Math 347 C1
HOUR EXAM I
30 June 2009
SOLUTIONS
1. Let
a
be a nonzero real number. Show that
a
2
+
1
a
2
≥
2
.
SOLUTION
Assume the opposite:
a
2
+
1
a
2
<
2
.
Then, since
a
2
>
0 we can multiply by
a
2
without changing the direction of the
inequality. Thus
a
4
+ 1
<
2
a
2
,
whence
a
4

2
a
2
+ 1
<
0
.
But
a
4

2
a
2
+ 1 = (
a
2

1)
2
≥
0 for all values of
a
, a contradiction. Hence the
opposite of our assumption is true, that is,
a
2
+
1
a
2
≥
2
.
2. Let
p
(
x
) be a polynomial of degree
r >
0 with real coefficients,
p
(
x
) =
a
0
+
a
1
x
+
a
2
x
2
+
. . .
+
a
r
x
r
with
a
i
∈
R
.
Let
A
be the sum of the coefficients of
p
(
x
) with even subscripts and let
B
be the
sum of the coefficients of
p
(
x
) with odd subscripts. Show that
A
=
1
2
(
p
(1) +
p
(

1))
and
B
=
1
2
(
p
(1)

p
(

1))
.
SOLUTION
p
(1) =
a
0
+
a
1
+
a
2
+
. . .
+
a
r
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and
p
(

1) =
a
0

a
1
+
a
2

. . .
±
a
r
.
Adding these two equations we get
p
(1) +
p
(

1) = 2
a
0
+ 2
a
2
+
. . .
+ 2
a
r
when
r
is even and
p
(1) +
p
(

1) = 2
a
0
+ 2
a
2
+
. . .
+ 2
a
r

1
when
r
is odd.
In each case, the right hand side is just 2
A
and so
A
=
1
2
(
p
(1) +
p
(

1))
.
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 Spring '09
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 Math, Addition, Negative and nonnegative numbers, Natural number

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