AP CALCULUS BC 2015 SCORING GUIDELINES

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2 05 SCORING GUIDELINES Question 5 Consider the function f =, where k is a nonzero constant. The derivative of f is given by k f = k ( k). (a) Let k =, so that f =. Write an equation for the line tangent to the graph of f at the point whose -coordinate is 4. (b) Let k = 4, so that f =. Determine whether f has a relative minimum, a relative maimum, or 4 neither at =. Justify your answer. (c) Find the value of k for which f has a critical point at = 5. (d) Let k = 6, so that f =. Find the partial fraction decomposition for the function f. 6 Find f d. 4 5 (a) f( 4) = = f (4) = = An equation for the line tangent to the graph of f at the point 5 whose -coordinate is 4 is y = ( 4 ) (b) f = f = = f changes sign from positive to negative at =. Therefore, f has a relative maimum at =. { : slope : : tangent line equation : considers f : : answer with justification k ( 5) (c) f ( 5) = = 0 k = 0 ( 5) k ( 5) (d) A B A( 6) B 6 = ( 6) = + 6 = + = 0 = A ( 6) A = 6 = 6 = B ( 6) B = = + ( 6) f d = + d 6 6 = ln + ln 6 + C = ln + C : answer 05 The College Board. Visit the College Board on the Web: { : partial fraction decomposition 4: : general antiderivative

3 0 SCORING GUIDELINES (Form B) Question 4 The graph of the differentiable function y = f with domain 0 0 is shown in the figure above. The area of the region enclosed between the graph of f and the -ais for 0 5 is 0, and the area of the region enclosed between the graph of f and the -ais for 5 0 is 7. The arc length for the portion of the graph of f between = 0 and = 5 is, and the arc length for the portion of the graph of f between = 5 and = 0 is 8. The function f has eactly two critical points that are located at = and = 8. (a) Find the average value of f on the interval (b) Evaluate ( f + ) d. Show the computations that lead to 0 your answer. 5 (c) Let g = f( t) dt. On what intervals, if any, is the graph of g both concave up and decreasing? Eplain your reasoning. (d) The function h is defined by h = f. The derivative of h is h f. the graph of y = h from = 0 to = 0. = Find the arc length of 5 0 (a) Average value = f d 5 = = : answer (b) ( f + ) d = f d + f d = ( 0 + 7) + 0 = 7 : answer (c) g = f g < 0 on 0 < < 5 g is increasing on < < 8. The graph of g is concave up and decreasing on < < 5. : : g = f : analysis : answer and reason 0 ( ) 0 (d) Arc length = + h d = + f d 0 0 Let u =. Then du = d and 0 ( ) + f d = + ( f u ) du = ( + 8) = : : integral : substitution : answer 0 The College Board. Visit the College Board on the Web:

4 00 SCORING GUIDELINES (Form B) Question 5 4 Let f and g be the functions defined by f = and g =, for all > (a) Find the absolute maimum value of g on the open interval ( 0, ) if the maimum eists. Find the absolute minimum value of g on the open interval ( 0, ) if the minimum eists. Justify your answers. (b) Find the area of the unbounded region in the first quadrant to the right of the vertical line =, below the graph of f, and above the graph of g (a) g = = ( + 4 ) + 4 For > 0, g = 0 for =. g > 0 for 0 < < g < 0 for > g = 5 : : g : critical point : answers : justification Therefore g has a maimum value of at =, and g has no minimum value on the open interval ( 0, ). = lim ( ln ln ( + 4 )) (b) ( f g ) d = lim ( f g ) d b b = b = b ( ( b) ( b ) ) = lim ln ln ln 5 b b 5 = lim ln b + 4b 5b = lim ln b + 4b 4 : : integral : antidifferentiation : answer 5b = lim ln b + 4b = 5 ln 4 00 The College Board. Visit the College Board on the Web:

