2017 AP CALCULUS AB FREE-RESPONSE QUESTIONS

Size: px

Start display at page:

Download "2017 AP CALCULUS AB FREE-RESPONSE QUESTIONS"

Caren Sullivan
5 years ago
Views:

1 07 AP CALCULUS AB FREE-RESPONSE QUESTIONS 6. Let f be the function defined by sin x f() x cos ( x ) e. Let g be a differentiable function. The table above gives values of g and its derivative g at selected values of x. Let h be the function whose graph, consisting of five line segments, is shown in the figure above. (a) Find the slope of the line tangent to the graph of f at x p. (b) Let k be the function defined by k() x h( f( x)). Find k(p ). (c) Let m be the function defined by m() x g( x ) hx (). Find m(). (d) Is there a number c in the closed interval 5, 3 such that g() c 4? Justify your answer. STOP END OF EXAM 07 The College Board. Visit the College Board on the Web: -7-

2 AP CALCULUS AB 07 SCORING GUIDELINES Question 6 (a) f = sin ( x) + cos xe sin x : f ( ) sin f ( ) = sin ( ) + cos e = (b) k = h ( f ) f k ( ) = h ( f( ) ) f ( ) = h ( ) ( ) = ( )( ) = 3 3 : : k : k ( ) (c) m = g ( x) h + g( x) h m ( ) = g ( 4) h( ) + g( 4) h ( ) = ( )( ) + 5( ) = : : m : m ( ) (d) g is differentiable. g is continuous on the interval [ 5, 3]. g( 3) g( 5) 0 = = 4 3 ( 5) Therefore, by the Mean Value Theorem, there is at least one value c, 5 < c < 3, such that g ( c) = 4. g( 3) g( 5) : 3 ( 5) : : justification, using Mean Value Theorem 07 The College Board. Visit the College Board on the Web:

3 AP CALCULUS AB 06 SCORING GUIDELINES Question 6 x f ( x ) f g( x ) g The functions f and g have continuous second derivatives. The table above gives values of the functions and their derivatives at selected values of x. (a) Let k = f ( g ). Write an equation for the line tangent to the graph of k at x = 3. g (b) Let h = f. Find h (. ) (c) Evaluate f ( x) dx. 3 (a) k( 3) = f ( g( 3) ) = f ( 6) = 4 k ( 3) = f ( g( 3) ) g ( 3) = f ( 6) = 5 = 0 { : slope at x = 3 3: : equation for tangent line An equation for the tangent line is y = 0( x 3) + 4. (b) h ( ) = = f ( ) g ( ) g( ) f ( ) ( f ( ) ) ( 6) = = ( 6) 36 { : expression for 3: : answer h ( ) 3 3 (c) f ( x) dx = ( ) ( 6) ( ) f x = f f 7 = 5 ( ) = 3: { : antiderivative : answer 06 The College Board. Visit the College Board on the Web:

9 AP CALCULUS AB 008 SCORING GUIDELINES (Form B) Question 4 The functions f and g are given by f = 4 + t dt and g = f( sin x). (a) Find f and g. (b) Write an equation for the line tangent to the graph of y = g at x = π. 3x 0 (c) Write, but do not evaluate, an integral expression that represents the maximum value of g on the interval 0 x π. Justify your answer. (a) f = ( 3x) g = f ( sin x) cos x 4 : : : f g = ( 3sin x) cos x (b) g( π ) = 0, g ( π ) = 6 Tangent line: y = 6( x π ) : g( π) or g ( π) : : tangent line equation (c) For 0 < x < π, g π = 0 only at x =. g( 0) = g( π ) = 0 ( ) π 3 = + > 0 g 4 t dt 0 0, π is The maximum value of g on [ ] t dt. 3 : : sets g = 0 π : justifies maximum at : integral expression for g π ( ) 008 The College Board. All rights reserved. Visit the College Board on the Web: 6

