5 Steps to a 5 AP Calculus BC 2019

AP Calculus BC Practice Exam 2 419

Solutions to BC Practice Exam 2---Section I

Section I Part A

1. 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.

2. 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.


3. The correct answer is (C).
   Sincefis an increasing function,f′>0.
   The only graph that is greater than 0 is
   choice (C).


4. 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.

5. 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).


6. 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

.

7. 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).