AP Calculus BC Practice Exam 2 431
5. (A)
dy
dx
=
y
2 x^2
; (2,1)
dy
dx
∣∣
∣∣
x=2,y= 1
=
1
2(2)^2
=
1
8
Equation of tangent:
y− 1 =
1
8
(x−2) or
y=
1
8
(x−2)+ 1.
(B) f(2.5)≈
1
8
(2. 5 −2)+ 1 = 1. 0625
≈ 1 .063 or
17
16
.
(C)
dy
dx
=
y
2 x^2
dy
y
=
dx
2 x^2
and
∫
dy
y
=
∫
dx
2 x^2
ln|y|=
∫
1
2
x−^2 dx=
1
2
(x−^1 )
(−1)
+C
=−
1
2 x
+C
eln|y|=e
(
− 21 x+C
)
y=e−
21 x+C
; f(2)= 1
1 =e−
2(2)^1 +C
⇒ 1 =e−
1
4 +C
Sincee^0 =1, −
1
4
+C= 0 ⇒C=
1
4
.
Thus, y=e
(
− 21 x+^14
)
.
(D) f(2.5)=e
(
−2(2^1 .5)+^14
)
=e
(
−^15 +^14
)
=e
1
(^20).
- (A) The first four non-zero terms of the
Maclaurin series forf(x) are
x−
x^3
3
+
x^5
5
−
x^7
7
. Sinceg(x)= f′(x), the
first four non-zero ofg(x) are
1 −x^2 +x^4 −x^6.
(B) The Maclaurin series for
h(x)=(2x)−
(2x)^3
3
+
(2x)^5
5
−
(2x)^7
7
...
=(2x)−
8 x^3
3
+
32 x^5
5
−
128 x^7
7
...and the
general term is (−1)n+^1
(2x)^2 n−^1
2 n− 1
.
(C) The ratio test tells you that limn→∞
|an+ 1 |
|an|
< 1
for a series to converge. Thus, you have
nlim→∞
(∣∣
∣∣(2x)
2(n+1)− 1
2(n+1)− 1
∣∣
∣∣·
∣∣
∣∣^2 n−^1
(2x)^2 n−^1
∣∣
∣∣
)
< 1
or limn→∞
∣∣
∣
∣
(2x)^2 n+^1
2 n+ 1
·
2 n− 1
(2x)^2 n−^1
∣∣
∣
∣<1or
nlim→∞
(
2 n− 1
2 n+ 1
)
(2x)^2 <1. Since
nlim→∞
(
2 n− 1
2 n+ 1
)
=1, (2x)^2 <1, which
implies− 1 < 4 x^2 <1or−
1
4
<x^2 <
1
4
.
And sincex^2 ≥0, 0<x^2 <^14 and
−
1
2
<x<
1
2
.
Atx=
1
2
the series is 1−
1
3
+
1
5
−
1
7
+···,
which is also a convergent alternating
series with limn→∞an=0 and
a 1 ≥a 2 ≥a 3 ≥ ···and thus the series
converges. Atx=−
1
2
, the series is
− 1 +
1
3
−
1
5
+
1
7
−···, which is also a
convergent alternating series. Thus, the
interval of convergences is−
1
2
≤x≤
1
2
.
(D) Since the series is a convergent alternating
series,
∣∣
∣∣h
(
1
4
)
−
11
24
∣∣
∣∣<
(
32
(
1
4
)) 5
5
=
1
160
,
which is less than