5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 16:32

AP Calculus AB Practice Exam 1 367

83. The correct answer is (A).

Note the following:

–6

0

f ́

f incr decr decr decr incr decr

+–

–4 –2 0 2 4 6

––0+ 0 –

Also note thatf′(x)=0atx=−4, 2, 4, so the

graph offhas a horizontal tangent at each of

these points. The only graph that satisfies these

conditions is choice (A).

84. The correct answer is (A).

Letxbe the length of the wire used to make a

circle. Therefore, the circumference of the

circle is 2πr=x,orr=

x

2 π

. The area of the

circle isπr^2 =π

(

x

2 π

) 2

. The remaining wire
has a length of( 40 −x)and is made into a
square. Therefore, a side of the square is(
40 −x
4

)

, and the area of the square is

(

40 −x

4

) 2

. LetAbe the total area of the two

figures. Thus,A=π

(

x

2 π

) 2

+

(

40 −x

4

) 2

.

Enter this equation into the graphing calculator

by lettingA=y 1. Examine the graph ofy 1.

ToolsF1 TraceF3

MAIN RAD EXACT

Minimum

xc:17.596 yc:56.0099

FUNC

ZoomF2 ReGraphF4 MathF5 DrawF6 F7Pen

[0, 40] by [0, 140]

Use the [Minimum] function on the calculator,

and obtainx= 17 .596 centimeters.

85. The correct answer is (A).

Remember that

d

dx

arccos(x)=

− 1

√

1 −x^2

.

Using the chain rule,

f′(x)=

√ −^1

1 −(2x)^2

· 2 =

√−^2

1 − 4 x^2

, and

f′(0)=−2. Therefore, the slope of the tangent

to the graph off atx=0is−2. Note that you

could also find f(x) and f′(0) using your

graphing calculator.

86. The correct answer is (B).
    ToolsF1 TraceF3

MAIN RAD EACT FUNC

ZoomF2 ReGraphF4 MathF5 DrawF6 F7Pen

[−3, 3] by [−3, 5]

Using the TI-89 graphing calculator, graph the

two given equations. Find their intersection

points in the second quadrant:x=− 1. 5321

andx=− 0 .3473. Find the volume of the solid

of revolution:

V=π

∫− 0. 3473

− 1. 5321

((x^3 − 2 x+1)^2 −(x+2)^2 )dx

≈ 2. 2405 π≈ 7. 039

87. The correct answer is (A).
    Sincef(x)=x^3 +1,f′(x)= 3 x^2 , and
    f′(c)= 3 (c)^2. The average rate of change off
    on[1, 3]is
    f( 3 )−f( 1 )
    3 − 1

=

28 − 2

2

=13. Set

3 c^2 =13, and obtainc=±

√

13

3

≈± 2 .082.

Note that only

√

13

3

is on the interval[1, 3].

Thus,c=

√

13

3

.

ToolsF1 TraceF3

MAIN RAD EXACT FUNC

X =^133

ZoomF2 ReGraphF4 MathF5 DrawF6 F7Pen

[−1, 4] by [−10, 35]

See the figure above for a geometric

representation of the problem.