10 AP CALCULUS AB 007 SCORING GUIDELINES Question 3 x f ( x ) f g( x ) g The functions f and g are differentiable for all real numbers, and g is strictly increasing. The table above gives values of the functions and their first derivatives at selected values of x. The function h is given by hx ( ) = f( gx ( )) 6. (a) Explain why there must be a value r for < r < 3 such that hr ( ) = 5. (b) Explain why there must be a value c for < c < 3 such that h ( c) = 5. g (c) Let w be the function given by wx ( ) = f( t) dt. Find the value of w ( 3. ) (d) If g is the inverse function of g, write an equation for the line tangent to the graph of y = g at x =. (a) h() = f( g() ) 6 = f( ) 6 = 9 6 = 3 h( 3) = f( g( 3) ) 6 = f( 4) 6 = 6 = 7 Since h( 3) < 5 < h( ) and h is continuous, by the Intermediate Value Theorem, there exists a value r, < r < 3, such that h( r ) = 5. (b) h( 3) h( ) 7 3 = = Since h is continuous and differentiable, by the Mean Value Theorem, there exists a value c, < c < 3, such that h ( c) = 5. (c) w ( 3) = f ( g( 3) ) g ( 3) = f ( 4) = (d) g () =, so g ( ) =. ( g ) ( ) = = = g ( g ( ) ) g () 5 An equation of the tangent line is y = ( x ). 5 : h() and h( 3) : : conclusion, using IVT h( 3) h( ) : : 3 : conclusion, using MVT : apply chain rule : { : answer : g ( ) 3 : : ( g ) ( ) : tangent line equation 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents). 08

11 Let f be a twice-differentiable function such that gx f f x. AP CALCULUS AB 007 SCORING GUIDELINES (Form B) Question 6 f 5 (a) Explain why there must be a value c for c 5 such that f c. and f 5. Let g be the function given by (b) Show that g g5. Use this result to explain why there must be a value k for k 5 such that g k 0. (c) Show that if f x 0 for all x, then the graph of g does not have a point of inflection. (d) Let hx fx x. Explain why there must be a value r for r 5 such that hr 0. (a) The Mean Value Theorem guarantees that there is a value c, with c 5, so that f5 f 5 f c. 5 5 f5 f : : 5 : conclusion, using MVT (b) g x f f x f x g f f f f5 f g5 f f5 f5 f f5 Thus, g g5. Since f is twice-differentiable, g is differentiable everywhere, so the Mean Value Theorem applied to g on, 5 guarantees there is a value k, with k 5, such that gk g5 g : : g x : g f5 f g5 : uses MVT with g (c) g x f f x f x f x f f x f x If f x 0 for all x, then g x 0 f x f x f f x 0 0 for all x. Thus, there is no x-value at which g x changes sign, so the graph of g has no inflection points. OR If f x 0 for all x, then f is linear, so g f f is linear and the graph of g has no inflection points. (d) Let h x f x x. h f 5 3 h5 f Since h 0 h5, the Intermediate Value Theorem guarantees that there is a value r, with r 5, such that hr 0. : considers g : : gx 0 for all x OR : f is linear : : g is linear : h and h5 : : conclusion, using IVT 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

12 AP CALCULUS AB 006 SCORING GUIDELINES Question 6 The twice-differentiable function f is defined for all real numbers and satisfies the following conditions: f ( 0) =, f ( 0) = 4, and f ( 0) = 3. ax (a) The function g is given by gx ( ) = e + f for all real numbers, where a is a constant. Find g ( 0) and g ( 0) in terms of a. Show the work that leads to your answers. (b) The function h is given by hx ( ) = cos( kx) f for all real numbers, where k is a constant. Find h and write an equation for the line tangent to the graph of h at x = 0. (a) g = ae ax + f g ( 0) = a 4 ax g = a e + f g ( 0) = a : : g : g ( 0) : g : g ( 0) (b) h = f cos( kx) ksin ( kx) f h ( 0) = f ( 0) cos( 0) ksin( 0) f( 0) = f ( 0) = 4 h( 0) = cos( 0) f( 0) = The equation of the tangent line is y = 4x +. 5 : : h : h ( 0) 3 : : h( 0) : equation of tangent line 006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 7 0

