AP Calculus AB Sample Student Responses and Scoring ...€¦ · AP Calculus AB Sample Student Responses and Scoring Commentary from the 2018 Exam Administration: Free-Response Question

2018

AP Calculus ABSample Student Responses and Scoring Commentary

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Free Response Question 5

R Scoring Guideline

R Student Samples

R Scoring Commentary

AP® CALCULUS AB 2018 SCORING GUIDELINES

© 2018 The College Board. Visit the College Board on the Web: www.collegeboard.org.

Question 5

(a) The average rate of change of f on the interval 0 x π≤ ≤ is( ) ( )0 1.0

f f eπππ π− − −=−

1 : answer

(b) ( ) cos sinx xf x e x e x′ = −

( ) ( ) ( )3 2 3 2 3 23 3 3cos sin2 2 2f e e eπ π ππ π π′ = − =

The slope of the line tangent to the graph of f at 32x π= is 3 2.e π

( ) 1 : 2 :

1 : slopef x′

(c) ( ) 50 cos sin 0 ,4 4f x x x x xπ π′ = ⇒ − = ⇒ = =

x ( )f x0 1

4π 41

2eπ

54π 5 41

2e π−

2π 2e π

The absolute minimum value of f on 0 2x π≤ ≤ is 5 41 .2

e π−

( ) 1 : sets 05 1 : identifies ,3 : 4 4

as candidates 1 : answer with justification

f x

x xπ π′ =

= =

(d) ( )2

lim 0x

f xπ→

=

Because g is differentiable, g is continuous.

( ) ( )2lim 02x

g x gπ

π→

= =

By L’Hospital’s Rule, ( )( )

( )( )

2

2 2lim lim .2x x

f x f x eg x g x

π

π π→ →

′ −= =′

1 : is continuous at 2and limits equal 03 :

1 : applies L’Hospital’s Rule 1 : answer

g x π =

Note: max 1 3 [1-0-0] if no limit notation attached to a ratio of derivatives

•·

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3n- ,; (b) What is the slope .of the line tangent to the graph off at x =

2 ?·

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(c) Find the absolute minimum value off on the interval O ,$; x :s; 2n. Justify your answer.

��-x ='Si'-'--X

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(d) Let g be a differentiable function such that g( 1) == 0. The graph of g'. the derivative of R, is shown

below. Find t�e value of lim f((x)) or state that it dQes not exist. Justify your answer.

x➔rr/2 g X

y

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5. Letf be the function defined by f(x) = ex cos x.

( a) Find the average rate of change off on the interval O !:> x < n.

� L-n-)- �lo) -:: -i' ,;,\ '° t f-�\

%"-0

(b) What is the slope of the line tangent to the graph off at x = 3; ?

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"Cf

h. � \ � r=- �e 1-(;\) 1- e - l O)

�· -=u

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5B

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b l F. d th al f l!.... f(x) th . ,., . · "J 'f e ow. m e v ue o lim _2 g(-):) orstate at it uoes not exist.- usti y your answer,

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5. Let/be the function defined by f(x) = ex cos x.

(a) Find the average rate of change of.J•on the interval O ::; x :;.1e.

it .- e, - I

(b) What is tbe slope of the line tangent to the graph off at x == 3; ?

ft7l')�e � to.\{�)

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,,, ..,J - ...... , ·"·'·��" .

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50'. 5'·l ... . '

(c) Find the absolute minimum value of j.on the interval .0 ,S; x :;:;'21r .. J.ustify your.answer.,. . f 1 (x)� e.�,t��i�), e a(s·:_, .. (11) �!()-

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c d) _Let g be.a differentiable function such that .i:( 1� . =}· '.·�e;gra;h, �f g<�rd_eri�a�ve,of ·�.){�.ib:�;iI'·. . 't(x) !. '. \�?,·:-. · {. ··•··: .. : ._:-.. ... - · - . �tI.:'.·;;·,.:,.F):

..

below. Find the value of 1im -�)· or- s'tate'tfiafif �o'es not exlst .. Justi'fy _your-answet.,

x➔n/2 gxx · ' · " · ' ' . ·. , .; , I ,-�-- . . . ' •,i '. i

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Question 5

Overview

In this problem the function f is defined by cos .xf x e x In part (a) students were asked for the average rate of

change of f on the interval 0 .x A correct response should demonstrate that the average rate of change is

found using a difference quotient, 0

.0

f f

In part (b) students were asked for the slope of the line tangent to

the graph of f at 3 .2

x A correct response should use the fact that the slope of the tangent line is the value of the

derivative of f at the indicated point. The given expression for f x can be differentiated using the product rule. In

part (c) students were asked to find, with justification, the absolute minimum value of f on the interval 0 2 .x The Extreme Value Theorem guarantees that f attains a minimum on the interval 0 2 .x Candidates for locations of this minimum value are the endpoints of the interval and critical points for f inside the interval. In this

case, f is differentiable, so critical points are the two zeros of ,f 4

x and 5 .4

x The absolute minimum

value of f is the least of the values of f at the four candidates. In part (d) it is given that g is a differentiable

function such that 0,2

g and the graph of the derivative of g for 0 2x is supplied. Students were asked

to find the value of 2

lim ,x

f xg x

if it exists, and to justify their answer. A correct response should start with a

confirmation that the requested limit is an indeterminate form to which L’Hospital’s Rule applies. The numerator,

