AP Calculus BC Practice Exam 2 421- The correct answer is (C).
Whent=4, you havex=4 andy=16, and
thus the point (4, 16) is on the curve. The
slope of the tangent line is
dy
dx
=
dy
dt
dx
dt=
2 t
1
= 2 tand whent=4,
dy
dx
=8. The equation of the
tangent line isy− 16 =8(x−4) ory= 8 x−16.- The correct answer is (C).
Sincef(x) andg(x) are inverse functions,
f(3)=9 implies thatg(9)=3. Also for inverse
functions,f′(a)=
1
g′(f(a)). Thus,
f′(3)=1
g′(9)=
1
(1/6)
=6.
- The correct answer is (C).
Write1
(n+2)(n+3)as partial fractionsA
n+ 2−
B
n+ 3=
A(n+3)+B(n+2)
(n+2)(n+3). Set
A(n+3)+B(n+2)=1. Letn=−2 and obtain
A=1. Similarly, letn=−3 and obtainB=−1.
Thus,1
(n+2)(n+3)=
1
n+ 2−
1
n− 3
and∑∞n= 01
(n+2)(n+3)=
∑∞n= 0(
1
n+ 2−
1
n− 3)
=
(
1
2−
1
3
)
+(
1
3−
1
4
)
+(
1
4−
1
5
)
+···
(
1
n+ 2−
1
n+ 3)
+···Note that limn→∞(
1
2−
1
n+ 3)
=1
2
. Therefore,
∑∞
n= 01
(n+2)(n+3)=
1
2
.
- The correct answer is (C).
First, the graph off(x) is above thex-axis on
the interval [0, 2] thusf(x)≥0, and
∫x0f(t)dt>0 on the interval [0, 2].Secondly,f(x)≤0 on the interval [2, 8] and∫ 82f(x)dx<0, and thus∫ 20f(t)dtis arelative maximum. Note that the area of the
region bounded byf(x) and thex-axis on
[2, 8] is greater than the sum of the areas of the
two regions above thex-axis. Therefore,
∫ 20f(x)dx+∫ 82f(x)dx+
∫ 98f(x)dx<0 and thus∫ 90f(x)<0.Consequently,∫ 20f(x)dxis the absolute
maximum value. Alternatively, you could also
use the first derivative test noting that
g′(x)=f(x).- The correct answer is (D).
The arc length of a curve fromt=atot=bis
L=
∫ba√[
x′(t)^2 +y′(t)^2]
dtand in this casex′(t)=1
2
t−(^12)
andy′(t)=e′, and thus,
L=
∫ba√√
√√[[
1
2
t−(^12)
] 2
+(et)^2
]
dt=
∫ 9
4
√
1
4 t
+e^2 tdt.
- The correct answer is (B).
Asx→∞the denominator 1+ 2 xincreases
much faster than the numerator 3x. Thus,
xlim→∞3 x
1 + 2 x
=0, andy=0 is a horizontal
asymptote. Secondly, asx→−∞,3x
approaches−∞, and (1+ 2 x) approaches 1.
Thus, limx→∞
3 x
1 + 2 x=−∞. Therefore,y=0is
the only horizontal asymptote. Note that you
could also applyL’Hoˆpital’sRule.