AP Calculus BC Practice Exam 2 419
Solutions to BC Practice Exam 2---Section I
Section I Part A
- The correct answer is (B).
Substituting 0 into the numerator and
denominator leads to
0
0
.
ApplyL'Hopital'sRule and obtain
limx→ 0
−2 sinx
2 x
=xlim→ 0 −
sinx
x
. Since limx→ 0
sinx
x
=1,
limx→ 0 −
2 cosx− 2
x^2
=−1. Note that you could
also applyL'Hopital'sRule again to limx→ 0 −
sinx
x
and obtain limx→ 0 −cosx=−1.
- The correct answer is (D).
f(x)=x^3 + 3 x^2 +cx+ 4
⇒f′(x)= 3 x^2 + 6 x+c⇒f′′(x)= 6 x+ 6.
Set 6x+ 6 =0sox=−1.f′′>0if
x>−1 andf′′<0ifx<−1. Thus, fhas a
point of inflection atx=−1, and f′(−1)
=3(−1)^2 +6(−1)+c= 0 ⇒ 3 − 6 +c= 0
⇒− 3 +c= 0 ⇒c=3. - The correct answer is (C).
Sincefis an increasing function,f′>0.
The only graph that is greater than 0 is
choice (C). - The correct answer is (B).
∑∞
k= 1
(
x
2
)k
=
x
2
+
(
x
2
) 2
+
(
x
2
) 3
. This is a
geometric series with a ratio of
x
2
. Thus,
∣∣
∣∣x
2
∣∣
∣∣<1or− 1 <x
2
<1or− 2 <x< 2.
- The correct answer is (C).
Since limx→ 1 +f(x)=xlim→ 1 − f(x)=4, limx→ 1 f(x)
exists. The graph shows that atx=1,f(x)= 1
and thus f(1) exists. Lastly, limx→ 1 f(x)=f(1). - The correct answer is (C).
The area of the region can be obtained as
follows:
Area=blim→∞
∫b
4
1
x^2
dx=blim→∞
∫b
4
x−^2 dx
=blim→∞
[
−
1
x
]b
4
=blim→∞
[
−
1
b
−
(
−
1
4
)]
=blim→∞
[
−
1
b
+
1
4
]
= 0 +
1
4
=
1
4
.
- The correct answer is (A).
Letu= 3 x,du= 3 dx⇒dx=
1
3
du,
x=
a
3
⇒u=a, andx=
b
3
⇒u=b. Then
∫b/ 3
a/ 3
g(3x)dx=
∫b
a
1
3
g(u)du
=
1
3
∫b
a
g(u)du
=
1
3
[G(b)−G(a)]
=
1
3
∫b
a
g(x)dx.
Note thatG(x) is the antiderivative ofg(x).