AP Calculus AB AP Calculus BC

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1 AP Calculus AB AP Calculus BC Free-Response Questions and Solutions 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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4 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.

5 969 AB Consider the following functions defined for all x: f ( x) = x f ( x) = xcosx f3 ( x ) = 3 e x f 4 ( x) = x x Answer the following questions (a, b, c, and d) about each of these functions. Indicate your answer by writing either yes or no in the appropriate space in the given rectangular grid. No justification is required but each blank space will be scored as an incorrect answer. Questions Functions f f f 3 f 4 Does f ( x) = f( x) Does the inverse function exist for all x? (c) Is the function periodic? (d) Is the function continuous at x =? Copyright 4 by College Entrance Examination Board. All rights reserved.

6 969 AB Solution Questions Functions f f f 3 f 4 Does f ( x) = f( x) Yes Yes No No Does the inverse function exist for all x? Yes No ** No (c) Is the function periodic? No No No No (d) Is the function continuous at x =? Yes Yes Yes Yes f ( x) = x f ( x) = xcosx 3 ( ) 3 x f x = e f 4 ( x) = x x ** It is not clear how this question was meant to be interpreted and answered for the function f 3 during the 969 exam. Reasonable arguments can be made for both an answer of No, and for an answer of Yes. No student lost credit for either answer. If one assumes that the question asks if the inverse function exists and has a domain of all real numbers (i.e., exists for all x), then the answer is "No" since the range of f consists only of the positive numbers. 3 If one assumes that the question simply asks if an inverse function g exists for which g( f ( x)) = xfor all x, then the answer is "Yes." 3 Copyright 4 by College Entrance Examination Board. All rights reserved.

7 969 AB/BC A particle moves along the x-axis in such a way that its position at time t is given by 4 3 x = 3t 6t + 4t for 5 t 5. Determine the velocity and acceleration of the particle at time t. (c) (d) At what values of t is the particle at rest? At what values of t does the particle change direction? What is the velocity when the acceleration is first zero? Copyright 4 by College Entrance Examination Board. All rights reserved.

8 969 AB/BC Solution dx v= = t 48t + 48t = t( t 4t+ 4) = t( t ) dt 3 dv a= = 36t 96t+ 48 = (3t 8t+ 4) = (3t )( t ) dt The particle is at rest when v =. This occurs when t = and t =. (c) The particle changes direction at t = only. (d) a = when t = and t = 3 v 4 8 = = The acceleration is first zero at t =. 3 Copyright 4 by College Entrance Examination Board. All rights reserved.

9 969 AB3/BC3 Given f ( x) = + lnx, defined only on the closed interval x e x e. Showing your reasoning, determine the value of x at which f has its (i) absolute maximum, (ii) absolute minimum. For what values of x is the curve concave up? (c) On the coordinate axes provided, sketch the graph of f over the interval x e e. (d) Given that the mean value (average ordinate) of f over the interval is, state in e words a geometrical interpretation of this number relative to the graph. Copyright 4 by College Entrance Examination Board. All rights reserved.

10 969 AB3/BC3 Solution x f ( x) = + = x x x f ( x) = at x =. The three candidates are f () = f = e e f() e e + = + = e e Therefore the absolute maximum is at x = and the absolute minimum is at x =. e (c) x f ( x) = = > for x 3 3 x x x e <. Therefore the curve is concave up for x e <. (d) e is the height of a rectangle of width e and having the same area as that e enclosed by the graph of y = f( x), the vertical lines x = and x = e, and the e x-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.

11 969 AB4/BC4 The number of bacteria in a culture at time t is given approximately by t / y = (5 + te ) for t. Find the largest number and the smallest number of bacteria in the culture during the interval. At what time during the interval is the rate of change in the number of bacteria a minimum? Copyright 4 by College Entrance Examination Board. All rights reserved.

12 969 AB4/BC4 Solution t t t t t y = e + e = e y = at t = At t =, y = (5) At t =, At t =, e y = (5 + ) 5 e y = (5 + ) The minimum number of bacteria is (5). 5 4 Since e > e because e > 5, the maximum number is (5 + e ). y t t t t = e + = e + t t 5e = y = at t = 4 At t =, y = At t = 4, y = ( e ) 5 At t =, y = ( 4 e ) The rate of change is a minimum at t = 4. Copyright 4 by College Entrance Examination Board. All rights reserved.

13 969 AB5 Let R denote the region enclosed between the graph of y = x and the graph of y = x. Find the area of region R. Find the volume of the solid obtained by revolving the region R about the y-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.

14 969 AB5 Solution Intersection: x = x when x = and x = Area or Area = ( x x ) dx 3 x 8 4 = x = 4 = y = y dy 3 y 6 4 = y = 4 = Shells: 4 3 x 6 8 Volume = π x( x x ) dx= π x = π 4 = π Disks: 4 4 y y =π y dy =π y dy 4 Volume ( ) 4 3 y y 64 8 =π =π 8 = π 3 Copyright 4 by College Entrance Examination Board. All rights reserved.

15 969 AB6 An arched window with base width b and height h is set into a wall. The arch is to be either an arc of a parabola or a half-cycle of a cosine curve. (c) If the arch is an arc of a parabola, write an equation for the parabola relative to the coordinate system shown in the figure. If the arch is a half-cycle of a cosine curve, write an equation for the cosine curve relative to the coordinate system shown in the figure. Of these two window designs, which has the greater area? Justify your answer. Copyright 4 by College Entrance Examination Board. All rights reserved.

16 969 AB6 Solution y = Ax + Bx+ C h= A + B + C = C = Ab + Bb + C = Ab + Bb + h B = = Ab Bb + C = Ab Bb + h h Ab + h = A = b Therefore = h + = h y x h ( b x b b ). b The parabola could also be written as x = ( y h). h y = Acos Bx h= Acos = A = AcosBb π π Bb = B = = Acos B( b) = AcosBb b πx Therefore y = hcos b (c) For parabola, area b 3 3 b x dx b x b 4 = b h h x h hb ( ) = = = b b 3 b 3 3 For cosine design, area b πx b πx 4bh = hcos dx= h sin = b π b π b The parabola design has the greater area since 3 < π. Copyright 4 by College Entrance Examination Board. All rights reserved.

17 969 AB7 On the coordinate axes provided, sketch the graph of x x e + e y =. Let R be a point on the curve and let the x-coordinate of R be r ( r ). The tangent line to the curve at R crosses the x-axis at a point Q. Find the coordinates of Q. (c) If P is the point ( r,), find the length of PQ as a function of r and the limiting value of this length as r increases without bound. Copyright 4 by College Entrance Examination Board. All rights reserved.

18 969 AB7 Solution Method x x e e y = The equation of the tangent line at the point R is r r r r e + e e e y = ( x r) or r r r r r r e e e + e e e y = x+ r. The line crosses the x-axis when y =. r r e + e This happens at x= r. r r e e r r e + e The coordinates of Q are r, r r. e e Method (hyperbolic functions) y = cosh( x) y = sinh( x) The equation of the tangent line at the point R is y cosh( r) = sinh( r) ( x r) or y = sinh( r) x+ cosh( x) rsinh( r) The line crosses the x-axis when y =. This happens at x = r coth( r). The coordinates of Q are ( r coth( r),). (c) r r r e + e + e PQ = = e r e r r e r + e lim PQ = lim = r r r e PQ = coth( r) lim PQ = lim coth( r) = r r Copyright 4 by College Entrance Examination Board. All rights reserved.

19 969 BC πx A closed region R of the plane has y = + sin as its upper boundary, x y = as its lower boundary, and the y-axis as its left-hand boundary. Sketch the region R on the axes provided. Set up but do not evaluate an integral expression in terms of the single variable x for each of the following: (i) the area A of R, (ii) the volume V of the solid obtained by revolving R about the x-axis, (iii) the total perimeter P of R. Copyright 4 by College Entrance Examination Board. All rights reserved.

20 969 BC Solution (i) Area πx x = + sin dx (ii) Volume πx x =π + sin dx πx (iii) Let y = + sin be the curve for the upper boundary. Then x y π π = cos. The total perimeter is π πx P= AC+ AB+ BC = cos dx Copyright 4 by College Entrance Examination Board. All rights reserved.

