MA 3972-MA-Book April 11, 2018 16:20
AP Calculus AB Practice Exam 2 393
- The correct answer is (A).
f(x)=
∣∣
x^3
∣∣
=
{
x^3 if x≥ 0
−x^3 if x< 0
f′(x)=
{
3 x^2 if x≥ 0
− 3 x^2 if x< 0
xlim→− 1 f′(x)=xlim→− 1 (−^3 x^2 )=−^3
- The correct answer is (B).
V=
4
3
πr^3 ;
dV
dt
= 4 πr^2
dr
dt
Since
dV
dt
= 4
dr
dt
⇒ 4 = 4 πr^2 or
r^2 =
1
π
orr=
1
√
π
.
- The correct answer is (D).
dy
dx
=
x^2
y
; ydy=x^2 dx
∫
ydy=
∫
x^2 dx
y^2
2
=
x^3
3
+C.Substituting (0, 4)
42
2
= 0 +C⇒C= 8.
Thus, a solution is
y^2
2
=
x^3
3
+ 8.
- The correct answer is (D).
x = y^2 – 1
x
y
1
A=
∣∣
∣∣
∫ 1
− 1
(
y^2 − 1
)
dy
∣∣
∣∣=
∣∣
∣
∣∣
[
y^3
3
−y
] 1
− 1
∣∣
∣
∣∣
=
∣∣
∣∣
(
1
3
− 1
)
−
(
−
1
3
−(− 1 )
)∣∣
∣∣=^4
3
- The correct answer is (D).
Atx=x 3 , f′=0. Thus, the tangent to the
graph off atx=x 3 is horizontal.
- The correct answer is (D).
The functionfis not differentiable atx=3,
has an absolute maximum atx=3, and has no
point of inflection. Thus, all three statements
are not true.
Section I Part B
- The correct answer is (D).
s(t)= 5 + 4 t−t^2 ;v(t)=s′(t)= 4 − 2 t
[–2, 12] by [–30, 5]
Sincev(t) is a straight line with a negative
slope, the maximum speed for 0≤t≤ 10
occurs att=10 wherev(t)= 4 −2(10)=
−16. Thus, the maximum speed=16.