5 009 SCORING GUIDELINES (Form B) Question A continuous function f is defined on the closed interval 4 6. The graph of f consists of a line segment and a curve that is tangent to the -ais at =, as shown in the figure above. On the interval 0 < < 6, the function f is twice differentiable, with f > 0. (a) Is f differentiable at = 0? Use the definition of the derivative with one-sided limits to justify your answer. (b) For how many values of a, 4 a 6, equal to 0? Give a reason for your answer. (c) Is there a value of a, 4 a 6, < is the average rate of change of f on the interval [ a,6] < for which the Mean Value Theorem, applied to the interval [ a ] guarantees a value c, a < c < 6, at which f ( c) =? Justify your answer. (d) The function g is defined by g = f( t) dt for 4 6. On what intervals contained in [ 4, 6] is the graph of g concave up? Eplain your reasoning. f( h) f( 0) (a) lim = h 0 h f( h) f( 0) lim < 0 + h 0 h Since the one-sided limits do not agree, f is not differentiable at = 0. f( 6) f( a) (b) = 0 when f( a) = f( 6. ) There are 6 a two values of a for which this is true. (c) Yes, a =. The function f is differentiable on the interval < < 6 and continuous on 6. f( 6) f 0 Also, = =. 6 6 By the Mean Value Theorem, there is a value c, < c < 6, such that f ( c) =. (d) g = f, g = f g > 0 when f > 0 This is true for 4 < < 0 and < < 6. 0,6, : sets up difference quotient at = 0 : { : answer with justification : epression for average rate of change : { : answer with reason : answers yes and identifies a = : { : justification : g = f : : considers g > 0 : answer 009 The College Board. All rights reserved. Visit the College Board on the Web:

6 008 SCORING GUIDELINES Question 5 The derivative of a function f is given by f = ( ) e for > 0, and f () = 7. (a) The function f has a critical point at =. At this point, does f have a relative minimum, a relative maimum, or neither? Justify your answer. (b) On what intervals, if any, is the graph of f both decreasing and concave up? Eplain your reasoning. (c) Find the value of f (. ) (a) f < 0 for 0 < < and f > 0 for > Therefore, f has a relative minimum at =. : minimum at = : : justification (b) f = e + ( ) e = ( ) e f > 0 for > : f : : answer with reason f < 0 for 0 < < Therefore, the graph of f is both decreasing and concave up on the interval < <. (c) f = f + f d = 7 + ( ) e d u = dv = e d du = d v = e f = 7 + ( ) e e d = 7 + (( ) e e ) = 7 + e e 4: : uses initial condition : integration by parts : answer 008 The College Board. All rights reserved. Visit the College Board on the Web:

7 007 SCORING GUIDELINES Question 4 Let f be the function defined for > 0, with f( e ) = and f, the first derivative of f, given by f = ln. (a) Write an equation for the line tangent to the graph of f at the point ( e,). (b) Is the graph of f concave up or concave down on the interval < <? Give a reason for your answer. (c) Use antidifferentiation to find f ( ). (a) f ( e) = e An equation for the line tangent to the graph of f at the point ( e,) is y = e ( e). : f ( e) : : equation of tangent line (b) f = + ln. For < <, > 0 and ln > 0, so f > 0. Thus, the graph of f is concave up on (, ). : f : : answer with reason f = ln d, we consider integration by (c) Since parts. u = ln dv = d du = d v = ( ) d = : antiderivative 4 : : uses f( e) = : answer Therefore, = ( ln ) f d = ln d = ln + C. 9 e e Since f( e ) =, = + C and C 9 Thus, f = ln + e. 9 9 = e The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

8 006 SCORING GUIDELINES (Form B) Question The figure above is the graph of a function of, which models the height of a skateboard ramp. The function meets the following requirements. (i) At = 0, the value of the function is 0, and the slope of the graph of the function is 0. (ii) At = 4, the value of the function is, and the slope of the graph of the function is. (iii) Between = 0 and = 4, the function is increasing. (a) Let f = a, where a is a nonzero constant. Show that it is not possible to find a value for a so that f meets requirement (ii) above. (b) Let g = c, where c is a nonzero constant. Find the value of c so that g meets requirement (ii) 6 above. Show the work that leads to your answer. (c) Using the function g and your value of c from part (b), show that g does not meet requirement (iii) above. n (d) Let h =, where k is a nonzero constant and n is a positive integer. Find the values of k and n so that k h meets requirement (ii) above. Show that h also meets requirements (i) and (iii) above. (a) f ( 4) = implies that implies that a = and f ( 4) = a( 4) = 6 a =. Thus, f cannot satisfy (ii). 8 : a = or a = : 6 8 : shows a does not work (b) g( 4) = 64c = implies that c =. When, 4 c = g ( 4) = c( 4) = ( 6) = 6 (c) g = = ( 4) 8 g 4 < 0 for 0 < <, so g does not satisfy (iii). n 4 n (d) h( 4) = = implies that 4 = k. k n n n4 n4 n h ( 4) = = = = gives n = 4 and k n h = h( 0) = h = h ( 0) = 0 and h > 0 for 0 < < k = 4 = 56. : value of c : g : : eplanation n 4 : = k n- 4 : n4 : = k : values for k and n : verifications 006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 4