21 969 BC5 Prove that the area of the region in the first quadrant between the curve the x-axis is divided into two equal parts by the line x = ln. y = e x and If the two regions of equal area described in part are rotated about the x-axis, are the resulting volumes equal? If so, prove it. If not, determine which is larger and by how much. Copyright 4 by College Entrance Examination Board. All rights reserved.

22 969 BC5 Solution Method A = e dx= e = e = = Left ln x x ln ln x R x x R R ln ln R ln R ln R ARight = e dx= lim e dx= lim ( e ) = lim ( e + e ) = Hence the line x = ln divides the region into two parts of area both equal to. Method Left k x x k k x R x x R R k k k R k R k R A = e dx= e = e A Right = e dx = lim e dx = lim ( e ) = lim ( e + e ) = e k k The two areas will be equal if e = e. k k e = e = k = ln Hence the line x = ln divides the region into two parts of equal area. ln x x ln ln π π 3π VLeft =π ( e ) dx=π e = ( e ) = = 4 8 Right x R x ( ) lim ln R ln R x V =π e dx=π e dx ln R ln ( e e ) e π π =πlim = lim + = R R 8 The volumes are not equal. The volume of the left region is larger by the amount π VLeft VRight =. 4 Copyright 4 by College Entrance Examination Board. All rights reserved.

23 969 BC6 Assume that the graph of a function f passes through the origin and has slope at that point. Determine f ( x ) if f ( x) + f ( x) = for all x. Determine f ( x ) if f ( x) + f ( x) = for all x. Copyright 4 by College Entrance Examination Board. All rights reserved.

24 969 BC6 Solution m + m= m=, x The general solution is f ( x) = C+ Ce. Then function passes through the origin with slope, = f () = C + C = f () = C x f ( x) = C e. Since the Therefore C = and C =. f ( x) = x e We need to find a particular solution. Since a constant function is already a solution to the homogeneous equation given in part, we try a function of the form y = Ax. Substituting into the differential equation gives p + A = and so A =. Hence yp ( ) f x = C + C e + x. Using the initial conditions, x = f = C+ C C =, C = f C () = () = + = x and the general solution to the differential equation is Hence x f ( x) = ( e ) + x (Note: there are other methods that can be used to find a particular solution or the general solution, such as variation of parameters or reduction of order.) Copyright 4 by College Entrance Examination Board. All rights reserved.

25 969 BC7 Given the finite sum S n n =. + k k= (c) By comparing S n with an appropriate integral, prove that Use part to deduce that exists. Show your reasoning. k= + k π π Prove that. 4 + k k= Sn arctan n for n. Copyright 4 by College Entrance Examination Board. All rights reserved.

26 969 BC7 Solution S n is a lower (right) Riemann sum for S n dx = arctan n. + x n n S n is monotone increasing (by inspection). dx + x dx. Therefore dx dx π dx = lim dx = lim arctan n = n + x n + x n So by part, the sequence ( S ) is bounded above. Since the sequence is increasing and bounded, k= n = lim S n exists. + k n (c) By considering S n as an upper (left) Riemann sum and also using part, n+ n dx dx dx S n dx. + x + x π Therefore arctan( n+ ) Sn arctan n. 4 π lim arctan( n+ ) lim Sn lim arctan n. n 4 n n Hence π π π which gives 4 + k k= π π. 4 + k k= Copyright 4 by College Entrance Examination Board. All rights reserved.

27 97 AB/BC Given the parabola y = x x+ 3: Find an equation for the line L, which contains the point (,3) and is perpendicular to the line tangent to the parabola at (,3). Find the area of that part of the first quadrant which lies below both the line L and the parabola. Copyright 4 by College Entrance Examination Board. All rights reserved.

28 97 AB/BC Solution y = x The slope of the line tangent to the parabola is m =. Therefore the slope of the line L that is perpendicular to the tangent line is. The equation of the line L is y 3 = ( x ), or y = x+ 4, or x+ y = 8. 3 A = x + 4 ( x x + 3) = + dx x x dx = x+ x x = A = ( x x+ 3) dx = + 3 = x x x A3 = x+ 4 dx= 9 (or A 3 = 63 = 9 using area of the triangle) Area of triangle ( A+ A + A 3) is 84 6 = 4 Method : area = A + A3 = Method : area =6 A = 6 = 3 3 Copyright 4 by College Entrance Examination Board. All rights reserved.

29 97 AB A function f is defined on the closed interval from 3 to 3 and has the graph shown below. On the axes provided sketch the entire graph of y = f( x). On the axes provided sketch the entire graph of y = f( x ). Copyright 4 by College Entrance Examination Board. All rights reserved.

30 (c) On the axes provided sketch the entire graph of y = f( x). (d) On the axes provided sketch the entire graph of y = f x. (e) On the axes provided sketch the entire graph of y = f( x ). Copyright 4 by College Entrance Examination Board. All rights reserved.

31 97 AB Solution y f( x) = y = f ( x) = (d) y = f ( x) (c) y f( x) (e) y = f( x ) Copyright 4 by College Entrance Examination Board. All rights reserved.

32 97 AB3/BC Consider the function f given by f ( x) = x + 4x on the interval 8 x 8. (c) (d) (e) Find the coordinates of all points at which the tangent to the curve is a horizontal line. Find the coordinates of all points at which the tangent to the curve is a vertical line. Find the coordinates of all points at which the absolute maximum and absolute minimum occur. For what values of x is this function concave down? On the axes provided, sketch the graph of the function on this interval. Copyright 4 by College Entrance Examination Board. All rights reserved.

33 97 AB3/BC Solution f( x) = x + 4 x = x ( x+ 4) 3 3 x f ( x) = x + x = x f ( x) = at x =. There is a horizontal tangent at (, 3). There is a vertical tangent at (,). (c) The absolute maximum and absolute minimum must occur at a critical point or an endpoint. The candidates are ( 8,8), (, 3), (,), and (8, 4) So the absolute maximum is at (8, 4) and the absolute minimum is at (, 3). (d) 3 53 x f ( x) = x x = x The graph is concave down for < x <. (e) Copyright 4 by College Entrance Examination Board. All rights reserved.

34 97 AB4 A right circular cone and a hemisphere have the same base, and the cone is inscribed in the hemisphere. The figure is expanding in such a way that the combined surface area of the hemisphere and its base is increasing at a constant rate of 8 square inches per second. At what rate is the volume of the cone changing at the instant when the radius of the common base is 4 inches? Show your work. Note: The surface area of a sphere of radius r is S = 4π r and the volume of a right circular cone of height h and base radius r is V = π r h. 3 Copyright 4 by College Entrance Examination Board. All rights reserved.

35 97 AB4 Solution Method : The combined surface area of the hemisphere and its base is (4 ) 3 S = π r +π r = πr ds dr dr 3 8 = = 6πr = dt dt dt π r Since the height of the cone is h = r, the volume of the cone is V = π r 3 3 dv dr 3 =π r =π r = 3r dt dt πr dv At r = 4, = dt Method : As in method, S = 3π r and so V 3 S = π 3 3π. dv 3 S ds S ds = π = dt 3 3π 3π dt 6 3π dt When r = 4, S = 48π and so dv dt = 48 = 6 Copyright 4 by College Entrance Examination Board. All rights reserved.

36 97 AB5 A particle moves along the x-axis in such a way that at time t > its position coordinate is x = sin( e ). t Find the velocity and acceleration of the particle at time t. (c) At what time does the particle first have zero velocity? What is the acceleration of the particle at the time determined in part? Copyright 4 by College Entrance Examination Board. All rights reserved.

37 97 AB5 Solution t x = sin( e ) dx t t v = = e cos( e ) dt dv t t t t a = = e (cos( e ) e sin( e )) dt vt ( ) = when cos( e ) =. Hence π zero, and so t = ln. t t e π = gives the first time when the velocity is (c) π π π π π π a ln = cos sin = 4 Copyright 4 by College Entrance Examination Board. All rights reserved.

