AP Calculus BC Practice Exam 1 389
Solutions to BC Practice Exam 1---Section I
Section I Part A
- 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.
- 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).
- 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
.
- 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.
- 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