5 Steps to a 5 AP Calculus BC 2019

AP Calculus BC Practice Exam 1 389

Solutions to BC Practice Exam 1---Section I

Section I Part A

1. The correct answer is (C).

x→lim(ln 2)+(ex)=eln 2=2 and limx→(ln 2)−(4−ex)

= 4 −eln 2= 4 − 2 = 2

Since the two one-sided limits are the same,

x→lim(ln 2)g(x)=^2.

2. The correct answer is (A).

incr. decr. incr.

f ′ ++ 00 –

ab

f

f ′ decr. incr.

0

concave

downward

concave

upward

f

The only graph that satisfies the behavior off

is (A).

3. The correct answer is (C).
   The definition off′(x)is

f′(x)= lim

Δx→ 0

f(x+Δx)− f(x)

Δx

.

Thus, lim

Δx→ 0

sin((π/3)+Δx)−sin(π/3)

Δx

=

d(sinx)

dx

∣∣

∣

∣x=π/ 3

=cos

(

π

3

)

=

1

2

.

4. The correct answer is (B).
   x=cost⇒

dx

dt

=−sintand

y=sin^2 t⇒

dy

dt

=2 sintcost.

Then,

dy

dx

=

(

dy

dt

)/(

dx

dt

)

=

2 sintcost

−sint

=−2 cost.

Then,

d^2 y

dx^2

=

(

dy′

dt

)/(

dx

dt

)

=

2 sint

−sint

=−2.

Evaluate att=

π

4

for

d^2 y

dx^2

∣∣

∣∣

t=π 4

=− 2

∣∣

t=π 4 =−^2.

5. The correct answer is (D).

Sincef(x)=

∫

xe−x^2 dx, letu=−x^2 ,

du=− 2 xdxor

−du

2

=xdx.

Thus, f(x)=

∫

eu

(

−

du

2

)

=−

1

2

eu+C

=−

1

2

e−x^2 +C

and f(0)= 1 ⇒−

1

2

(e^0 )+C= 1

⇒−

1

2

+C= 1

⇒C=

3

2

.

Therefore, f(x)=−

1

2

e−x

2

+

3

2

and

f(1)=−

1

2

e−^1 +

3

2

=−

1

2 e

+

3

2

.