38 97 AB6/BC5 A parabola P is symmetric to the y-axis and passes through b >. Write an equation for P. b (,) and ( b, e ) where (c) The closed region bounded by P and the line y = e b is revolved about the y-axis to form a solid figure F. Compute the volume of F. For what value of b is the volume of F a maximum? Justify your answer. Copyright 4 by College Entrance Examination Board. All rights reserved.

39 97 AB6/BC5 Solution y = Ax + Bx+ C. Since y = when x =, must have C =. Since the graph is symmetric to the y-axis, must have B =. Since e b = Ab, must have x b Therefore y = e or x b b e A =. b b = yb e Disks: b e b b y Volume =π x dy =π yb e dy b e b e =π = π e e b b b Shells: b b 4 x x x Volume = π x e e dx= πe = πb e b 4b b b b b (c) ( ) = π Vb be b V ( b) b( b ) e b =π V ( b) = when b =,, or. Since b >, the only viable solution is b =. b 4 V ( b) =πe (b 5b + ) < when b =. Therefore there is a local maximum at b =. Since this is the only critical point for b >, it is also the absolute maximum for b >. Another way to justify the absolute maximum at b = is to observe that V ( b) > for < b <, so that Vb ( ) is increasing on this interval, and V ( b) < for b >, so that Vb ( ) is decreasing for b >. Copyright 4 by College Entrance Examination Board. All rights reserved.

40 97 AB7 From the fact that sin t t for all t, use integration repeatedly to prove the following inequalities. Show your work. 4 x cos x x + x for all x!! 4! Copyright 4 by College Entrance Examination Board. All rights reserved.

41 97 AB7 Solution Given sin t t for all t, integrate both sides over the interval t x for x. x sin tdt cost x x o x t tdt x cos x + x cos x for x Now integrate both sides with respect to t over the interval t x for x. x sin t t costdt dt x x 3 t t 6 x sin x x for x 6 3 x Now integrate both sides with respect to t over the interval t x for x. x 3 t sin tdx t dt 6 4 x t t cost 4 4 x x cos x + 4 x 4 x x x cos x + for x 4 Copyright 4 by College Entrance Examination Board. All rights reserved.

42 97 BC3 A particle moves along the x-axis in such a way that at time t >, its position coordinate is x = sin( e ). t Find the velocity of the particle at time t. (c) At what time does the particle first have zero velocity? Show that the length of time between successive instants at which the particle has zero velocity decreases as t increases. Copyright 4 by College Entrance Examination Board. All rights reserved.

43 97 BC3 Solution x= sin( e ) t v= e cos( e ) t t vt ( ) = when cos( e ) =. Hence π zero, and so t = ln. t t e π = gives the first time when the velocity is t π π (c) vt ( ) = when e = (n ), that is, at tn = ln (n ) for each integer n. Let n + Tn ( ) be the time between instants n and n +. Then Tn ( ) = tn+ tn = ln n. Method : x + Let T( x) = ln x. Then 4 T ( x) = < for x > and so T( x ) is a (x+ )(x ) decreasing function of x for x >. Therefore Tn ( ) decreases as n increases. Method : Since n + = is a decreasing function of n and ln x is an increasing n n function, then the composition Tn ( ) is a decreasing function. Copyright 4 by College Entrance Examination Board. All rights reserved.

44 97 BC4 Find the general solution of the differential equation Find the general solution of the differential equation d y 5y dx + =. d y dy 6 5y dx dx + =. (c) d y dy For the differential equation 6 + 5y =, find the solution that has a dx dx graph passing through the point (,) with slope 6. Copyright 4 by College Entrance Examination Board. All rights reserved.

45 97 BC4 Solution y + 5y = The auxiliary equation m + 5 = has roots m= ± 5i. The general solution is y = Csin 5x+ Ccos5x This can also be written as y = Csin 5x+ Ccos5x or y = C5cos(5 x+ C6). The auxiliary equation m 6m+ 5= has roots m= 3± 4i. 3x The general solution is y = e ( C cos4x+ C sin4 x). This can also be written as 3x 3 4 y = C e sin(4 x+ C ) or 3x 5 6 y = C e cos(4 x+ C ). (c) 3x y = e ( C cos4x+ C sin4 x) 3x y = e ((4C + 3 C )cos4 x+ ( 4C + 3 C )sin 4 x ) From the given conditions, = C 6 = 4C + 3C This has the solution C = and C =. Therefore 3x y = e cos4x. Copyright 4 by College Entrance Examination Board. All rights reserved.

46 97 BC6 Let ak k+ π k = ( ) sinkxdx. Evaluate a k. (c) Show that the infinite series 3 Show that ak. k= ak converges. k= Copyright 4 by College Entrance Examination Board. All rights reserved.

47 97 BC6 Solution π k π k sin kx dx = cos kx = ( cos π ) =, so k k k a k = ( ) k + k The series k ( ) k + a = k= k= k converges by the Alternating Series Test since (i) the terms alternate (ii) the terms are decreasing in absolute value, i.e. ak+ (iii) lim a = k k Alternatively, the series converges because (i) lim a = k k (ii) The partial sums S n are bounded above (iii) The partial sums S n+ are bounded below < a for all k k (c) Method k+ ak = ( ) k k= = + + = + + = 3 4 k+ ak = ( ) + k= 3 4 k = = < < Method ak = ln ( ).38 (recognizing the alternating harmonic series) k= Method 3 8 k= a k ( ) (.634 ±.) <. The error stopping at the 8 th term satisfies error 9. Copyright 4 by College Entrance Examination Board. All rights reserved.

48 97 BC7 The function f is defined for all x in the interval < x < by x sin x f( x) =. sin x How should f () be defined in order that f be continuous for all x in the interval < x <? With f () defined as in part, use the definition of the derivative to determine whether f () exists. Show your work. Copyright 4 by College Entrance Examination Board. All rights reserved.

49 97 BC 7 Solution x sin x x f ( x) = = cosx, so sin x sin x lim f( x) = = x Or Or 3 8x x x x x + f( x) = = 3 5 as x. 3 4 x x x x cosx lim f( x) = lim = cos x x x Therefore defining f () = will make f continuous for all x in the interval < x <. f () = lim h f ( + h) f() h h sin h + sin h = lim h h h sin h+ sin h = lim h hsin h = cos h+ cos h = lim = h hcos h+ sin h 4sinh sinh = lim = = h hsin h+ cos h Therefore f () exists and has the value. Copyright 4 by College Entrance Examination Board. All rights reserved.

50 97 AB Let f ( x) = lnx for all x >, and let gx ( ) = x 4 for all real x. Let H be the composition of f with g, that is H( x) = f( g( x)). Let K be the composition of g with f, that is K( x) = g( f( x)). Find the domain of H. Find the range of H. (c) Find the domain of K. (d) Find the range of K. (e) Find H (7). Copyright 4 by College Entrance Examination Board. All rights reserved.

51 97 AB Solution H( x) = f( g( x)) = ln( x 4) The domain of H is the set of x for which x 4>, that is, x > or x <. There are other equivalent ways to write this set. The range of H is the set of all real numbers. K( x) = g( f( x)) = (ln x) 4 (c) The domain of K is the same as the domain of the natural logarithm, that is, x >. (d) The range of K is the set of all real numbers 4. (e) x H ( x) = f ( g( x)) g ( x) = x= gx ( ) x 4 4 H (7) = 45 Copyright 4 by College Entrance Examination Board. All rights reserved.

52 97 AB Let R be the region in the first quadrant that lies below both of the curves 3 y = and to the left of the line x = k where k >. x y = 3x and Find the area of R as a function of k. When the area of R is 7, what is the value of k? (c) If the area of R is increasing at the constant rate of 5 square units per second, at what rate is k increasing when k = 5? Copyright 4 by College Entrance Examination Board. All rights reserved.

53 97 AB Solution Let Ak ( ) be the area of R to the left of the line x = k. k 3 3 k Ak ( ) = 3x dx+ dx= x + 3ln x = + 3ln k x 7 = Ak ( ) = + 3ln k ln k = k = e (c) da da dk 3 dk = = dt dk dt k dt 3 dk 5 = kdt dk 5k dt = 3 dk When k = 5, 5 dt = Copyright 4 by College Entrance Examination Board. All rights reserved.

54 97 AB3 Consider f ( x) = cos x+ cos x over one complete period beginning with x =. Find all values of x in this period at which f( x ) =. (c) Find all values of x in this period at which the function has a minimum. Justify your answer. Over what intervals in this period is the curve concave up? Copyright 4 by College Entrance Examination Board. All rights reserved.

55 97 AB3 Solution f ( x) = cos x+ cos x in [, π ) f( x) = cos x(cos x+ ) = when cos x =. So π x = or 3π x =. f ( x) = cos x sin x sin x= sin x( + cos x) f ( x) = when x =, π (and π). Possible justifications for the location of the minimum: First derivative test: f ( x) < for < x<π so the graph is decreasing on this interval. f ( x) > for π< x< π so the graph is increasing on this interval. Therefore there is a minimum at x = π. Test the critical points: f () = 3 f ( π ) = f ( π ) = 3 Therefore there is a minimum at x = π Second derivative test: f ( x) = cos x+ sin x cosx = ( cos x)(+ cos x) f ( π ) = so provides no conclusion. But f ( x) > for x just less than π and just greater than π, thus the graph is concave up in an interval containing x =π, so x =π gives a local minimum. Since it is the only interior critical point, it must be the location of the absolute minimum. Non-calculus reasoning: f( x) = (cosx+ ) Because of the square, the minimum will occur when cos x + =, i.e. when x =π. (c) f ( x) = ( cos x)( + cos x) π 5π f ( x) = at x = and x =. 3 3 π 5π f ( x) > for < x <. Therefore the graph is concave up for 3 3 π 5π < x <. 3 3 Copyright 4 by College Entrance Examination Board. All rights reserved.

56 97 AB4/BC Find the area of the largest rectangle (with sides parallel to the coordinate axes) that can be inscribed in the region enclosed by the graphs of f( x) = 8 x and g( x) = x 9. Copyright 4 by College Entrance Examination Board. All rights reserved.

57 97 AB4/BC Solution Let Ax ( ) be the area of the rectangle with side located at the coordinate x on the x-axis for x. Then 3 A= x((8 x ) (x 9)) = x(7 3 x ) = 6(9 x x ) for x 3. A = 6(9 3 x ) = at x = 3 A = 6( 6 x) < at x = 3. Therefore there is a local maximum at x = 3. But since this is the only critical value for x >, it must be an absolute maximum. Alternatively, the absolute maximum must occur at x = 3 or at one of the endpoints. But A = at x =, 3. Therefore the absolute maximum is at x = 3. The maximum area is 6( ) = Copyright 4 by College Entrance Examination Board. All rights reserved.

58 97 AB5 Let R be the region in the first quadrant bounded by the x-axis and the curve y = x x. Find the volume produced when R is revolved about the x-axis. Find the volume produced when R is revolved about the y-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.

59 97 AB5 Solution 3 4 Volume =π ( x x ) dx=π (4x 4 x + x ) dx =π x x + x = π Shells: Volume = π xy dx = π x( x x ) dx = π 3 ( x x ) dx π = π x x = π 3 4 = Disks: Volume =π (( + y) ( y) ) dy =π (( + y + y) ( y + y)) dy =π 4 ydy 3 8π 8π = 4 π ( y) = ( ) = Copyright 4 by College Entrance Examination Board. All rights reserved.

60 97 AB6 A particle starts at the point (5,) at t = and moves along the x-axis in such a way that t at time t > its velocity vt ( ) is given by vt () =. + t Determine the maximum velocity attained by the particle. Justify your answer. Determine the position of the particle at t = 6. (c) (d) Find the limiting value of the velocity as t increases without bound. Does the particle ever pass the point (5,)? Explain. Copyright 4 by College Entrance Examination Board. All rights reserved.

61 97 AB6 Solution t v () t = = ( + t ) at t =. v () t > for < t < and v ( t) > for t >. Therefore the velocity is a maximum at t = and the maximum velocity is v () =. ( tt 3) Alternatively, v () t = and so v () = <. Therefore v has a local 3 ( + t ) maximum at t =, but since this is the only critical point for t >, it must also be an absolute maximum. Therefore the maximum velocity is v () =. (One could also justify that since vt () = t + t then vt ( ) has a maximum value of.) and t + has a minimum value of, t (c) (d) st () = vtdt () = ln( + t ) + C 5 = s() = ln() + C = C Hence st () = 5+ ln( + t ) and so s (6) = 5 + ln(37) = 5 + ln 37. t lim vt ( ) = lim = lim t = t t + t t + t or t lim vt ( ) = lim = lim = using L Hôpital s rule. t t + t t t Yes, the particle does pass the point (5, ) since the natural logarithm is an unbounded function ln( + t ) for 99 t e Copyright 4 by College Entrance Examination Board. All rights reserved.

62 97 AB7/BC3 Let f be the function defined by f ( x) = x e x for all real numbers x. Describe the symmetry of the graph of f. (c) Over what intervals of the domain is this function increasing? Sketch the graph of f on the axes provided showing clearly: (i) behavior near the origin, (ii) maximum and minimum points, (iii) behavior for large x. Copyright 4 by College Entrance Examination Board. All rights reserved.

63 97 AB7/BC3 Solution The graph is symmetric with respect to the y-axis since ( x) x f ( x) = x e = x e = f( x) For x >, x x x x f ( x) = x e x xe = x e ( 4 x ) = x e ( x)(+ x) Therefore the function is increasing on the intervals, and,. (c) (i) there is a vertical tangent at (, ) (ii) the maximum points are at x = ±, where The minimum point is (, ). y = e (iii) lim f( x) = x Copyright 4 by College Entrance Examination Board. All rights reserved.

64 97 BC Let R be the region of the first quadrant bounded by the x-axis and the curve y = kx x, where k >. (c) In terms of k, find the volume produced when R is revolved about the x-axis. In terms of k, find the volume produced when R is revolved about the y-axis. Find the value of k for which the volumes found in parts and are equal. Copyright 4 by College Entrance Examination Board. All rights reserved.

65 97 BC Solution k k 3 4 Volume =π ( kx x ) dx =π ( k x kx + x ) dx k k x kx x πk =π + = k k k 3 kx x πk Volume = π x( kx x ) dx= π ( kx x ) dx = π = or k 4 k k 4y k k 4y + Volume =π π k 4 ( ) k k 3 4 dy πk =π k k 4 y dy =π ( k 4 y) = (c) 5 4 πk πk = when k = Copyright 4 by College Entrance Examination Board. All rights reserved.

66 97 BC4 Write the first three nonzero terms in the Taylor series expansion of cos x about π x =. (c) (d) What is the interval of convergence of the Taylor series mentioned in part? Show your method. Use the first two nonzero terms of the series in part to approximate π cos +.. Estimate the accuracy of the approximation found in part (c). Show your method. Copyright 4 by College Entrance Examination Board. All rights reserved.

67 97 BC4 Solution f ( x) = cos x f ( π ) = f ( x) = sinx f ( π ) = f ( x) = cosx f ( π ) = f ( x) = sinx f ( π ) = f (4) ( x) = cosx f (5) ( x) = sinx f (4) ( π ) = f (5) ( π ) = 3 5 π π x x π f( x) = cosx= x + + 3! 5! n+ π π x x (n )! R = = (n+ )! n π (n+ )( n) x lim R = < for all x and so the series converges for all x. n or For all x, the series is strictly alternating with lim un and un+ n < u for all n sufficiently large. The series therefore converges for all x by the Alternating Series Test. n (c) 3 π (.)..599 cos +. (.) + =.+ = ! 6 6 (d) Since this is an alternating series with the terms decreasing to in absolute value 5 (.) 8 starting with the first term, the error satisfies error < = ! or 4 (.) 6 Could use remainder form to get error < = 4.6 4! Copyright 4 by College Entrance Examination Board. All rights reserved.

68 97 BC5 Determine whether or not your reasoning. x xe dx converges. If it converges, give its value. Show Copyright 4 by College Entrance Examination Board. All rights reserved.