9 005 SCORING GUIDELINES (Form B) Question 5 Consider the curve given by y = + y. dy y (a) Show that =. d y (b) Find all points (, y ) on the curve where the line tangent to the curve has slope. (c) Show that there are no points (, y ) on the curve where the line tangent to the curve is horizontal. (d) Let and y be functions of time t that are related by the equation dy value of y is and = 6. Find the value of d at time t = 5. dt dt y = + y. At time t = 5, the (a) yy = y + y ( y ) y = y y y = y : implicit differentiation : : solves for y (b) y y = y = y = 0 y =± ( 0, ), ( 0, ) y : = : y : answer y (c) 0 y = y = 0 The curve has no horizontal tangent since for any. : y = 0 : : eplanation (d) When y =, dt t = 5 = + so = dy dy d y d = = dt d dt y dt d 9 d At t = 5, 6 = = 7 dt dt 6 d = 7. : : solves for : chain rule : answer Copyright 005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6

10 004 SCORING GUIDELINES Question 4 Consider the curve given by + 4y = 7 + y. dy y (a) Show that =. d 8y (b) Show that there is a point P with -coordinate at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P. d y (c) Find the value of at the point P found in part (b). Does the curve have a local maimum, a d local minimum, or neither at the point P? Justify your answer. (a) + 8yy = y + y ( 8y ) y = y y y = 8y : implicit differentiation : : solves for y (b) y 8y = 0; y = 0 When =, y = 6 y = + 4 = 5 and 7 + = 5 Therefore, P = (, ) is on the curve and the slope is 0 at this point. dy : = 0 d : : shows slope is 0 at (, ) : shows (, ) lies on curve (c) d y ( 8y )( y ) ( y )( 8y ) = d ( 8y ) d y ( 6 9)( ) At P = (, ), = =. d ( 6 9) 7 Since y = 0 and y < 0 at P, the curve has a local maimum at P. 4 : d y : d d y : value of at (, ) d : conclusion with justification Copyright 004 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 5

11 AP CALCULUS BC 00 SCORING GUIDELINES Question 4 Let h be a function defined for all L 0 such that h(4) Г and the derivative of h is given Г by h= for all L 0. (a) Find all values of for which the graph of h has a horizontal tangent, and determine whether h has a local maimum, a local minimum, or neither at each of these values. Justify your answers. (b) On what intervals, if any, is the graph of h concave up? Justify your answer. (c) Write an equation for the line tangent to the graph of h at = 4. (d) Does the line tangent to the graph of h at = 4 lie above or below the graph of h for 4? Why? (a) h= 0 at h= 0 + und 0 + Г 0 4 : : : analysis : conclusions Г > not dealing with discontinuity at 0 Local minima at Г and at (b) h== 0 for all L 0. Therefore, the graph of h is concave up for all L 0. : : h== : h== 0 : answer (c) 6 Г 7 h= (4) 4 7 y ( Г 4) : tangent line equation (d) The tangent line is below the graph because the graph of h is concave up for 4. : answer with reason Copyright 00 by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 5

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13 998 Calculus BC Scoring Guidelines Copyright 998 by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board.

AP CALCULUS AB 2017 SCORING GUIDELINES

AP CALCULUS AB 2017 SCORING GUIDELINES AP CALCULUS AB 07 SCORING GUIDELINES 06 SCORING GUIDELINES Question 6 f ( ) f ( ) g( ) g ( ) 6 8 0 8 7 6 6 5 The functions f and g have continuous second derivatives. The table above gives values of the