394 STEP 5. Build Your Test-Taking Confidence
- The correct answer is (D).
sinx=x−
x^3
3!
+
x^5
5!
−
x^7
7!
+···
x^2 sinx=x^2
(
x−
x^3
3!
+
x^5
5!
−
x^7
7!
+···
)
=x^3 −
x^5
3!
+
x^7
5!
−
x^9
7!
+···
- The correct answer is (B).
To enclose the area,θmust sweep through the
interval from 0 to 2π. The area of the region
enclosed byr= 3 −sinθis
A=
1
2
∫ 2 π
0
( 3 −sinθ)^2 dθ.
- The correct answer is (D). Using the Ratio
Test,
nlim→∞
∣∣
∣∣an+^1
an
∣∣
∣∣=lim
n→∞
∣
∣∣xn+^1
(n+ 1 )!
∣
∣∣
∣∣
∣∣x
n
n!
∣∣
∣∣
=nlim→∞
∣∣
∣∣ x
n+ 1
(n+ 1 )!·
n!
xn
∣∣
∣∣
=nlim→∞
∣∣
∣
∣
x
n+ 1
∣∣
∣
∣=^0
for all values ofx.
- The correct answer is (B).
g(x)=
∫x
a
f(t)dt⇒g′(x)= f(x)
[]
+0+
incr.
a 0 b
incr.
g′ = f
g
The graph in (B) is the only one that satisfies
the behavior ofg.
- The correct answer is choice A.
Since the series
∑∞
n= 1
(− 1 )n+^1
1
n
satisfies the
hypotheses of the alternating series test,
|S−sn|<an+ 1 , and in this case,an+ 1 =
1
n+ 1
.
Settingan+ 1 ≤ 0 .1, you have
1
n+ 1
≤ 0 .1or
n≥9. The minimum value ofnis 9.
Section I Part B
- The correct answer is (C). Since
v(t)= <t,4t^3 >,a(t)=<1, 12t^2 >and at
t=2,a(t)=<1, 48>.
- The correct answer is (B).
y=e^2 x;
dy
dx
=(e^2 x)2= 2 e^2 x
Set
dy
dx
= 2 ⇒ 2 e^2 x= 2 ⇒e^2 x= 1 ⇒
ln(e^2 x)=ln 1⇒ 2 x=0orx= 0.
Atx=0, y=e^2 x=e2(0)=1; (0, 1) and the
normal line isy=−
1
2
x+ 1.
- The correct answer is (B).
decr. incr.
x 2
–+ 0
concave
downward
point of inflection
concave
upward
f′
f ′′
f
The graph off has a point of inflection at
x=x 2.
