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

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

362 STEP 5. Build Your Test-Taking Confidence

8. The correct answer is (A).
   Note that the notation
   d
   dx
   h(k(x))

∣∣

∣

∣x= 4 is

equivalent to (h◦k)′(4), the derivative of a

composite function, atx=4. Begin with the

Chain Rule:

d

dx

h(k(x))=h′(k(x))k′(x).

Therefore,

d

dx

h(k(x))

∣∣

∣∣

x= 4

=h′(k(4))k′(4)=

h′(4)k′(4)=(1)(3)=3.

9. The correct answer is (A).
   Note that
   f(x)=

∫ 1

x^2

1

2 +t^2

dt=−

∫x 2

1

1

2 +t^2

dt.

Applying the Fundamental Theorem of

Calculus, you have

f′(x)=

− 1

2 +(x^2 )^2

(2x)=

− 2 x

2 +x^4

. Thus,

f′(2)=

−2(2)

2 + 24

=

− 4

18

=

− 2

9

.

10. The correct answer is (C).
    Remember that iff is increasing, then
    f′(x)>0, and that if fis decreasing, then
    f′(x)<0. Examining the graph of f, you see
    thatf(x) is increasing on the interval (−2, 0).
    Therefore,f′(x)>0 on the interval (−2, 0).


11. The correct answer is (C).

The speed of a moving particle decreases when

its velocity and acceleration have opposite

signs, e.g.,v(t)>0 anda(t)<0, orv(t)< 0

anda(t)>0. The velocity and acceleration

functions arev(t)=t^2 − 4 t+3 anda(t)= 2 t−4.

Setv(t)=0, and obtaint^2 − 4 t+ 3 =0or

(t−3)(t−1)=0, which yieldst=1 andt=3.

Settinga(t)=0, you have 2t− 4 =0, ort=2.

0

0

v (t)

+–

123

–+ 0

0

a (t)

––

123

0+ +

Note thatv(t) anda(t) have opposite signs on

(0, 1) and (2, 3). Thus, the speed of the

particle decreases on (0, 1) and (2, 3).

12. The correct answer is (B).
    Note that asxapproaches−∞,

√

x^2 =−x.

Therefore,

xlim→−∞

√

4 x^2 + 3

2 x− 1

=x→lim−∞

√

4 x^2 + 3

√

x^2

2 x− 1

√

x^2

=xlim→−∞

√

4 x^2 + 3

√

x^2

2 x− 1

−x

=xlim→−∞

√

4 +

3

x^2

− 2 +

1

x

.

Also, limx→−∞

3

x^2

=0 and limx→−∞

1

x

=0.

Thus, limx→−∞

√

4 +

3

x^2

− 2 +

1

x

=

√

4

− 2

=−1.

13. The correct answer is (D).
    Note that the formula∫
    1
    1 +x^2
    dx=tan−^1 (x)+c. Therefore,
    ∫ 1

− 1

2

1 +x^2

dx=2 tan−^1 (x)

] 1

− 1

=

2 tan−^1 ( 1 )−2 tan−^1 (− 1 )= 2

(

π

4

)

− 2

(

−

π

4

)

=π.

14. The correct answer is (B).
    Using the Chain Rule, you have
    f′=24 sin^2 (x)cosx. Therefore,
    f′

(

π

3

)

=24 sin^2

(

π

3

)

cos

(

π

3

)

=

24

(√

3

2

) (^2) (
1
2
)
=9.

15. The correct answer is (C).
    The first and second derivatives offare
    f′(x)= 3 x^2 − 6 x+3 and f′′(x)= 6 x−6.
    Setf′(x)=0, and you have 3x^2 − 6 x+ 3 =0.
    Solve forxand obtain 3(x^2 − 2 x+1)=0, or
    3(x−1)(x−1)=0, which yieldsx=1. Setting
    f′′(x)=0, you have 6x− 6 =0, orx=1.