively, let Ft ( ) = f( d ). By the Mean Value Theorem, there is a c satisfying Fb ( ) Fa ( ) b a< c< b such that = F () c. But Fb ( ) Fa ( ) f( d ) b a =, and F () c = f() c by a the Fundamental Theorem of Calculus. This again gives option E as the answer. This result is called the Mean Value Theorem for Integrals. 5. D This is an infinite geometric series with a first term of sin and a ratio of sin. sin The series converges to tan sin = for (k+ ), k an integer. The answer is therefore tan =.6. AP Calculus Multiple-Choice Question Collection 6 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
997 Calculus BC Solutions: Part A. C. E. A. C 5. C 6. A 5 6 ( + ) d= + d= + = 5 5 t dy cos( t) cos( t) = e, y = sin( t); = = d t t e e 5 f( ) = ; f ( ) = 5 = (5 + )( ) = (5 + )( + )( ) ; f changes from positive to negative only at =. d ln ln e = ; so e = and ( ) = d f( ) = ( ) + e ; f ( ) = ( ) + e ; f () = + = y = (6 ) ; y = (6 ) ; y () = ; The slope of the normal line is 8. 8 7. C The slope at = is. The equation of the tangent line is y 5 = ( ). 8. E Points of inflection occur where f changes from increasing to decreasing, or from decreasing to increasing. There are si such points. 9. A f increases for 6 and decreases for 6 8. By comparing areas it is clear that f increases more than it decreases, so the absolute minimum must occur at the left endpoint, =.. B. C y = y+ + ; y = y + y+ ; at =, y = ; y = y + y = L d L ( + ) = lim ( + ) = lim = L ( + L ). A f changes from positive to negative once and from negative to positive twice. Thus one relative maimum and two relative minimums.. B at ( ) = t 7 and v() = 6; so vt ( ) = t 7t+ 6 = ( t )( t 6). Movement is right then left with the particle changing direction at t =,, therefore it will be farthest to the right at t =. AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
997 Calculus BC Solutions: Part A. C Geometric Series. r = < convergence. a= so the sum will be S = =. 8 8 5. D = cos t, y = sin t for t. d dy L= dt + dt dt ( cos sin ) (sin cos ) 9 cos sin 9sin cos L= t t + t t dt = t t+ t tdt 6. B 7. B 8. C 9. D. E h h h e e e e lim = lim = f (), where f( ) = e and f () =. lim = h h h h h h f( ) = ln( ); f ( ) =, f ( ) =, f ( ) = ; ( ) ( ) f() =, f () =, f () =, f () = ; a =, a =, a =, a = ( ) ( ) f( ) ( ) dy t + t 8 t + t 8 = t t, y = t + t 8 t; = =. Vertical tangents at d t t t(t ) f ( d ) f( d ) = ( A A) ( A) = A+ A n t =, ( ). The endpoints of the interval of convergence are when ( ) =± ; =, 5. n n= n Check endpoints: = gives the alternating harmonic series which converges. = 5 gives the harmonic series which diverges. Therefore the interval is < 5.. A Area = ((cos ) cos ) d cos d θ θ θ= θ θ. C g ( ) = f( ). The only critical value of g on ( ad, ) is at = c. Since g changes from positive to negative at = c, the absolute maimum for g occurs at this relative maimum. AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
997 Calculus BC Solutions: Part A. E. D = 5sin θ ; d = 5cos θ dθ ; When =, cos θ= ; d = 5 () = dt dt 5 dt 5 ( ) 6 7 f ( ) = sin( ) = + = + f( ) = + The! 6 7 coefficieint of is. 5. A This is the limit of a right Riemann sum of the function f ( ) n b b lim i = d= = b a n a = i ( ) a = on the interval [ ab,, ] so AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
997 Calculus BC Solutions: Part B 76. D Sequence 5 I ; sequence II ; sequence III. Therefore I and III only. 77. E Use shells (which is no longer part of the AP Course Description.) rh where r = and h= Volume = ( ) d= ( ) d 78. A ln( e+ h) ln( e+ h) ln e lim = lim = f ( e) where f( ) = ln h h h h 79. D Count the number of places where the graph of yt ( ) has a horizontal tangent line. Si places. 8 B Find the first turning point on the graph of y = f ( ). Occurs at =.9. 8. D f assumes every value between and on the interval (,6). Thus f( c ) = at least once. 8. B 8. E ( t t) dt tdt ; the interval [,] that satisfies this inequality is found to occur at =.887.. Using the calculator, the greatest value on dy k ln ln ( ln d ) ; ln y ln k ln k; y ee y Ce y = + = + + = + = =. Since y = when =, C =. Hence ln y = e. AP Calculus Multiple-Choice Question Collection 5 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
997 Calculus BC Solutions: Part B 8. C sin d; Use the technique of antiderivatives by parts with u = and dv = sin d. It will take iterations with a different choice of u and dv for the second iteration. sin d = cos + cos d ( ) = cos + sin sin d = cos + sin + cos + C 85. D f() f() 5 I. Average rate of change of f is =. True II. Not enough information to determine the average value of f. False III. Average value of f is the average rate of change of f. True 86. A Use partial fractions. A B = + ; = A ( + ) + B ( ) ( )( + ) + Choose = A= and choose = B=. d = d d = ln + C ( )( + ) + + 87. B Squares with sides of length. Volume = dy = ( y) dy 88. C For the average rate of change of f we need to f ( ) = sin tdt; f ( ) = sin( ); determine f () and f ( ). () f = and f ( ) = sin tdt =. The average rate of change of f on the interval is. See how many points of intersection there are for the graphs of y = sin( ) and y = on the interval,. There are two. AP Calculus Multiple-Choice Question Collection 6 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
997 Calculus BC Solutions: Part B 89. D ( ) t 5 ; () t f = dt f = dt = 5 + t + t.76 Or, () () f = f + d= 5 +.76 Both statements follow from the Fundamental Theorem of Calculus. 9. B 5 6 65 5 F( ) = k; = k k = ; Work = F( ) d d 5 inch-lbs = = = 6 AP Calculus Multiple-Choice Question Collection 7 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
998 Calculus BC Solutions: Part A. C f will be increasing when its derivative is positive. f ( ) = + 6 9= ( + ) f ( ) = ( + )( ) > for < or >.. A d dy dy = 5 and = = dt dt d 5. D Find the derivative implicitly and substitute. ( ) y + (()( ) + ) (() y + ( )) = ; y y + ( y+ ) ( y + y) = ; y + 6 y = ; y =. A Use partial fractions. = = 6+ 8 ( )( ) d = ( ln ln ) + C = ln + C 6+ 8 h ( ) = f g( ) g ( ) ; h ( ) = f g( ) g ( ) g ( ) + f g( ) g ( ) 5. A ( ) ( ) ( ) ( ) ( ) ( ) h ( ) = f g( ) g ( ) + f g( ) g ( ) 6. E The graph of h has turning points and one point of inflection. The graph of h will have -intercepts and one turning point. Only (C) and (E) are possible answers. Since the first turning point on the graph of h is a relative maimum, the first zero of h must be a place where the sign changes from positive to negative. This is option (E). 7. E e e d = d ln e e = = = e y ( ) = cos + C; Let =, = cos + C C= 8. B ( ) y () = cos =. ( ) 9. D Let rt ( ) be the rate of oil flow as given by the graph, where t is measured in hours. The total number of barrels is given by rtdt (). This can be approimated by counting the squares below the curve and above the horizontal ais. There are approimately five squares with area 6 barrels. Thus the total is about, barrels.. E ( ) ( ) vt ( ) = t,6(t ) and at ( ) = 6 t, (t ) a() = (6, ) AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
998 Calculus BC Solutions: Part A. A Since f is linear, its second derivative is zero and the integral gives the area of a rectangle with zero height and width ( b a). This area is zero.. E lim f ( ) = ln ln= lim f( ). Therefore the limit does not eist. +. B At = and = only. The graph has a non-vertical tangent line at every other point in the interval and so has a derivative at each of these other s.. E 5 5 sin + ; sin + = +! 5!! 5! 6 5. B Use the technique of antiderivatives by parts. Let u = and dv = cos d. cos d= sin sin d= sin + cos + C 6. C Inflection point will occur when f changes sign. f ( ) = 5. f ( ) = 6 6 = 6 ( ). The only sign change is at =. 7. D From the graph f () =. Since f () represents the slope of the graph at =, f () >. Also, since f () represents the concavity of the graph at =, f () <. 8. B I. Divergent. The limit of the nth term is not zero. II. Convergent. This is the same as the alternating harmonic series. III. Divergent. This is the harmonic series. 9. D Find the points of intersection of the two curves to determine the limits of integration.. E 5 sin when sin.5; this is at and. 6 6 ( ) θ= θ= θ= 6 ( ) ( ) = 8 = 8 5 Area = sin θ () dθ d d d d d = = (8) = = k = dt dt dt dt dt 6 C The length of this parametric curve is given by d dy ( ) + = + dt t t dt dt dt.. A This is the integral test applied to the series in (A). Thus the series in (A) converges. None of the others must be true. AP Calculus Multiple-Choice Question Collection Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
998 Calculus BC Solutions: Part A. E I. False. The relative maimum could be at a cusp. II. True. There is a critical point at = c where f ( c) eists III. True. If f ( c) >, then there would be a relative minimum, not maimum. C All slopes along the diagonal y = appear to be. This is consistent only with option (C). For each of the others you can see why they do not work. Option (A) does not work because all slopes at points with the same coordinate would have to be equal. Option (B) does not work because all slopes would have to be positive. Option (D) does not work because all slopes in the third quadrant would have to be positive. Option (E) does not work because there would only be slopes for y >. 5. C = lim = lim = b e d e d e b b. b 6. E dp As lim = for a population satisfying a logistic differential equation, this means that t dt P,. 7. D n n = n = n = n n n= n= n= n n n= n= If f( ) a, then f( ) na na. f () = na n = na 8. C Apply L Hôpital s rule. t e dt lim lim e e = = AP Calculus Multiple-Choice Question Collection 5 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
998 Calculus BC Solutions: Part B 76. D The first series is either the harmonic series or the alternating harmonic series depending on whether k is odd or even. It will converge if k is odd. The second series is geometric and will converge if k <. 77. E () ( t t f t e, sin t) ; f () t ( e, cost) = =. 78. B da dr A = r r dt = dt. However, C dr da = r and =.. Thus.C dt dt =. 79. A None. For every positive value of a the denominator will be zero for some value of. 8. B The area is given by ( ) + ln(cos ) d=.99 dy ; d y d = y = ( y ) = y ( y) dy = y d d d d 8. B ( ) 5 5 5 5 8. B [ ] [ ] f( ) + g ( ) d= g ( ) + 7 d= gd ( ) + (7)() = gd ( ) + 8. C Use a calculator. The maimum for..7 occurs at =.. ln ( ) ( ) ( ) + on the interval 8. B You may use the ratio test. However, the series will converge if the numerator is ( ) n and diverge if the numerator is n. Any value of for which + > in the numerator will make the series diverge. Hence the interval is <. 85. C There are trapezoids. ( f() + f(5) ) + ( f(5) + f(7) ) + ( f(7) + f(8) ) 86. C Each cross section is a semicircle with a diameter of y. The volume would be given by 8 8 y 8 d = d = 6.755 8 AP Calculus Multiple-Choice Question Collection 6 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.
998 Calculus BC Solutions: Part B 87. D Find the for which f ( ) =. f ( ) = + = only for =.7. Then f (.7) =.5. So the equation is y.5 =.7. This is equivalent to option (D). 88. C From the given information, f is the derivative of g. We want a graph for f that represents the slopes of the graph g. The slope of g is zero at a and b. Also the slope of g changes from positive to negative at one point between a and b. This is true only for figure (C). 89. A The series is the Maclaurin epansion of e. Use the calculator to solve e =. 9. A Constant acceleration means linear velocity which in turn leads to quadratic position. Only the graph in (A) is quadratic with initial s =. t ( ) = + ( ) + 5 + + 8 = ft/sec. 9. E vt a d [ ] 9. D f ( ) =, f () =, and f() =, so an equation for the tangent line is y =. The difference between the function and the tangent line is represented by ( ). Solve ( ) <.5. This inequality is satisfied for all such that.5 < < +.5. This is the same as.9 < <.77. Thus the largest value in the list that satisfies the inequality is.7. AP Calculus Multiple-Choice Question Collection 7 Copyright 5 by College Board. All rights reserved. Available at apcentral.collegeboard.com.