390 STEP 5. Build Your Test-Taking Confidence
- The correct answer is (C).
I. f′(0)/=0 since the tangent to f(x)at
x=0 is not parallel to thex-axis.
II. f has an absolute maximum atx=a.
III. f′′is less than 0 on (0,b) since fis
concave downward.
Thus, only statements II and III are true. - The correct answer is (A).
∑∞
n= 1
1
(2n−1)(2n+1)
=
∑∞
n= 1
1
4 n^2 − 1
The sequence of partial sums
{
1
3
,
2
5
,
3
7
,
4
9
,···,
n
2 n+ 1
,···
}
and
=nlim→∞
[
n
2 n+ 1
]
=
1
2
.
Another approach is to use partial fractions to
obtain a telescoping sum.
- The correct answer is (A).
Which of the following series are convergent?
I. 12 − 8 +
16
3
−
32
9
+···=
∑∞
n= 0
12
(
− 2
3
)n
is a geometric series withr=
− 2
3
. Since
|r|<1, the series converges.
II. 5 +
5
√
2
2
+
5
√
3
3
+
5
2
+
√
5 +
5
√
5
6
+···
= 5 +
5
√
2
+
5
√
3
+
5
√
4
+
5
√
5
+
5
√
6
+···
=
∑∞
n= 1
5
√
n
is ap-series, withp=
1
2
. Since
p<1, the series diverges.
III. 8 + 20 + 50 + 125 +···=
∑∞
n= 0
8
(
5
2
)n
is
also a geometric series, but since
r=
5
2
>1, the series diverges.
Therefore, only series I converges.
- The correct answer is (A).
I. f(−1)= 0
II.Sincef is increasing,f′(−1)>0.
III. Sincef is concave upward,f′′(−1)>0.
Thus,f(−1) has the smallest value. - The correct answer is (D).
The velocity vectorv(t)=〈 2 − 3 t^2 ,
πsin(πt)〉=(2− 3 t^2 )i+(πsin(πt))j.
Integrate to find the position.
s(t)=(2t−t^3 )i+
(−π
π
cos(πt)
)
j+C.
Evaluate att=2 to find the constant.
s(2)=(4−8)i+(−1 cos(2π))j+C
= 4 i+ 3 j
s(2)=(−4)i−j+C= 4 i+ 3 j
C= 8 i+ 4 j
Therefore,s(t)=
(
8 + 2 t−t^3
)
i+
(4−cos(πt))j=
〈
8 + 2 t−t^3 ,4−cos(πt)
〉
.
Evaluate att=3.
s(3)=(8+ 6 −27)i+(4−cos(3π))j
s(3)=− 13 i+ 5 j
The position vector is
〈
−13, 5
〉
.
- The correct answer is (C).
Sincedy=∫ 3 e^2 xdx⇒
1 dy=
∫
3 e^2 xdx⇒
y=
3 e^2 x
2
+C.
Atx=0,
5
2
=
3
(
e^0
)
2
+c⇒
5
2
=
3
2
+C
⇒C= 1.
Therefore,y=
3 e^2 x
2