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

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

364 STEP 5. Build Your Test-Taking Confidence

21. The correct answer is (B).

x

y

(–1, –1) y = –x^2

0

y = x

Find the intersection points by settingx=−x^2.

Rewritex=−x^2 asx^2 +x=0orx(x+ 1 )= 0

and obtainx=0 andx=−1. The area of the

region bounded by the curves is

∫ 0

− 1

(−x^2 −x)dx=

−x^3

3

−

x^2

2

] 0

− 1

=

0 −

[

−(−1)^3

3

−

(−1)^2

2

]

=

1

6

.

22. The correct answer is (B).

Rewrite

∫

4

e^2 x

dxas 4

∫

e−^2 xdx. Letu=− 2 x

and obtaindu=− 2 dx,ordx=

−du

2

.

Therefore, 4

∫

e−^2 xdx= 4

∫

eu

(

−

du

2

)

=− 2

∫

eudu=− 2 eu+c=− 2 e−^2 x+c=

− 2

e^2 x

+c.

23. The correct answer is (D).

Applying the quotient rule for derivatives, you

havef′(x)=

( 2 )( 2 x− 1 )^2 − 2 ( 2 x− 1 )( 2 )( 2 x)

( 2 x− 1 )^4

=

2 ( 2 x− 1 ) [(( 2 x− 1 )− 4 x)]

( 2 x− 1 )^4

=

2 ( 2 x− 1 )(− 2 x− 1 )

( 2 x− 1 )^4

=

− 2 ( 2 x− 1 )( 2 x+ 1 )

( 2 x− 1 )^4

=

− 2 ( 2 x+ 1 )

( 2 x− 1 )^3

.

24. The correct answer is (C).

Sincefandgare inverse functions, if (a,b)is

on the graph off, then (b,a) is on the graph

ofg, andg′(b)=

1

f′(a)

. Note thatf(4)=2,
which implies thatg(2)=4. Therefore,
g′(2)=

1

f′(4)

=−

1

3

.

25. The correct answer is (B).
    Note that the slopes of the tangents depend on
    they-coordinates and only they-coordinates.
    Also, note that wheny>1, the slopes are
    positive, and wheny<1, the slopes are
    negative. Of the given choices, only the
    differential equation in choice B satisfies these
    conditions.


26. The correct choice is (D).
    Sincef′′(x)>0,f′(x)is increasing, and in
    this case, f′(4.5)<f′(4.6)< f′(4.7). Note
    thatf′( 4. 5 )≈

10. 2 − 10

4. 6 − 4. 5

≈2, and

f′(4.7)≈

10. 8 − 10. 2

4. 7 − 4. 6

≈6. Also note that

slope of tangent=f′(4.6); therefore,

2 <f′(4.6)<6.

(4.7, 10.8)

Slope = 6

Slope = 2

(4.6, 10.2)

(4.5, 10)

Slope

= f ́ (4.5)

f

27. The correct choice is (C).
    Note thatg(− 3 )=

∫− 3

0

f(t)dt

=−

∫ 0

− 3

f(t)dt

=−

(∫− 1

− 3

f(t)dt+

∫ 0

− 1

f(t)dt

)

.