,f x has limit 0 as ,2

x using limit properties and continuity of the exponential and cosine functions. Because

g is differentiable, it is continuous, so the limit of the denominator, ,g x can be computed by substitution, yielding

2

lim 0.x

g x

After the indeterminate form is confirmed, applying L’Hospital’s Rule leads to the limit

2

lim ,x

f xg x

which can be evaluated using properties of limits and the provided graph of .g

For part (a) see LO 2.1A/EK 2.1A1. For part (b) see LO 2.1C/EK 2.1C3, LO 2.3B/EK 2.3B1. For part (c) see LO 1.2B/EK 1.2B1, LO 2.2A/EK 2.2A1. For part (d) see LO 1.1C/EK 1.1C3, LO 2.2B/EK 2.2B2. This problem incorporates the following Mathematical Practices for AP Calculus (MPACs): reasoning with definitions and theorems, connecting concepts, implementing algebraic/computational processes, connecting multiple representations, building notational fluency, and communicating.

Sample: 5A Score: 9

The response earned all 9 points: 1 point in part (a), 2 points in part (b), 3 points in part (c), and 3 points in part (d). In part (a) the response presents a difference quotient and would have earned the point for

0cos cos 0e e

in line 1 with no simplification. In this case, correct simplification to 1e

earned the

point. In part (b) the response would have earned the first point for cos sinx xdfe x e x

dx in line 1.

Responses that go on to factor must do so correctly. The response earned the point with the derivative expression

factored correctly to cos sin .xe x x The response would have earned the second point for the slope

AP® CALCULUS AB 2018 SCORING COMMENTARY


Question 5 (continued)

3 2 0 1e in line 3 with no simplification. In this case, correct simplification to 3 2e earned the point. In

part (c) the response earned the first point with cos sin 0xf x e x x in line 1. The response earned the

second point for identifying 4

x and 54

x as candidates in line 2 on the right. The response earned the

third point for the absolute minimum value of 5 422e and the global argument appealing to the behavior of

f x on the entire interval. The candidates test is used to justify that the minimum value is the only negative

extreme value on the interval. In part (d) the response earned the first point by stating that g x is continuous in

line 1 and showing in lines 1 and 3 that 2

lim 0x

g x

and 2

lim 0.x

f x

The response earned the second

point for a limit attached to a ratio of derivatives 2

limx

f xg x

in the last line. The third point was earned for the

answer 2

.2e

Sample: 5B Score: 6

The response earned 6 points: 1 point in part (a), 2 points in part (b), 1 point in part (c), and 2 points in part (d). In

part (a) the response presents a difference quotient and would have earned the point for 1e

in line 1. In this

case, the rewriting of the expression as 1e

earned the point. In part (b) the response earned the first point

for sin cos .x xf x e x e x The response would have earned the second point for the slope

3 2 3 23 1 02

f e e in line 2 with no simplification. In this case, correct simplification to 3 2e

earned the point. In part (c) the response earned the first point with sin cos 0.x xf x e x e x Because the

response does not identify both 4

x and 54

x as candidates, the second point was not earned. The third

point was not earned because 42

2e

is not the absolute minimum value of f on 0 2 .x Additionally,

4x is not in the interval. In part (d) the response did not earn the first point because the conditions that g is

continuous and that 2 2

lim lim 0x x

f x g x

are not verified. The response earned the second point in line 1

for a limit attached to a ratio of derivatives 2

lim .x

f xg x

The third point was earned for the answer

2.

2e

Sample: 5C Score: 3

The response earned 3 points: 1 point in part (a), 1 point in part (b), 1 point in part (c), and no points in part (d). In

part (a) the response presents a difference quotient and would have earned the point for 0cos cos 0e e



Question 5 (continued)

in line 1 with no simplification. In this case, correct simplification to 1e

earned the point. In part (b) the

response earned the first point for cos sinx xf x e x e x in line 1 on the right. The response would have

earned the second point for the slope 3 2 3 23 3 3cos sin 2 2 2

f e e in line 2 on the right with no

simplification. In this case, incorrect simplification to 3 2e did not earn the point. In part (c) the response

earned the first point with cos sin 0.x xf x e x e x Because the response does not identify both

4x and 5

4x as candidates, the second point was not earned. The third point was not earned because an

absolute minimum value with a global argument is not presented. In part (d) the response did not earn the first point because the conditions that g is continuous and that

2 2lim lim 0x x

f x g x

are not verified. The

second point was not earned because there is no limit attached to a ratio of derivatives. A maximum of 1 point can be earned in part (d) for responses with no limit notation attached to a ratio of derivatives. As a result, the response is not eligible for the third point.


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AP Calculus AB Sample Student Responses and Scoring ...€¦ · AP Calculus AB Sample Student Responses and Scoring Commentary from the 2018 Exam Administration: Free-Response Question