Economics 2P30
Foundations of Economic Analysis
Lester M.K. Kwong
Department of Economics
Brock University
Winter 2013
Final Examination  Suggested Solutions
Time: 3 Hours
This examination will not be deposited in the library reserve.
Section A: Definitions
∗ ∗ ∗ ