AP Calculus AB AP Calculus BC Free-Response Questions and Solutions 979 988 Copyright 4 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of 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.
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Notes About AP Calculus Free-Response Questions The solution to each free-response question is based on the scoring guidelines from the AP Reading. Where appropriate, modifications have been made by the editor to clarify the solution. Other mathematically correct solutions are possible. Scientific calculators were permitted, but not required, on the AP Calculus Exams in 983 and 984. Scientific (nongraphing) calculators were required on the AP Calculus Exams in 993 and 994. Graphing calculators have been required on the AP Calculus Exams since 995. From 995 to 999, the calculator could be used on all six free-response questions. Since the exams, the free-response section has consisted of two parts -- Part A (questions - 3) requires a graphing calculator and Part B (questions 4-6) does not allow the use of a calculator. Always refer to the most recent edition of the Course Description for AP Calculus AB and BC for the most current topic outline, as earlier exams may not reflect current exam topics. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB Given the function f defined by 3 f( x) = x 3x x+. (a) Find the zeros of f. Write an equation of the line normal to the graph of f at x =. (c) Find the x- and y-coordinates of all points on the graph of f where the line tangent to the graph is parallel to the x-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB Solution (a) f( x) = x 3x x+ = ( x )(x + x ) = ( x ) x+ 5 The zeros of f are at x = and x =. 3 5 f ( x) = 6x 6x ; f () = The slope of the normal line is m= = ; f () = f () The equation of the normal line is y = ( x ), or y = x+, or y = x+ 4 (c) The tangent line will be parallel to the x-axis if the slope is. f x = x x = ( ) 6 6 6( x )( x+ ) = x=, f ( ) = 7 and f () = The coordinates are (,7) and (,). Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB A function f is defined by x f( x) = xe with domain x. (a) Find all values of x for which the graph of f is increasing and all values of x for which the graph is decreasing. Give the x- and y-coordinates of all absolute maximum and minimum points on the graph of f. Justify your answers. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB Solution (a) x x x f ( x) = e xe = e ( x) f ( x) > when x>. The graph of f is increasing for x <. f ( x) < when x<. The graph of f is decreasing for < x. x f ( x) = e ( x) = There is a critical point when x =, hence only at x =. < x< f ( x) > < x< f ( x) < The graph of f increases and then decreases on the interval x. Therefore the absolute maximum point is at, e. The absolute minimum value must be at an endpoint. f() =, f() = e Therefore the absolute minimum point is at (,) The absolute maximum can also be justified by using the second derivative test to show that there is a relative maximum at x =, then observing that the absolute maximum also occurs at this x value since it is the only critical point in the domain. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB3/BC3 Find the maximum volume of a box that can be made by cutting out squares from the corners of an 8-inch by 5-inch rectangular sheet of cardboard and folding up the sides. Justify your answer. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB3/BC3 Solution 3 V( x) = x(8 x)(5 x) = 4x 46x + x V ( x) = x 9x+ 3x 3x+ 3 = (3x 5)( x 6) = 5 x=, x= 6 3 Since we must have x 4, we pick 5 x =. 3 V max 5 5 4 35 45 = 8 5 = = = 9 9.7 3 3 3 3 3 3 7 7 Justification using the st derivative test: 5 x< V ( x) > 3 5 x> V ( x) < 3 or There is therefore a relative maximum at the absolute maximum is at 5 x =. 3 5 x =. But since V () = and V (4) =, 3 Justification using the nd derivative test: 5 V < and so there is a relative maximum at 3 5 x =. There is only one critical 3 5 x =. 3 point in the domain x 4, so there is an absolute maximum at Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB4/BC A particle moves along a line so that at any time t its position is given by x() t = π t + cosπ t. (a) Find the velocity at time t. Find the acceleration at time t. (c) (d) What are all values of t, t 3, for which the particle is at rest? What is the maximum velocity? Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB4/BC Solution (a) vt ( ) = π πsin π t= π( sin π t) at () = 4π cosπ t (c) vt ( ) = π( sin π t) = sin π t = The particle is at rest for 5 9 t =,,. 4 4 4 (d) at () = 4π cosπ t= 3 5 t =,,, 4 4 4 The maximum velocity is or 3 v = 4π. 4 Since sin π t = is the minimum of sin π t, the maximum of vt ( ) is π( ( ) ) = 4π. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB5/BC5 Let R be the region bounded by the graph of y = ln x, the x-axis, and the line x = e. x (a) Find the area of the region R. Find the volume of the solid formed by revolving the region R about the y-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB5/BC5 Solution (a) ln x = x = x Area = ln xdx ( ln x) e e = = x e Volume = π x ln x dx= π ln xdx x e Use integration by parts with u = ln x dv = dx du = dx x v = x Volume ln e e ( ln ) e = π x x dx = π x x x = π Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB6 The curve in the figure represents the graph of f, where numbers x. f ( x) = x x for all real (a) On the axes provided, sketch the graph of y = f( x). Determine whether the derivative of f ( x ) exists at x =. Justify your answer. (c) On the axes provided, sketch the graph of y = f ( x). (d) Determine whether y f ( x) = is continuous at x =. Justify your answer. Axes for (a) Axes for (c) Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB6 Solution (a) y f( x) = (c) y = f ( x) d d x < f( x) = x x f( x) x lim f( x) dx = dx = x < d d x < f( x) = x + x f( x) x lim f( x) dx = + dx = + x If f ( x ) were differentiable at x =, then since both limits above exist, they would have to be equal. They are not equal, so the derivative of f ( x ) does not exist at. Alternatively, the derivative of f ( x ) does not exist at x = because f( + h) f() h h lim = lim = lim ( h ) = h h h h h f( + h) f() h + h lim = lim = lim ( h + ) = + h h h x h (d) x f ( x) At =, = ( ) ( ) lim f x = lim x x = x x ( ) Therefore f x is continuous at x=. or x f ( x) ( ) At =, = lim f x = lim f( x) = + + x x ( ) lim f x = lim f( x) = x x ( ) ( ) lim f x = lim f x = + x x ( ) Therefore f x is continuous at x=. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB7 Let f be the function defined by properties. 3 y = f ( x) = x + ax + bx + c and having the following (i) The graph of f has a point of inflection at (, ). (ii) The average (mean) value of f ( x ) on the closed interval [,] is 3. (a) Determine the values of a, b, and c. Determine the value of x that satisfies the conclusion of the Mean Value Theorem for f on the closed interval [,3]. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 AB7 Solution (a) = f () = c f ( x) = 3x + ax+ b f ( x) = 6x+ a = f () = a, so a = 3 f( x) = x + bx 4 x bx 3 = f ( xdx ) = + x = ( 4+ b 4) = b 4 So a =, b = 3, and c =. By the Mean Value Theorem, there is an x satisfying < x < 3 such that f ( x) = f (3) f () 3 6 ( ) 3x 3= = 6 3 x = 3 x= 3 Copyright 4 by College Entrance Examination Board. All rights reserved.
979 BC Given the differential equation py + y y = qx (a) Find the general solution of the differential equation when p= and q=. Find the general solution of the differential equation when p= and q=. (c) Find the general solution of the differential equation when p= and q=. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 BC Solution (a) p=, q= y y = dy = dx y ln y = x+ ln C x lnc y = e + The general solution is y = Ce x. p=, q= y + y y = m + m = ( m+ )( m ) = m =, The general solution is x x y = C e + C e. (c) p=, q= y + y y = x x yh = Ce + Ce from Let x yp = Ax+ B. Then y p = A and y p =. + A Ax B= x A B= A = The solution is A=, B= x x Hence the general solution is y = yh + yp = Ce + Ce x. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 BC4 Let f be the function defined by f( x) =. x (a) (c) Write the first four terms and the general term of the Taylor series expansion of f ( x ) about x =. What is the interval of convergence for the series found in part (a)? Show your method. Find the value of f at x =. How many terms of the series are adequate for 4 approximating f with an error not exceeding one per cent? Justify your 4 answer. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 BC4 Solutions (a) ( n) f () The Taylor series has the form ( x ) n= n! n= : f( x) = ; f() = x n= : f ( x) = ; f () = x ( x) 8 n= : f ( x) = ; f () = 8 4x 3 ( x) 48 3 n= 3; f ( x) = ; f () = 48 8x 4 ( x) n The general term is n x n. n+ n+ x = x so series converges for n n x n n x = : ( ) = () ; divergent n n= n= n n x < n x = : ( ) = ( ) ; divergent n= n= Interval of convergence < x < (c) f = =. 4 + / 3 n n n = ( ) 4 n Alternating series with absolute value of terms a decreasing to. n S aj < an j= = n 3 Since n <, 8 terms suffice. 56 3 or n ( n) n f c j j= n () S a for < c<! 4 4 n n n ( ) n!( c) ( ) = n! 4 n ( c) n 4 n = n n 4 3 Copyright 4 by College Entrance Examination Board. All rights reserved. n n
979 BC6 A particle moves in the xy-plane so that at any time t its position ( x, y ) is given by t t x e e = + and t t y e e =. (a) Find the velocity vector for any t. (c) Find lim t dy dt dx dt. The particle moves on a hyperbola. Find an equation for this hyperbola in terms of x and y. (d) On the axes provided, sketch the path of the particle showing the velocity vector for t =. Copyright 4 by College Entrance Examination Board. All rights reserved.
979 BC6 Solution (a) dx dy = e e = e + e dt dt t t t t t t t t t t t t vt () = ( e e, e + e ) = ( e e ) i+ ( e + e ) j Method Method Method 3 Method 4 + lim e t e t t = t e lim + t t e e t = e lim t e t t + t e = lim = t t e lim coth( t) t = (c) t t x = cosh x x = e + + e or t t y = e + e y = sinh x Therefore x y = 4 (d) v() = j = (,) Copyright 4 by College Entrance Examination Board. All rights reserved.
979 BC7 Let f be a function with domain the set of all real numbers and having the following properties. (i) f ( x+ y) = f( x) f( y) for all real numbers x and y. f( h) (ii) lim = k, where k is a nonzero real number. h h (a) Use these properties and a definition of the derivative to show that f ( x) exists for all real numbers x. ( n) Let f denote the nth derivative of f. Write an expression for of f ( x ). ( n f ) ( x ) in terms (c) Given that f () =, use the Mean Value Theorem to show that there exists a 7 number c such that < c < 3 and f () c =. 3 Copyright 4 by College Entrance Examination Board. All rights reserved.
979 BC7 Solution (a) f( x+ h) f( x) f( x) f( h) f( x) f( h) f ( x) = lim = lim = lim f( x) = k f( x) h h h h h h f ( x) = k f ( x) = k f( x) By induction, ( n ) n f ( x) = k f( x) (c) Property (i) gives f () = f( + ) = f() f() Therefore f () = By Property (i), f() = f() f() = 4 By Property (i), f(3) = f() f() = 8 By the Mean Value Theorem, there is a c satisfying < c < 3 such that f(3) f() 8 7 f () c = = = 3 3 3 Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB Let R be the region enclosed by the graphs of 3 y = x and y = x. (a) Find the area of R. Find the volume of the solid generated by revolving R about the x-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB Solution (a) The intersection of the two graphs is at (, ) and (, ). Area = 3 4 3 3 4 ( x x ) dx= x x = = 5 3 4 or 3 43 3 y y dy = y y Area = ( ) = 3 = 5 4 3 3 4 3 Volume = = π ( x x ) dx=π x x 7 6 7 5 π = π 7 4 or Volume = π ( 3 ) = π ( 43 3 ) = = y y ydy y y dy 3 π y y 7 4 7/3 4 3 5 π = π 7 4 4 Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB A rectangle ABCD with sides parallel to the coordinate axes is inscribed in the region enclosed by the graph of y = 4x + 4 and the x-axis as shown in the figure above. (a) Find the x- and y-coordinates of C so that the area of rectangle ABCD is a maximum. The point C moves along the curve with its x-coordinate increasing at the constant rate of units per second. Find the rate of change of the area of rectangle ABCD when x =. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB Solution (a) 3 Ax ( ) = x( 4x + 4) = 8( x x) da = 8( 3 x ) dx da = when x = dx 3 The maximum area occurs when x = and 3 8 y = 4 =. 3 3 3 Ax ( ) = 8( x x) da dx = 8( 3 x ) dt dt dx When x = and dt =, da = 8 3 = 4. dt 4 or A= xy da x dy y dx = + dt dt dt When da dt dx x = and dt =, y = 4x + 4= 4 + 4= 3 dy dx = 8x = 8 = 8 dt dt = ( 8) + 3 = 4 Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB3 x Let ln( x ) for x > and gx ( ) = e for x. Let H be the composition of f with g, that is, H( x) = f( g( x )), and let K be the composition of g with f, that is K( x) = g( f( x)). (a) Find the domain of H and write an expression for H( x ) that does not contain the exponential function. Find the domain of K and write an expression for K( x ) that does not contain the exponential function. (c) Find an expression for f find the domain of f. ( x), where f denotes the inverse function of f, and Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB3 Solution (a) The domain of H consists of x for which x and domain is x. x gx ( ) = e >. Hence the x 4x H( x) = f( g( x)) = ln(( e ) ) = ln( e ) = 4x for x The domain of K consists of x for which x > and domain is x. f( x) = ln( x ). Hence the 4 ln( x ) ln x 4 K( x) = g( f( x)) = e = e = x for x (c) y y = ln x e = x y x = e = e y x f ( x) = e The domain of f is the range of f which is the set of all real numbers. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB4/BC The acceleration of a particle moving along a straight line is given by t a = e. (a) Write an expression for the velocity v, in terms of time t, if v= 5 when t =. (c) During the time that the velocity increases from 5 to 5, how far does the particle travel? Write an expression for the position s, in terms of time t, of the particle if s = when t =. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB4/BC Solution (a) a = e t t t v= e dt = 5e + C v() = 5 C = Therefore v= 5e t v= 5 when t = t v= 5 5e = 5 t = ln 3 ln 3 ln 3 t 5 t e dt e Distance = 5 = or 5 ln 3 5 = e = 5 t 5 t s = 5e dt = e + C Distance = ln 3 ln 3 ln 3 ( ) 5 5 s s = e + C + C = 5 e 5 = 5 (c) t 5 t s = 5e dt = e + C 5 s() = C = Therefore 5 t 5 s = e Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB5/BC Given the function f defined by f( x) = cos x cos x for π x π. (a) Find the x-intercepts of the graph of f. (c) (d) Find the x- and y-coordinates of all relative maximum points of f. Justify your answer. Find the intervals on which the graph of f is increasing. Using the information found in parts (a),, and (c), sketch the graph of f on the axes provided. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB5/BC Solution (a) f ( x) = cos x ( cos x) π π Either cos x = or cosx =, so the x-intercepts are x =, x =, and x =. f ( x) = sin x+ sin xcos x = sin x ( + cos x) π Either sin x = or cos x =, so the candidates are x = ±π, x =, and x =±. 3 π The relative maximum points are at ±, 3 4. Justification: (i) f ( x) = cos x+ cos x f ( ±π ) = 3 relative minimum f () = relative minimum π 3 f ± = relative maximum 3 or (ii) Selecting critical values: x or (iii) Sign chart: π π 3 f ( x ) /4 /4 π 3 π (c) (d) π π Graph of f increases on the intervals π< x < and < x <. 3 3 Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB6/BC4 4 4 Let y = f( x) be the continuous function that satisfies the equation x 5x y + 4y = and whose graph contains the points (,) and (, ). Let be the line tangent to the graph of f at x =. (a) Find an expression for y. Write an equation for line. (c) Give the coordinates of a point that is on the graph of f but is not on line. (d) Give the coordinates of a point that is on line but is not on the graph of f. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB6/BC4 Solution Solution : (a) 3 3 4x xy x yy + 6y y = 3 3 4x xy x 5xy y = = x y 6y 5x y 8y 3 3 6 The slope is the value of y at the point (,), so m = =. 8 The equation of is therefore y = ( x ) or y = x. (c) The point (, ) is one example. Any point of the form ( aa, ) for a < will be on the graph of f but not on the line. (For reason, see solution.) a (d) The point (, ) is one example. Any point of the form a, for a < will be on the line but not on the graph of f. (For reason, see solution.) Solution : (a) The equation can be rewritten as( x y)( x+ y)( x y)( x+ y ) =. Four different lines passing through the origin satisfy this implicit equation. Because y = f( x) is continuous, only one line can be used for x < and x, respectively. Which line is used is determined by the two points that are given as being on the graph. So we must have x, x y = x, x<, x > y =, x < y = x (c) ( aa, ) for a < (d) a a, for a < Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB7 Let p and q be real numbers and let f be the function defined by: + px + x x ( ) ( ), for f( x) = qx + p, for x >. (a) Find the value of q, in terms of p, for which f is continuous at x =. Find the values of p and q for which f is differentiable at x =. (c) If p and q have the values determined in part, is f a continuous function? Justify your answer. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB7 Solution (a) (c) Must have f() = lim f( x) and f() = x lim f( x) = p+ q and lim f( x) = + x x Therefore p+ q= q= p lim f ( x) = p x lim f ( x) = q + x So for f () to exist, p = q f () exists implies that f is continuous at x =. Therefore q= p. Hence p = p. p =, q =. 3 3 No, f is not a continuous function because it is not continuous at x =. This is because f is not defined at x =, or because lim f ( x) = and lim f ( x) =. + x x or Yes, f is a continuous function because f is continuous at each point of its domain. (Note: Different answers were accepted on the 98 grading standard because students might have interpreted the question either as asking if f is a continuous function for all real numbers, or a continuous function on its domain.) Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC3 (a) Determine whether the series answer. 4n A = converges or diverges. Justify your n + n= (c) If S is the series formed by multiplying the nth term in A by the nth term in, write an expression using summation notation for S. n n= Determine whether the series S found in part converges or diverges. Justify your answer. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC3 Solution (a) Comparison or limit comparison test 4n > since 4n > n + or n + n Since the series Integral test b diverges, so does series A. n n= 4 lim x lim ln( ) dx = x + = x + b b Since the integral diverges, so does series A. b 4n 4n lim lim 4 n n n = = + n n + (c) n= n + Comparison or limit comparison test < n + n or n lim lim n = = n + n n n + Since the series converges, so does series S. n n= Integral test lim b b π dx = lim tan x = x + b Since the integral converges, so does series S. b Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC5 (a) Find the general solution of the differential equation xy + y =. Find the general solution of the differential equation xy + y = x y. (c) Find the particular solution of the differential equation in part that satisfies the condition that y = e when x =. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC5 Solution (a) xy + y = Solution using separation of variables y = ln y = ln x+ C y x Solution using integrating factor y + y =, P( x) = x x y = Ae x dx In either case, y = Ae ln x or A y = or xy = A or ln y = ln x + C x xy + y = x y Solution using separation of variables y x = = x y x x ln y = x ln x+ C Solution using integrating factor x y + y =, P( x) = x x x ( x x ) dx y = Ae In either case, x Ae y = or x x ln x y = Ae or ln y = x ln x + C (c) e = Ae e= A Therefore x x ee e y = = or x x + ln y = x ln x+ Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC6 Let R be the region enclosed by the graphs of y e x, x k ( k ) axes. = = >, and the coordinate (a) (c) Write an improper integral that represents the limit of the area of the region R as k increases without bound and find the value of the integral if it exists. Find the volume, in terms of k, of the solid generated if R is rotated about the y-axis. Find the volume, in terms of k, of the solid whose base is R and whose cross sections perpendicular to the x-axis are squares. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC6 Solution (a) Area = x k x = lim k e dx e dx x = lim ( e ) k k k = lim ( e + ) = k Volume = k x x π xe dx ( u = x; v' = e ) x k k x = π xe e dx x x = π( xe e ) k k k = π( ke e + ) (c) Volume = ( ) = = = ( ) k x k x x k k e dx e dx e e Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC7 Note: This is the graph of the derivative of f, NOT the graph of f. Let f be a function that has domain the closed interval [,4] and range the closed interval [,]. Let f( ) =, f() =, and f (4) =. Also let f have the derivative function f that is continuous and that has the graph shown in the figure above. (a) (c) (d) Find all values of x for which f assumes a relative maximum. Justify your answer. Find all values of x for which f assumes its absolute minimum. Justify your answer. Find the intervals on which f is concave downward. Give all the values of x for which f has a point of inflection. (e) On the axes provided, sketch the graph of f. Note: The graph of f has been slightly modified from the original on the 98 exam to be consistent with the given values of f at x =, x =, and x = 4. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC7 Solution (a) f ( x) = at x=, There is a relative maximum at x =, since f () = and f ( x) changes from positive to negative at x =. There is no minimum at x =, since f ( x) does not change sign there. So the absolute minimum must occur at an endpoint. Since f ( ) < f (4), the absolute minimum occurs at x =. (c) The graph of f is concave down on the intervals [, ) and (,3) because f is decreasing on those intervals. (d) The graph of f has a point of inflection at x =,, and 3 because f changes from decreasing to increasing or from increasing to decreasing at each of those x values. (e) Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB 4 Let f be the function defined by f( x) = x 3x +. (a) Find the zeros of f. Write an equation of the line tangent to the graph of f at the point where x =. (c) Find the x-coordinate of each point at which the line tangent to the graph of f is parallel to the line y = x+ 4. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB Solution (a) f( x) = ( x )( x ) = ( x+ )( x )( x+ )( x ) The zeros are x =±, ±. 3 f ( x) = 4x 6x f () = Point: (,) The equation of the tangent line is y = ( x ). (c) 3 4x 6x= 3 4x 6x+ = One solution is x =. ( x )(4x + 4x ) = The other two solutions are ( 3) x = ±. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB Let R be the region in the first quadrant enclosed by the graphs of and the y-axis. y = 4 x, y = 3 x, (a) Find the area of region R. Find the volume of the solid formed by revolving the region R about the x-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB Solution (a) Intersection: 4 x = 3x x + 3x 4= x = 4, x = 3 Area = ( ) or 3 3 4 x 3x dx= 4x x x = 3 6 3 ydy+ y dy= 3 6 3 Area = ( 4 ) Disks: Volume =π ( 4 ) ( 3 ) or 4 x x dx ( 6 7 ) =π x + x dx 7 3 5 58π =π 6x x + x = 3 5 5 y y dy y y dy 3 3 4 Shells: Volume = π + π ( 4 ) 3 3 3 4 3 5 y 4 58π = π + π y( 4 y) ( 4 y) = 9 3 5 5 3 Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB3/BC Let f be the function defined by 3 f ( x) = x 4x. (a) (c) (d) (e) Find the intervals on which f is increasing. Find the x- and y-coordinates of all relative maximum points. Find the x- and y-coordinates of all relative minimum points. Find the intervals on which f is concave downward. Using the information found in parts (a),, (c), and (d), sketch the graph of f on the axes provided. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB3/BC Solution (a) /3 /3 f( x) = x 4 x; f ( x) = 8x 4 /3 (8x 4) >, x> x< 8 /3 (8x 4) >, x< no x satisfies this or Critical numbers: x = 8, x = Therefore f is increasing on the interval < x < 8. nd Derivative Test: 8 4/3 f ( x) = x 3 f (8) < relative maximum at (8,6) (c) (d) The nd Derivative Test cannot be used at x = where the second derivative is undefined. Since f ( x) < for x just less than, and f ( x) > for x just greater than, there is a relative minimum at (,). 8 4/3 f ( x) = x < if x 3 The graph of f is concave down on (,) and (, + ). (e) Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB4 Let f be the function defined by f( x) = 5 x. (a) Is f an even or odd function? Justify your answer. Find the domain of f. (c) Find the range of f. (d) Find f ( x). Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB4 Solution (a) f is even since ( x) x f ( x) = 5 = 5 = f( x) (c) x x Domain is,, x x 5 Range is, [ ) (d) ( ) x f x = 5 ln5 4x x or y = 5 x ln y = x ln 5 y = 4x ln5 y x y y = 4x ln5 x x 5 y = 4x ln5 x Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB5/BC x+, for x Let f be the function defined by f( x) = x + k,for x> (a) For what value of k will f be continuous at x =? Justify your answer. Using the value of k found in part (a), determine whether f is differentiable at x =. Use the definition of the derivative to justify your answer. (c) Let k = 4. Determine whether f is differentiable at x =. Justify your answer. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB5/BC Solution (a) (c) f () = 5 lim (x + ) = 5 x lim x k k + + = + x For continuity at x =, we must have + k = 5, and so k = 3. f( x) f() We compute lim : x x x + 5 lim = x x x + 3 5 lim = + x x So f () exists and f () = When k = 4, lim f( x) = lim x 4 6 + + + = x x Hence f is not continuous at x = and so is not differentiable at x =. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB6/BC4 A particle moves along the x-axis so that at time t its position is given by ( t ) x() t = sin π for t. (a) Find the velocity at time t. Find the acceleration at time t. (c) (d) For what values of t does the particle change direction? Find all values of t for which the particle is moving to the left. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB6/BC4 Solution (a) (c) vt () = x () t = πtcosπ t at () = v () t = πcosπt 4π t sinπ t vt () = t = or cosπ t = π π t =± The particle changes direction at t =±,. (d) The particle is moving to the left when vt ( ) <. t + + cos π t + + vt () + + Particle moves to the left when < t < or < t <. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB7 Let f be a continuous function that is defined for all real numbers x and that has the following properties. (i) 3 5 f( x) dx= (ii) 5 f( x) dx= (a) Find the average (mean) value of f over the closed interval [,3]. Find the value of ( f x + ) 5 ( ) 6 dx. 3 (c) Given that f ( x) = ax+ b, find the values of a and b. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB7 Solution (a) Average (mean) value 3 5 5 ( ) 3 f x dx 4 = = = 5 ( f( x) + 6) 3 dx 5 5 = f ( x) dx + 6dx 3 3 5 3 5 = f ( xdx ) f( xdx ) + 6dx 3 5 = + 6(5 3) = 5 + = 7 (c) 3 5 ax = ( ax b) dx bx 4a b + = + = + 5 ax = ( ax + b) dx = + bx = a + 4b Solving these two simultaneous equations yields 3 5 5 5 a=, b=. 4 4 Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC3 Let S be the series t S = where t. + t n= n (a) Find the value to which S converges when t =. Determine the values of t for which S converges. Justify your answer. (c) Find all the values of t that make the sum of the series S greater than. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC3 Solution (a) t t = = + t Then S = = = n n= t Since S is a geometric series, S converges if and only if r = <. + t (i) We must have t < t+. This means that the distance of t from is less than the distance of t from. Therefore t >. or (ii) We must have t < t + t+. Therefore or (iii) If t < : t+ < t < t t+ < t < no solution If t > : t < t < t+ < t < t Therefore t >. t >. (c) St () = = + t> for t> 9 t + t Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC5 (a) Find the general solution of the differential equation d y dy 6y dx + dx =. (c) Find the particular solution for the differential equation in part (a) that satisfies the dy conditions that y = and = when x=. dx d y dy x Find the general solution of the differential equation 6y e dx + dx =. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC5 Solution (a) (c) m + m 6= ( m+ 3)( m ) = m = 3, x 3x y = Ae + Be 3 x x y = Ae 3Be y = when x = gives A+ B= y = when x = gives A 3B= 3 Therefore A = and B =. 5 5 x 3 3x y = e + e 5 5 From part (a), the homogeneous solution is Try a particular solution of the form y y y Ce x p = p = p = x x x x Ce + Ce 6Ce = e 4C = C = 4 Therefore the general solution is yp h x = Ce. x 3x y = Ae + Be. x 3x x y = Ae + Be e. 4 Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC6 (a) A solid is constructed so that it has a circular base of radius r centimeters and every plane section perpendicular to a certain diameter of the base is a square, with a side of the square being a chord of the circle. Find the volume of the solid. If the solid described in part (a) expands so that the radius of the base increases at a constant rate of centimeters per minute, how fast is the volume changing when the radius is 4 centimeters? Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC6 Solution (a) The cross section at x has area Ax ( ) = ( y) = 4 r x. ( ) r Volume = 4( r ) = 4( ) r x dx r r x dx r x y = r 3 6 3 8 r x x = r 3 3 V = 6 3 r 3 dv dr = 6r = 6(4 ) = 8 dt dt or dr dt = for all t implies that r = t+ C. Choose t = when r =. Then C = and 3 6 3 6 3 V r = = t = t 3 3 3 dv = t dt When r = 4, then t = 8 and dv dt = 8 = 8. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC7 Let f be a differentiable function defined for all x > such that (i) f () =, (ii) f () =, and d f ( ) = f ( x ), for all >. dx (iii) [ ] (a) Find f (). Suppose f is differentiable. Prove that there is a number c, < c < 4, such that f () c =. 8 (c) Prove that f ( x) = f() + f( x) for all x >. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC7 Solution d f ( ) = f ( x ) dx by (iii) f ( x) = f ( x) f ( x) = f ( x) (*) f () = f () = by (ii) (a) [ ] There is a c, < c < 4, so that f () = from (a) f (4) = 4 from (*) Therefore f () c = 4 =. 4 8 d dx (c) [ ] f () c = d f ( x) = f( x) by (iii) dx Therefore f ( x) = f( x) + C f () = f() + C = C by (i) f ( x) = f( x) + f() f (4) f () 4 by the Mean Value theorem. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB A particle moves along the x-axis in such a way that its acceleration at time t for t > is 3 given by at () =. When t =, the position of the particle is 6 and the velocity is. t (a) Write an equation for the velocity, vt (), of the particle for all t >. Write an equation for the position, x( t ), of the particle for all t >. (c) Find the position of the particle when t = e. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB Solution (a) 3 3 v() t = a() t dt = dt = + C t t = v() = 3+ C 3 Therefore C = 5 and so vt () = + 5. t 3 x() t = v() t dt = ( + 5) dt = 3lnt+ 5t+ C t 6 = x() = 3ln+ 5+ C Therefore C = and so x( t) = 3ln t+ 5t+. (c) x( e) = 3ln e+ 5e+ = 3+ 5e+ = 5e Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB Given that f is the function defined by (a) Find the lim f ( x). x Find the zeros of f. f( x) = 3 x x. 3 x 4x (c) Write an equation for each vertical and each horizontal asymptote to the graph of f. (d) Describe the symmetry of the graph of f. (e) Using the information found in parts (a),, (c), and (d), sketch the graph of f on the axes provided. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB Solution (a) 3 x x x lim = lim = x 4x x 4 4 x 3 x 3 f( x) = for x x=, x x = The zeros are x = and x =. (c) Vertical asymptote: x=, x= Horiztonal asymptote: y = (d) The graph is symmetric with respect to the y-axis. 3 3 3 ( x) ( x) x + x x x (because f ( x) = = = = f( x) ) 3 3 3 ( x) 4( x) x + 4x x 4x (e) Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB3/BC Let R be the region in the first quadrant that is enclosed by the graph of π x-axis, and the line x =. 3 y = tan x, the (a) Find the area of R. Find the volume of the solid formed by revolving R about the x-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB3/BC Solution (a) Area = π/3 tan x dx = ln(cos x) = ln = ln π/3 π 3 Volume = πtan π 3 =π (sec x ) dx =π(tan x x) π =π 3 3 xdx π 3 Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB4 A ladder 5 feet long is leaning against a building so that end X is on level ground and end Y is on the wall as shown in the figure. X is moved away from the building at the constant rate of foot per second. (a) Find the rate in feet per second at which the length OY is changing when X is 9 feet from the building. Find the rate of change in square feet per second of the area of triangle XOY when X is 9 feet from the building. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB4 Solution (a) x + y = 5 dx dy Implicit: x + y = dt dt dy 9 + = dt dy 3 dt = 8 Explicit: y = ( 5 x ) dy x dx = dt dt dy dt ( 5 x ) 9 3 = = 8 A= xy Implicit: da x dy y dx = + dt dt dt da 3 = 9 dt + 8 da dt = 6 A= x 5 x da 5 x dx = dt 5 x dt Explicit: ( ) da dt = 6 ( ) Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB5/BC Let f be the function defined by f ( x) = ( x + ) e x for 4 x 4. (a) For what value of x does f reach its absolute maximum? Justify your answer. Find the x-coordinates of all points of inflection of f. Justify your answer. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB5/BC Solution (a) x f( x) = ( x + ) e 4 x 4 x x x f ( x) = xe ( x + ) e = e ( x+ ) f ( x) for all x and therefore f is decreasing for all x. or Since f is decreasing on the entire interval, the absolute maximum is at x = 4. or The absolute maximum is at a critical point or an endpoint. There is a critical point at x =. f ( 4) = 7e f () = e 7 f (4) = 4 e 4 Therefore the absolute maximum is at x = 4. x x x f ( x) = e ( x ) e ( x ) = e ( x )( x 3) f ( x) > 4< x< f "( x) < < x< 3 f ( x) > 3< x< 4 The points of inflection are at x = and x = 3. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB6/BC3 A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $ per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank? Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB6/BC3 Solution h 4 l Volume = 4lh = 36 Therefore lh = 9 ( ) Cost = C = (4 l) + 5 (4 h) + hl = 4l+ 4h+ hl Method : Direct solution 9 C = 4 + 4h+ 9 h 9 C = 4 + h C = when h=± 3 Thus h= 3 and l = 3 Method : Implicit Differentiation dl l = dh h dc dl = 4 + 4 dh dh dc l = 4 + 4 dh h l = h l = h= 3 When l = 3 and h = 3, then C = 4( 3) + 4( 3) + 9 = $33. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB7 For all real numbers x, f is a differentiable function such that f ( x) = f( x). Let f( p ) = and f ( p) = 5 for some p>. (a) Find f ( p). Find f (). (c) If and are lines tangent to the graph of f at ( p, ) and ( p, ), respectively, and if and intersect at point Q, find the x- and y-coordinates of Q in terms of p. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 AB7 Solution (a) f( x) = f( x) f ( x) = f ( x) f ( p) = f ( p) = 5 f ( ) = f () f () = (c) Equations of the tangent lines : y = 5( x+ p) : y = 5( x p) Solution : At the intersection, we must have 5x 5p= 5x 5p and so x =. The coordinates of Q are x =, y = 5p. Solution : Since f is even, the tangent lines and intersect on the y-axis since they are tangent at symmetric points on the graph with respect to the y-axis. Therefore x = at the point of intersection. The y-coordinate is y = 5p. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC4 A particle moves along the x-axis so that its position function x( t ) satisfies the differential equation dx 9. dt = d x dx 6x = and has the property that at time t =, x=, and dt dt (a) Write an expression for x( t ) in terms of t. (c) At what times t, if any, does the particle pass through the origin? At what times t, if any, is the particle at rest? Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC4 Solution (a) d x dx 6x = dt dt r r 6= r =,3 t 3t x= C e + C e dx = Ce + 3 dt t 3t Ce = C + C 9= C + 3C C = 3, C = t 3t xt () = 3e e The particle passes through the origin when xt ( ) =. t 3t 3e = e 3 = e t = 5t ln 3 5 dx (c) The particle is at rest when dt =. dx t 3t t 3t = 6e 3e = 3( e + e ) < dt dx Since < for all t, the particle is never at rest. dt Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC5 (a) Write the Taylor series expansion about x = for f ( x) = ln( + x). Include an expression for the general term. (c) (d) For what values of x does the series in part (a) converge? 3 Estimate the error in evaluating ln by using only the first five nonzero terms of the series in part (a). Justify your answer. Use the result found in part (a) to determine the logarithmic function whose Taylor n+ n ( ) x series is. n n= Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC5 Solution (a) f( x) = ln( + x) f() = f ( x) = ( x) f () = f ( x) = ( x) f () = ( n) n n f ( x) = ( ) ( n )!( + x) ( n) n f () = ( ) ( n )! 3 x x n x f( x) = x + + ( ) + 3 n n n n x n+ x = ( ) or ( ) n n n= n= The radius of convergence is R =. n at x = : ( ) converges n n= n ( ) at x = : ( ) = diverges n n n= n= The interval of convergence is < x. n n (c) 3 + x =, so x =. Because this is an alternating series with terms decreasing to in absolute value, the error satisfies 6 ( ) E5 =.6. 6 384 Or using a Lagrange error bound, there is a c with c such that E (6) 6 f ()() c 6 5!( ) + c 5 = = = 6 6 6 6 6! 6! ( + c) (6) 6 (d) ( ) ( x ) = ln( + x ) = ln + x n n= n+ n Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC6 Point Pxy (, ) moves in the xy-plane in such a way that dx dt = t + and dy = t for. dt t (a) Find the coordinates of P in terms of t if, when t =, x = ln and y =. Write an equation expressing y in terms of x. (c) Find the average rate of change of y with respect to x as t varies from to 4. (d) Find the instantaneous rate of change of y with respect to x when t =. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC6 Solution (a) x = dt = ln( t + ) + C t + ln = x() = ln + C, so C = x= ln( t+ ) y = tdt = t + C = y() = + C, so C = y = t (c) x x x x e = t+, so y = ( e ) = e e y(4) y() 5 ( ) 6 = = x(4) x() ln 5 ln ln 5 (d) At t =, dy t = = 4 dx t + or From, using x = ln when t = dy ln ln e e 8 4 4 dx = = = Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC7 Let f be the function given by x sin, for x f( x) = x, for x = (a) Using the definition of the derivative, prove that f is differentiable at x =. Find f ( x) for x. (c) Show that f is not continuous at x =. Copyright 4 by College Entrance Examination Board. All rights reserved.