13 AP CALCULUS AB 005 SCORING GUIDELINES Question 4 x 0 0 < x < < x < < x < < x < 4 f ( x ) Negative 0 Positive Positive 0 Negative f 4 Positive 0 Positive DNE Negative 3 Negative f Negative 0 Positive DNE Negative 0 Positive 0, 4. The function f is twice differentiable except at x =. The function f and its derivatives have the properties indicated in the table above, where DNE indicates that the derivatives of f do not exist at x =. (a) For 0 < x < 4, find all values of x at which f has a relative extremum. Determine whether f has a relative maximum or a relative minimum at each of these values. Justify your answer. (b) On the axes provided, sketch the graph of a function that has all the characteristics of f. (Note: Use the axes provided in the pink test booklet.) Let f be a function that is continuous on the interval [ ) (c) Let g be the function defined by g = f( t) dt on the open interval ( 0, 4 ). For x 0 < x < 4, find all values of x at which g has a relative extremum. Determine whether g has a relative maximum or a relative minimum at each of these values. Justify your answer. (d) For the function g defined in part (c), find all values of x, for 0 < x < 4, at which the graph of g has a point of inflection. Justify your answer. (a) f has a relative maximum at x = because f changes from : relative extremum at x = : positive to negative at x =. { : relative maximum with justification (b) : : points at x = 0,,, 3 and behavior at (, ) : appropriate increasing/decreasing and concavity behavior (c) g = f = 0 at x =, 3. g changes from negative to positive at x = so g has a relative minimum at x =. g changes from positive to negative at x = 3 so g has a relative maximum at x = 3. (d) The graph of g has a point of inflection at x = because g = f changes sign at x =. 3 : : g = f : critical points : answer with justification : { : x = : answer with justification Copyright 005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and 07 (for AP students and parents). 5

14 AP CALCULUS AB 00 SCORING GUIDELINES N BN Г# Г Question 6 Г# # # Г Г" Г$ Г% Г$ Г" Г B= N Г% Г# Г!! # % AJ B >A=BK?JEJD=JEI@EBBAHAJE=>ABH=HA=K>AHI6DAJ=>A=>LACELAIJDAL=KAIB B = BHIAA?JA@FEJI N EJDA?IA@EJAHL=Г # > N > # 6DAIA?@@AHEL=JELAB B D=IJDA FHFAHJOJD=J B== N BHГ # > N > # # = -L=K=JA! B = N > 9HEJA=AGK=JEBJDAEAJ=CAJJJDACH=FDB B =JJDAFEJMDAHA N 7IAJDEIEAJ =FFHNE=JAJDAL=KAB B /ELA=HA=IBHOKH=IMAH IJDEI=FFHNE=JECHA=JAHJD=HAIIJD=JDA=?JK=L=KAB B?.E@=FIEJELAHA=K>AH H D=LECJDAFHFAHJOJD=JJDAHAKIJANEIJ=L=KA?MEJD? # =@ B ==? H/ELA=HA=IBHOKH=IMAH N ГN Г% BH AJ C>AJDABK?JECELA>OCN N N Г% BH N F 6DACH=FDB C F=IIAIJDHKCDA=?DBJDAFEJI NBN CELAEJDAJ=>A=>LAIEJFIIE>AJD=JB =@ C =HAJDAI=ABK?JE/ELA=HA=IBHOKH=IMAH # # # =! B = N B = "@N #! BN " N! ГГ Г% " # " > O # N Г Г " B N # Г " Г! 6DA=FFHNE=JEEIAIIJD= B >A?=KIAJDA CH=FDBBEI??=LAKFJDAEJAHL= N? *OJDAA=8=KA6DAHAJDAHAEI=?MEJD? # IK?DJD=J B= # Г B=! Г B ==? $ H # Г E C = N E " N Г Г N Г N Г C N N N N E = E " 6DKI C = EIJ?JEKKI=JN>KJ B = EI?JEKKI=JNI B L C 4 C== N " BH= N L >KJEJM=IIDMEF=HJ?JD=J B ==? $ BHIA? L I B L C =JE@AHEL=JELA =IMAH J=CAJEA!?FKJAIOJ=CAJEA=JN =IMAHMEJDHA=I HABAHA?AJ86BHB = H@EBBAHAJE=>EEJO BB = L=KABHBHEJAHL= > N > # =IMAHI]^MEJDHABAHA?AJ C= HC==?HHA?JHA=I Copyright 00 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 7

15 99 AB5 Let f be a function that is even and continuous on the closed interval [ 3,3]. The function f and its derivatives have the properties indicated in the table below. x 0 0 < x < < x < < x < 3 f (x) Positive 0 Negative Negative f (x) Undefined Negative 0 Negative Undefined Positive f (x) Undefined Positive 0 Negative Undefined Negative (a) (b) (c) Find the x-coordinate of each point at which f attains an absolute maximum value or an absolute minimum value. For each x-coordinate you give, state whether f attains an absolute maximum or an absolute minimum. Find the x-coordinate of each point of inflection on the graph of f. Justify your answer. In the xy-plane provided below, sketch the graph of a function with all the given characteristics of f. y x 3 Copyright 003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 09

16 99 AB5 Solution (a) Absolute maximum at x = 0 Absolute minimum at x =± (b) Points of inflection at and f is even x =± because the sign of f changes at x = (c) y 3 3 x Copyright 003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 0

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