69 97 BC5 Solution Let x I = xe dx. Use integration by parts with x u = x dv= e dx du = dx v = e x x x x I = xe + e dx= e ( x ) x b x = lim b xe dx xe dx x x = lim ( xe e ) b b b b = lim ( be e + ) b b = lim + b b b e e = Therefore the integral converges and has the value. Note: x xe dx =Γ () = where Γ ( x) is the Gamma function. Knowledge of the gamma function could also be used to show that this improper integral converges and has the value. Copyright 4 by College Entrance Examination Board. All rights reserved.

70 97 BC6 Find a function f that has a continuous derivative on (, ) and that has both of the following properties: Note: (i) The graph of f goes through the point (,). (ii) The length L of the curve from (,) to any point ( x, f( x )) is given by L= ln x+ f( x). Recall that the arc length from ( a, f( a )) to ( b, f( b )) is given by b + ( f ( t)) dt. a Copyright 4 by College Entrance Examination Board. All rights reserved.

71 Copyright 4 by College Entrance Examination Board. All rights reserved.

72 97 BC7 Definition: A function f, defined for all positive numbers, is tractible if and only if for every ε>, there exists an integer M > such that x > M implies that f( x) x <ε. If hx ( ) = x 3 for all positive real numbers x, prove that h is not tractible. (c) If gx ( ) = x+ for all positive real numbers x, prove that g is tractible. x Discuss the graphical significance of tractibility. Copyright 4 by College Entrance Examination Board. All rights reserved.

73 97 BC7 Solution hx ( ) x = ( x+ 3) x = 3. It is not possible to make 3 < ε for every ε>. Therefore hx ( ) = x 3 is not tractible. (c) gx ( ) x= x+ x= <ε for x >. Choose M = + (where x x x ε ε is the greatest integer function), or choose M to be any integer greater than ε. This choice of M for every ε > will satisfy the condition for gx ( ) to be tractible. As x increases without bound, the vertical difference between a tractible function and the line y = x will approach zero. This means that the graph of a tractible function approaches the line y = x asymptotically. Copyright 4 by College Entrance Examination Board. All rights reserved.

74 97 AB Let 3 f ( x) = 4x 3x. Find the x-intercepts of the graph of f. Write an equation for the tangent line to the graph of f at x =. (c) Write an equation of the graph that is the reflection across the y-axis of the graph of f. Copyright 4 by College Entrance Examination Board. All rights reserved.

75 97 AB Solution 3 Must solve f( x) = 4x 3x =. The polynomial can be factored by using long division or synthetic division. Possible rational zeros are ±, ±, ±. Since 4 f () = 4 3 =, divide the polynomial by x. 4x + 4x+ 3 x 4 x 3x 3 4x 4x + 4x 3x + 4x 4x x Therefore x = and f( x) = ( x )(4x + 4x+ ) = ( x )(x+ ) and the x-intercepts are x =. f ( x) = x 3 f () = 48 3 = 45 f () = 5 The equation of the tangent line is y 5 = 45( x ) or y = 45x 65. (c) The reflection across the y-axis is given by f ( x). So the equation is 3 3 y = f( x) = 4( x) 3( x) = 4x + 3x Copyright 4 by College Entrance Examination Board. All rights reserved.

76 97 AB/BC A particle starts at time t = and moves on a number line so that its position at time t is given by xt () = ( t ) ( t 6). 3 (c) (d) When is the particle moving to the right? When is the particle at rest? When does the particle change direction? What is the farthest to the left of the origin that the particle moves? Copyright 4 by College Entrance Examination Board. All rights reserved.

77 97 AB/BC Solution xt () = ( t ) ( t 6) 3 3 vt ( ) = x ( t) = ( t ) + 3( t ) ( t 6) = ( t ) (( t ) + 3( t 6)) = ( t ) (4t ) = 4( t ) ( t 5) at ( ) = x ( t) = ( t )( t 4) The particle is moving to the right when t > 5. vt ( ) = 4( t ) ( t 5) >. This happens for The particle is at rest when t = 5. vt ( ) = 4( t ) ( t 5) =. This happens at t = and (c) The particle changes direction when the velocity changes sign. The velocity is negative just to the left and just to the right of t =. The velocity is negative for t < 5 and positive for t > 5. Therefore the particle only changes direction at t = 5. (d) The minimum value of x is at t = 5 since the particle moves to the left for t < 5 and moves back to the right for t > 5. 3 x (5) = 3 ( ) = 7 Therefore the farthest to the left of the origin that the particle moves is x = 7. t = 5 7 t = t = 48 x() t Copyright 4 by College Entrance Examination Board. All rights reserved.

78 97 AB3/BC Let f ( x) = ksin( kx) where k is a positive constant. Find the area of the region bounded by one arch of the graph of f and the x-axis. Find the area of the triangle formed by the x-axis and the tangents to one arch of f at the points where the graph of f crosses the x-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.

79 97 AB3/BC Solution Area π k π k = k sin( kx) dx= cos( kx) = cos π ( cos ) = + = f ( x) = k cos( kx) f () = k and so the equation of the left tangent line is y = k x. π f = k k π and so the equation of the right tangent line is y = k x k. The tangent lines intersect when π k x= k x k k x= kπ π x = k π kπ At this x value, y = k =. k The area of the triangle is therefore π kπ π = k 4. Copyright 4 by College Entrance Examination Board. All rights reserved.

80 97 AB4/BC3 A man has 34 yard of fencing for enclosing two separate fields, one of which is to be a rectangle twice as long as it is wide and the other a square. The square field must contain at least square yards and the rectangular one must contain at least 8 square yards. If x is the width of the rectangular field, what are the maximum and minimum possible values of x? What is the greatest number of square yards that can be enclosed in the two fields? Justify your answer. Copyright 4 by College Entrance Examination Board. All rights reserved.

81 97 AB4/ BC3 Solution Let y be the width of the square field. Then 6x+ 4y = 34. x y Since Since y, then y. x 8, then x. 3 4y = 34 6x y = 85 x 3 3 y 85 x x 75 x 5 x y Therefore the maximum value of x is 5; the minimum value of x is. Let A be the total area. Then A= x + y = x + 85 x A = 4x+ 85 x = 4x 55 + x= x 55 A = only when x = 55 = But A = > and so x = 3 gives a relative minimum for A. Hence the maximum area must occur at x = or at x = 5. At x =, At x = 5, A = () + (55) = 385 A = (5) + () = 5 Therefore, 5 is the greatest number of square yards that can be enclosed. Copyright 4 by College Entrance Examination Board. All rights reserved.

82 97 AB5 Let cos x y = e. Calculate dy dx and d y dx. If x and y both vary with time in such a way that y increases at a steady rate of 5 π units per second, at what rate is x changing when x =? Copyright 4 by College Entrance Examination Board. All rights reserved.

83 97 AB5 Solution cos x y = e dy cos x cos x e ( sin x) (sin x) e dx = = d y cos cos cos = (sin x) e x ( sin x) (cos x) e x = e x (sin x cos x) dx dy dy dx cos x dx = = (sin xe ) dt dx dt dt dy Substituting 5 dt = and π x = gives π cos( π ) dx dx dx 5= sin e = () e =. dt dt dt Therefore dx 5 dt = when π x =. Copyright 4 by College Entrance Examination Board. All rights reserved.

84 97 AB6 In the figure, the shaded region R is bounded by the graphs of xy =, x =, x =, and y =. Find the volume of the solid generated by revolving the region R about the x-axis. Find the volume of the solid generated by revolving the region R about the line x = 3. Copyright 4 by College Entrance Examination Board. All rights reserved.

85 97 AB6 Solution Volume π π =π y dx=π dx = π = +π= x x 3 Volume = π (3 x) y dx = π (3 x) dx = π dx x x = π(3ln x x ) = π((3ln ) ( )) = π(3ln ) Copyright 4 by College Entrance Examination Board. All rights reserved.

86 97 AB7 A function f is defined for all real numbers and has the following three properties: (i) f () = 5, (ii) f (3) =, and (iii) for all real values of a and b, f ( a+ b) f( a) = kab+ b where k is a fixed real number independent of a and b. Use a = and b= to find the value of k. Find f (3). (c) Find f ( x) and f ( x ) for all real x. Copyright 4 by College Entrance Examination Board. All rights reserved.