98 BC7 Solution x sin f ( x) f ( ) x (a) f () = lim = lim = lim xsin = x x x x x x since sin x is bounded implies that xsin x for all x. x For x, ( ) f ( x) = xsin + x cos x xsin cos x = x x x (c) lim f ( x) = lim xsin cos x x x x This limit does not exist since cos oscillates between and as x x. Therefore f is not continuous at x =. Copyright 4 by College Entrance Examination Board. All rights reserved.
983 AB Let f be the function defined by f ( x) = + ln( x ). (a) For what real numbers x is f defined? Find the zeros of f. (c) Write an equation for the line tangent to the graph of f at x =. Copyright 4 by College Entrance Examination Board. All rights reserved.
983 AB Solution (a) ln u is defined only for u >. x > except for x =. Therefore f ( x ) is defined for all x. f( x ) = when ln( x ) =. x = e x = e The zeros are x =± e. (c) x f ( x) = = x x f () = = f () = + ln( ) = The equation of the tangent line is y ( ) = ( x ) or y = x 4 Copyright 4 by College Entrance Examination Board. All rights reserved.
983 AB A particle moves along the x-axis so that at time t its position is given by 3 xt () = t 6t + 9t+. (a) What is the velocity of the particle at t =? During what time intervals is the particle moving to the left? (c) What is the total distance traveled by the particle from t = to t =? Copyright 4 by College Entrance Examination Board. All rights reserved.
983 AB Solution (a) vt () = x () t = 3t t+ 9 v() = 9 The particle is moving to the left when vt ( ) <. 3t t+ 9< t 4t+ 3< ( t )( t 3) < The particle is moving to the left on the interval < t < 3. (c) Distance = ( x() x()) + ( x() x()) = 5 + 5 3 = 6 or Distance = = vt () dt vt () dt vt () dt 3 3 t 6t + 9t t 6t + 9t = = 4 ( ) = 6 Copyright 4 by College Entrance Examination Board. All rights reserved.
983 AB3/BC Let f be the function defined for π 5π x by 6 6 f ( x) = x+ sin x. (a) Find all values of x for which f ( x) =. (c) Find the x-coordinates of all minimum points of f. Justify your answer. Find the x-coordinates of all inflection points of f. Justify your answer. Copyright 4 by College Entrance Examination Board. All rights reserved.
983 AB3/BC Solution (a) f ( x) = + sinxcosx= + sinx = + sinx π x = is the only solution in the interval π 5π x. 6 6 3π f ( x) = + sin x=, so x = 4 The minimum occurs at the critical point or at the endpoints. 3π 3π critical point: f = + =.856 4 4 π π endpoints: f = + =.774 6 6 4 5π 5π f = + =.868 6 6 4 Therefore the minimum is at or π x =. 6 Since f ( x) = + sin x for all x, the function f is increasing on the entire π interval. Therefore the minimum is at x =. 6 (c) f ( x) = cosx cosx = π 3π x =, 4 4 π 3π Therefore the inflection points occur at x = and x = since this is where f 4 4 changes sign from positive to negative and from negative to positive, respectively. Copyright 4 by College Entrance Examination Board. All rights reserved.
983 AB4 The figure above shows the graph of the equation region between the graph of x y x y + =. Let R be the shaded + = and the x-axis from x = to x =. (a) (c) Find the area of R by setting up and integrating a definite integral. Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid formed by revolving the region R about the x-axis. Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid formed by revolving the region R about the line x =. Copyright 4 by College Entrance Examination Board. All rights reserved.
983 AB4 Solution (a) ( ) y = x = 4 4x + x 8 3 x = x dx= 4 4x + x dx= 4x x + = 3 6 Area ( ) ( ) or 4 8 3 = dy + 4 4y y dy 4y y y + = + + = 3 6 Area = ( ) 4 Volume = π ( ) or x dx = π ( 6 3 / + 4 8 3/ + ) x x x x dx Volume = π + π 4 ( ) = π+ 4 3 ( y y + y ) dy ydy y y dy 4 4 4 (c) Volume = or π ( x ) ( xdx ) = π ( 4 4 3 + 4 3 ) x x x x dx Volume = π dy +π 4 ( ( x) ) dy = π+π 4 ( x x ) dy = π+π 4 ( ( y ) ( y ) 4 ) dy = π+π 4 ( 8+ 4 + 8 3 ) y y y y dy Copyright 4 by College Entrance Examination Board. All rights reserved.
983 AB5/BC3 At time t =, a jogger is running at a velocity of 3 meters per minute. The jogger is slowing down with a negative acceleration that is directly proportional to time t. This brings the jogger to a stop in minutes. (a) Write an expression for the velocity of the jogger at time t. What is the total distance traveled by the jogger in that -minute interval? Copyright 4 by College Entrance Examination Board. All rights reserved.