AP Calculus BC Practice Exam 2 4315. (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) ory=1
8
(x−2)+ 1.(B) f(2.5)≈1
8
(2. 5 −2)+ 1 = 1. 0625
≈ 1 .063 or17
16
.
(C)
dy
dx=
y
2 x^2
dy
y=
dx
2 x^2and∫
dy
y=
∫
dx
2 x^2ln|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 +CSincee^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)
=e1(^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 forh(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 thegeneral 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 havenlim→∞(∣∣
∣∣(2x)2(n+1)− 1
2(n+1)− 1∣∣
∣∣·∣∣
∣∣^2 n−^1
(2x)^2 n−^1∣∣
∣∣)
< 1or limn→∞∣∣
∣
∣(2x)^2 n+^1
2 n+ 1·
2 n− 1
(2x)^2 n−^1∣∣
∣
∣<1ornlim→∞(
2 n− 1
2 n+ 1)
(2x)^2 <1. Sincenlim→∞(
2 n− 1
2 n+ 1)
=1, (2x)^2 <1, whichimplies− 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)) 55
=
1
160
,
which is less than