87 97 AB7 Solution Let a = and b = in (iii): f(+ ) f() = k()() + () f (3) f() = k+ 8 From (i) and (ii), 5 = k + 8 Hence k = 4. Method (using (iii)) f(3 + h) f(3) 4 3 h+ h f (3) = lim = lim = lim( + h) = h h h h h Method (using (iii)) f (3) f( t) f( t+ (3 t)) f( t) 4 t(3 t) + (3 t) f (3) = lim = lim = lim t 3 3 t t 3 3 t t 3 3 t = lim (4t+ (3 t)) = t 3 (c) f ( x) f( x+ h) f( x) 4xh+ h = lim = lim = lim (4x + h) = 4x h h h h h f ( x) = 4xdx= x + C 5 = f () = () + C = + C, so C = 3. Therefore f( x) = x + 3. Alternate method: Let a =, b= x, and k = 4 in (iii). f( + ( x )) f() = 4( x ) + ( x ) = x Since f () = 5, this gives f( x) = x + 3 and so f ( x) = 4x. Copyright 4 by College Entrance Examination Board. All rights reserved.

88 97 BC4 Does the series converge? Justify your answer. 3 5 k= k + Determine all values of x for which the series answer. k x k converges. Justify your k k= Copyright 4 by College Entrance Examination Board. All rights reserved.

89 97 BC4 Solution The terms of the series satisfy the conditions needed to use the integral test. dx b dx = lim dx = lim (3x + 5) dx 3x+ 5 3x+ 5 b b b b = lim (3x+ 5) = lim (3b+ 5) (8) = b 3 b 3 3 Since the improper integral diverges, so does the series. Ratio Test k+ k+ x k+ k+ lim k + x k k = lim = lim x k k k k k k k x + x k k+ = x k The series converges for x < or < x <. At x =, the series is harmonic series. At x =, the series is alternating harmonic series. k k k= k= k k= k= = which diverges since this is the k k k k ( ) = which converges since this is the k k The interval of convergence is x <. Copyright 4 by College Entrance Examination Board. All rights reserved.

90 97 BC5 Determine the general solution of the differential equation x y 4y + 4y = e. x Assume that xe is a solution of the differential equation y + ay + by =. Determine the constants a and b. Copyright 4 by College Entrance Examination Board. All rights reserved.

91 97 BC5 Solution First find the solution to associated homogeneous equation y 4y + 4 =. m 4m+ 4= ( m ) = m = Therefore the homogeneous solution is y = C e + C xe. h x x Find a particular solution using the method of undetermined coefficients. x y = Ax e p x x x p = + = ( + ) x x x p y Ax e Axe Ae x x y = Ae (x + ) + 4 Ae ( x + x) = Ae (x + 4x + ) x Substituting into y 4y + 4y = e gives Ae x x = e and so x Therefore yp = x e. The general solution is y = yh + yp = C e + C xe + x e x x x A =. y = xe x x y = e ( x+ ) x y = e ( x+ ) Substitute into y + ay + by = x x x e ( x+ ) + ae ( x+ ) + bxe = x ( + ae ) + ( + a+ bxe ) = x + a = (since the equality must hold for all x) + a+ b= a = b = Copyright 4 by College Entrance Examination Board. All rights reserved.

92 97 BC6 Consider the function f defined by x if x > ln x f( x) = if x= x if x < ln( x) For what values of x is f continuous? Is the graph of f symmetric with respect to (i) (ii) the y-axis? the origin? (c) (d) Find the coordinates of all relative maximum points. Find the coordinates of all relative minimum points. Copyright 4 by College Entrance Examination Board. All rights reserved.

93 97 BC6 Solution x lim = = + x ln x Therefore f is not continuous at x =. f is continuous at all other values of x except for x = and x = where the function is not defined. f ( ) For x >, ( x) x f ( x) = = = f( x) ln( ( x)) ln x For x <, ( x) f ( x) = = f( x) ln( x) (i) The graph is symmetric with respect to the y-axis since f ( x) = f( x) for all x. (ii) The graph is not symmetric with respect to the origin since f ( x) f( x). (c) and (d) ln x f ( x) = for x > (ln x) ln( x) + f ( x) = (ln( x)) for x < (ln x) (ln x ) ln x x x f ( x) = for x >. 4 (ln x) f( x ) < for < x < and for < x <. Since f () =, the point (,) is a relative maximum. For x >, f ( x) = when ln x =, that is, at x = e. f () e = >. Therefore there is a relative minimum at ( ee, ). By symmetry, there e is also a relative minimum at ( ee, ). Copyright 4 by College Entrance Examination Board. All rights reserved.

94 97 BC7 Let C be the curve defined from t = to t = 6 by the parametric equations t + x =, y = t (6 t ). Let R be the region bounded by C and the x-axis. Set up but do not evaluate an integral expression in terms of a single variable for the length of C. (c) the volume of the solid generated by revolving R about the x-axis. the surface area of the solid generated by revolving R about the x-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.

95 97 BC7 Solution dx dy =, = 6 t dt dt dy y = 4( x 5x+ 4) for x 4, so 4(x 5) dx = 5 x = 9 y for y 9 (left half of curve), so dx dy Each of the integrals can be written in terms of dt, dx, or dy. = 4 9 y b dx dy Length = + dt dt a 6 = + (6 t ) 4 4 = + 6(x 5) 9 = dt ( y) dx dt dy Volume b =π y dx a 6 =π ((6 t t)) dt 4 =π 6( x 5x+ 4) dx 9 = π y 9 ydy (c) Surface area = b π yds a = π (6 ) + (6 ) 6 t t t dt = π 4( x 5x+ 4) + 6(x 5) dx = 4π y + dy 6(9 y ) Copyright 4 by College Entrance Examination Board. All rights reserved.

96 973 AB Given 3 f ( x) = x 6x + 9x and gx= ( ) 4. Find the coordinates of the points common to the graphs of f and g. Find all the zeros of f. (c) If the domain of f is limited to the closed interval [,], what is the range of f? Show your reasoning. Copyright 4 by College Entrance Examination Board. All rights reserved.

97 973 AB Solution 3 x 6x + 9x= 4 3 x 6x + 9x 4= ( x )( x 5x+ 4) = ( x )( x )( x 4) = The roots are x = and x = 4. The coordinates of the common points are therefore (,4) and (4,4). 3 x 6x + 9x= xx ( 6x+ 9) = xx ( 3) = The zeros of f are x = and x = 3. (c) f ( x) = 3x x+ 9 = 3( x 3)( x ) On the interval [,], the only critical point is at x =. f ( x) = 6x so f () = 6 <. Therefore there is a relative maximum at x =. Alternatively, the graph of f is increasing on the interval (,) and decreasing on the interval (,) since f ( x) > on (,) and f ( x) < on (,). Hence there is an absolute maximum at x = on the interval [,]. f () = f () = 4 f () = Since there are no other critical points in the restricted domain, the range is the closed interval [,4]. Copyright 4 by College Entrance Examination Board. All rights reserved.

98 973 AB A particle moves on the x-axis so that its acceleration at any time t > is given by t 9 5 a =. When t =, v=, and s =. 8 t 6 48 Find the velocity v in terms of t. Does the numerical value of the velocity ever exceed 5? Explain. (c) Find the distance s from the origin at time t =. Copyright 4 by College Entrance Examination Board. All rights reserved.

99 973 AB Solution t t v= dt = + + C 8 t 6 t At t =, 9 = + + C and so 6 6 C =. Therefore t v = +. 6 t Yes, the numerical value does exceed 5. This can be explained by any one of the following observations: (i) lim vt ( ) = (ii) t + lim vt ( ) = t + (iii) When t = 3, for example, 3 v = + = 64 + > (c) 3 t t s = + dt = + ln t t+ C 6 t 48 At t =, = 48 + C and so C =. Therefore t s= + ln t t+. 48 At t =, 8 s = + ln + = + ln Copyright 4 by College Entrance Examination Board. All rights reserved.

100 973 AB3/BC Given the curve x+ xy+ y = 6. Find an expression for the slope of the curve at any point ( x, y ) on the curve. Write an equation for the line tangent to the curve at the point (,). (c) Find the coordinates of all other points on this curve with slope equal to the slope at (,). Copyright 4 by College Entrance Examination Board. All rights reserved.

