Solution to Practice Test IIB
I:
a) Has a solution since
f
(
x
) is continuous,
f
(2
/π
) = 2
/π >
1
/
2 and
f
(3
/
(2
π
)) =

3
/
(2
π
)
<
1
/
2.
b) False
c) False, in fact we always have
(1 +
a
)
1
/n
≤
1 +
a/n .
d) True
e) True
II:
a) 1
/
2, b) 2, c) 0, d) 1
/e
2
, e) 0
III:
First start with some hypothetical considerations.
Solving the equation
c
=
1

(1

c
)
2
the limit will be presumably either 0 or 1.
a) The sequence
a
n
≤
1 and hence is bounded above. The sequence is also bounded
below by zero. Clearly
a
1
= 1
/
2
>
0. Assume that
a
n
≥
0 for some
n
then
a
n
+1
= 1

(1

a
n
)
2
≥
0
since 1
≥
a
n
≥
0.
b) Is the sequence monotone decreasing?
a
n

a
n
+1
=
a
n

1 + (1

a
n
)
2
= (
a
n

1)(2

a
n
)
≤
0
so
a
n
is not decreasing but increasing.
c) Therefor the sequence converges and has a limit which we call
c
. Clearly, since the
function 1

(1

x
)
2
is continuous we have that
c
= 1

(1

c
)
2
which has the solution
c
= 1 since
c >
1
/
2.
0
≤
1
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 Fall '08
 N/A
 Calculus, Practice Test IIB

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