101 973 AB3/BC Solution Implicit differentiation gives x+ xy+ y = 6 dy dy + x + y+ 4y = dx dx dy + y = dx x + 4y dy + At the point (,), = =. Therefore the equation of the tangent line is dx y = ( x ), or y = x+, or 3y+ x 5 = (c) + y = x+ 4y 3 x 44 y = 3 3 x y + = 3 3 y = 3 x Substituting this into the equation for the curve gives x+ x(3 x) + (3 x) = 6 x+ 3x x + 8 x+ x = 6 x 8x+ = ( x 6)( x ) = Therefore x = 6 is the only other point on the curve with the desired property. The coordinates at this point are (6, 3). Copyright 4 by College Entrance Examination Board. All rights reserved.

102 973 AB4/BC What is the set of all values of b for which the graphs of intersect in two distinct points? y = x+ b and y = 4x In the case b = 4, find the area of the region enclosed by y = x 4 and y = 4x. (c) In the case b =, find the volume of the solid generated by revolving about the x-axis the region bounded by y = x and y = 4x. Copyright 4 by College Entrance Examination Board. All rights reserved.

103 973 AB4/BC Solution The intersection occurs when ( x+ b) = 4x 4x + 4bx+ b = 4x 4x + 4( b ) x+ b = 4( b ) ± 6( b ) 6b The solutions to this equation are x =. There will be 8 two distinct roots when 6( b ) 6b >, this is, when 3b + 6 >. Hence there are two distinct roots when b <. With b = 4, the intersection occurs when y = x 4 y = 4x 4x 6x+ 6= 4x x 5x+ 4= ( x )( x 4) = Therefore the intersection points are (, ) and (4, 4). Area = y + 4 y y 64 8 y dy = + y = = (c) When b =, intersection points are at (,) and (,). Volume Or Volume 4 =π (4x 4 x ) dx =π x x π =π = 3 3 y y = π y 4 dy 3 y 4 y 8 6 π = π = π = Copyright 4 by College Entrance Examination Board. All rights reserved.

104 973 AB5/BC3 Find the coordinates of the absolute maximum point for the curve k is a fixed positive number. Justify your answer. y = xe kx where Write an equation for the set of absolute maximum points for the curves as k varies through positive values. y = xe kx Copyright 4 by College Entrance Examination Board. All rights reserved.

105 973 AB5/BC3 Solution kx kx kx y = kxe + e = e ( kx) kx Since e, y = when x =. k kx kx kx kx y = k xe ke ke = e ( k x k) At x =, y = e ( k k) = e ( k) < since k is positive. Therefore x = gives k k a relative maximum. Since x = is the only critical point, x = also gives the k k absolute maximum. Therefore the coordinates of the absolute maximum point are, k ke. By part, x = and k y =. Therefore ke x y = for x >. e Copyright 4 by College Entrance Examination Board. All rights reserved.

106 973 AB6 A manufacturer finds it costs him x + 5x+ 7 dollars to produce x tons of an item. At production levels above 3 tons, he must hire additional workers, and his costs increase by 3( x 3) dollars on his total production. If the price he receives is $3 per ton regardless of how much he manufactures and if he has a plant capacity of tons, what level of output maximizes profits? Copyright 4 by College Entrance Examination Board. All rights reserved.

107 973 AB6 Solution For x 3, For 3< x, Px ( ) = 3 x ( x + 5x+ 7) = x + 8x 7 Px ( ) = 3 x ( x + 5x+ 7+ 3( x 3)) = x + 5x + Profit function: Px ( ) x + x x = x + x + < x 8 7, 3 5, 3 x+ 8, < x< 3 P ( x) = x + 5, 3 < x < P ( x) for any x in the interval < x <. However P ( x) > for < x < 3 and so Px ( ) is increasing on this interval. Also P ( x) < for 3< x < and so Px ( ) is decreasing on this interval. Therefore P has an absolute maximum at x = 3. This can also be verified by checking the value at x = 3 against the values at the endpoints. P () = 7 P () = 48 P (3) = 8 Copyright 4 by College Entrance Examination Board. All rights reserved.

108 973 AB7 Find the area A, as a function of k, of the region in the first quadrant enclosed by the y-axis and the graphs of y = tan x and y = k for k >. What is the value of A when k =? (c) If the line y = k is moving upward at the rate of A changing when k =? units per second, at what rate is Copyright 4 by College Entrance Examination Board. All rights reserved.

109 973 AB7 Solution A tan k = ( k tan x) dx= kx+ ln(cos x) = ktan k + ln(cos(tan k )) = ktan k+ ln + k tan k or, using integration by parts, A k y = tan ydy= ytan y dy + y = ytan y ln( + y ) = ktan k ln( + k ) k k π = tan ln+ = ln. 4 When k =, A () () ( ) (c) By the chain rule, da da dk =. We are given that dt dk dt dk dt =. da k tan = ( ) k k dk + + k + k So at k =, da dk π π = + () = 4 4 Hence da dt π π = = 4 4 Copyright 4 by College Entrance Examination Board. All rights reserved.

110 973 BC4 A kite flies according to the parametric equations t 3 x=, y = t( t 8) 8 64 where t is measured in seconds and < t 9. How high is the kite above the ground at time t = 3 seconds? At what rate is the kite rising at t = 3 seconds? (c) At what rate is the string being reeled out at t = 3 seconds? (d) At what time does the kite start to lose altitude? Copyright 4 by College Entrance Examination Board. All rights reserved.

111 973 BC4 Solution 3 y = t ( t 8) At t = 3, y = (3)(3 8) = ( 96) = dy dt 3 3 = t + ( t 8) At t = 3, dy dt = + ( 96) = + = 3 64 (c) Let s be the length of the string. s = x + y ds ( ) dx dy = x + y x + y dt dt dt At t = 3, x = 4 and y = 44. At t = 3, dx dt = 8 and dy dt = 3. Therefore at t = 3, ds = (4) (44)(3) + = =. dt (6 9) 4 97 (d) The kite will start to lose altitude at the instant when at t = 64 seconds. dy 6 = = t + 6. This occurs dt 64 Copyright 4 by College Entrance Examination Board. All rights reserved.

112 973 BC5 The distance x of a particle from a fixed point x = is given by the differential equation d x dx x = for all time t. dt dt Find the general solution of this differential equation. Find the solution of this differential equation satisfying the conditions x = and dx dt = when t =. (c) Does the particle pass through the point x = after it begins moving? If so, determine the value of t at each such time; if not, explain. Copyright 4 by College Entrance Examination Board. All rights reserved.

113 973 BC5 Solution The general solution can be found by computing the roots of the auxiliary equation. d x dx x = dt dt m + 8m+ = ( m+ )( m+ 6) = m=, m= 6 Therefore the general solution is t 6t x = Ce + C e. dx When t =, x = and dt =. dx = Ce 6 dt t 6t Ce The initial conditions give the two simultaneous equations: = C+ C = C 6C Hence C = and C =. Therefore (c) Can x =? t 6t = e e 4t = e 4t e = 4t = ln t = ln 4 t 6t x = e e. But ln < and hence the particle does not pass through the origin for any 4 t >. Copyright 4 by College Entrance Examination Board. All rights reserved.

114 973 BC6 In each of the following cases, decide whether the infinite series converges. Justify your answers. (c) k= + k sin k k= k + k= 3 kln k k k Copyright 4 by College Entrance Examination Board. All rights reserved.

115 973 BC6 Solution k Since lim + = e, the series must diverge. It is also sufficient to observe k k that for each term, + > which would be enough to force the series to k diverge. sin k < k + k k + k k The series converges absolutely by comparison with a convergent p-series (p = ), and therefore the series converges. (c) The terms of the series satisfy the conditions needed to use the integral test. b dx = lim dx = lim = lim + = ln ln b ln ln x(ln x) b x(ln x) b x b Since the improper integral converges, so does the series. b Copyright 4 by College Entrance Examination Board. All rights reserved.

116 973 BC7 Given the definite integral closed interval [,]. π I = xf(sin x) dx, where f is a continuous function on the Use the substitution x =π y to show that Evaluate π xsin x dx. + cos x π π I = f(sin x) dx. Copyright 4 by College Entrance Examination Board. All rights reserved.

117 973 BC7 Solution Let x =π y π I = xf(sin x) dx = ( π y) f(sin( π y))( ) dy π π = ( π y) f(sin( π y)) dy π = ( π y) f(sin y) dy π =π f (sin y) dy yf(sin y) dy π =π f(sin x) dx I π Therefore π I =π f(sin x) dx and so π π I = f(sin x) dx. Using part with x f( x) =, we have x π xsin x π sin x dx = + cos x + cos π dx x Method π xsin x π sin x π π π π π π dx = dx = arctan(cos x) = = + + x cos x cos π Method π π xsin x π sin x dx = dx (let u = cos x) + cos x + cos x π π π π π π = du = arctan u = = + u Copyright 4 by College Entrance Examination Board. All rights reserved.

118 974 AB/BC Given f( x) = sin x, π x π, and gx ( ) On the axes provided sketch the graph of f. = x for all real x. Let H( x) = g( f( x)). Write an expression for H( x ). (c) Find the domain and range of H. (d) Find and equation of the line tangent to the graph of H at the point where π x =. 4 Copyright 4 by College Entrance Examination Board. All rights reserved.

119 974 AB/BC Solution ( ) ( ) = sin = sin H x x x (c) The domain of H is π x The range of H is y (d) H ( x) = sin xcos x H π = = 4 H π = 4 π Therefore the equation of the tangent line is π y = x 4, or π y = x +. 4 Copyright 4 by College Entrance Examination Board. All rights reserved.

120 974 AB 4 3 Let P( x) = x + ax + bx + cx+ d. The graph of y = P( x) is symmetric with respect to the y-axis, has a relative maximum at (,), and an absolute minimum at ( q, 3). Determine the values of a, b, c, and d, and using these values write an expression for Px. ( ) Find all possible values of q. Copyright 4 by College Entrance Examination Board. All rights reserved.

121 974 AB Solution = P() = d = P () = c P( x) = P( x) x ax + bx + = x + ax + bx + ax = a = Alternatively, symmetry with respect to the y-axis means that only even powers of x will appear in the polynomial, and hence a = c=. 4 Px ( ) = x + bx + 4 Pq ( ) = q + bq + = 3 3 P ( q) = 4q + bq = Method : 4 q + bq + 8 = 4 q 8 = q =± 4 4q + bq = Hence 6 + 4b = and so b = 4. Method : q( q + b) = q = b (note that q cannot equal.) Therefore b b b + = 3 = 4 b = 6 b= (since q = b shows that b < ) b q =. Therefore 4 Px ( ) = x 4x +. b q = = q =± Copyright 4 by College Entrance Examination Board. All rights reserved.

122 974 AB3 Let f ( x) = kx + c. Find x in terms of k such that the tangent lines to the graph of f at ( x, f( x )) and ( x, f( x)) are perpendicular. (c) Find the slopes of the tangent lines mentioned in. Find the coordinates, in terms of k and c, of the point of intersection of the tangent lines mentioned in. Copyright 4 by College Entrance Examination Board. All rights reserved.

123 974 AB3 Solution f ( x) = kx The tangent lines will be perpendicular if = f ( x ) f ( x ) = ( kx )( kx ) = 4k x x x 4k = =± k f = k = k k f = k = k k (c) The equations of the two tangent lines are y ( kx + c) = ( x x ) y ( kx + c) = ( x ( x )) The point of intersection is therefore at x = since the left sides are equal. At x =, y = kx + c x = k c c + =. 4k k 4k The point of intersection is, c 4k. Copyright 4 by College Entrance Examination Board. All rights reserved.

124 974 AB4/BC Let f be a function defined for all x > 5 and having the following properties. (i) f ( x) = for all x in the domain of f. 3 x + 5 (ii) The line tangent to the graph of f at (4,) has an angle of inclination of 45. Find an expression for f ( x ). Copyright 4 by College Entrance Examination Board. All rights reserved.

125 974 AB4/BC Solution f ( x ) ( 5) ( 5) 3 5 dx 3 x = = + x dx = 3 x C = tan 45 = f (4) = (9) + C = + C C = 3 Therefore f ( x) = ( x+ 5) f ( x) = ( x+ 5) dx= ( x+ 5) x+ C = f(4) = (9) 4+ C C = 6 9 Therefore 4 3 f ( x) = ( x+ 5) x 6. 9 Copyright 4 by College Entrance Examination Board. All rights reserved.

126 974 AB5/BC4 A ball is thrown from the origin of a coordinate system. The equation of its path is m e y = mx x, where m is positive and represents the slope of the path of the ball at the origin. For what value of m will the ball strike the horizontal axis at the greatest distance from the origin? Justify your answer. For what value of m will the ball strike at the greatest height on a vertical wall located feet from the origin? Copyright 4 by College Entrance Examination Board. All rights reserved.

127 974 AB5/BC4 Solution The ball will strike the horizontal axis when y =. m e mx x = x = me, m dx m m = e me = e m ( m) = when dm dx dm > for dx m < and dm < for m >. Therefore distance from the origin. m =. m = gives the maximum d y m m m Alternatively, = ( e + ( m)( ) e ) = 4 e ( m). At dm d x m =, = e <. Therefore m = gives a relative maximum. But dm since this is the only critical point, it is also an absolute maximum. m e m When x =, y = m () = m e. We want the value of m that makes y a maximum. dy m e dm = = 5 = e m = m ln 5 Copyright 4 by College Entrance Examination Board. All rights reserved.

128 974 AB6 Given two functions f and g defined by f ( x) = tan x and g( x) = cos x. Find the coordinates of the point of intersection of the graphs of f and g in the π interval < x <. Find the area of the region enclosed by the y-axis and the graphs of f and g. Copyright 4 by College Entrance Examination Board. All rights reserved.

129 974 AB6 Solution At the intersection tan x= cos x sin x = cos x= ( sin x) sin x+ sinx = ( sinx )(sinx+ ) = The only possibility is sin x =. The solution in π π the interval < x < is at x =. So the point of 4 π intersection is, 4. Area π 4 = ( cos x tan xdx ) = ( sinx+ lncos ) 4 x π = + ln = ln Copyright 4 by College Entrance Examination Board. All rights reserved.

130 974 AB7 The rate of change in the number of bacteria in a culture is proportional to the number present. In a certain laboratory experiment, a culture had, bacteria initially,, bacteria at time t minutes, and, bacteria at ( t + ) minutes. (c) In terms of t only, find the number of bacteria in the culture at any time t minutes, t. How many bacteria were there after minutes? How many minutes had elapsed when the, bacteria were observed? Copyright 4 by College Entrance Examination Board. All rights reserved.

131 974 AB7 Solution Let N = number of bacteria present at time t. Then dn = kn dt dn = kdt N ln N = kt+ C kt N = C e k At t =, = Ce = C. Therefore N = e kt. At t = t, = e = e kt At t = t +, ( + = e ) = e e = e Therefore At t =, ln 5 k = and so kt kt kt k k ln 5 t N = e. ln 5 ( ) ln 5 N = e = e = 5 (c) = e ln 5 t = e ln 5 t ln 5 ln = t ln t = ln 5 Copyright 4 by College Entrance Examination Board. All rights reserved.

132 974 BC3 Let R be the region bounded by the graph of y = ln x, the line x = e, and the x-axis. Find the volume generated by revolving R about the x-axis. Find the volume generated by revolving R about the y-axis. Copyright 4 by College Entrance Examination Board. All rights reserved.

133 974 BC3 Solution Shells: Volume = π ( e e ) ydy y y = π ( ey ye ) dy (integrate by parts) ey y y = π ( ye e ) e = π =π( e ) Disks: Volume e =π x dx (integrate by parts twice) (ln ) ( x(ln x) xln x x) e =π + = π( e e+ e ) =π( e ) Disks: Volume ( y ) e =π e e dy =π e y y e π = π + = ( e + ) Shells: Volume e x = π ln xdx (integrate by parts) x ln x x = π 4 e e e e π = π + = π + = ( e + ) Copyright 4 by College Entrance Examination Board. All rights reserved.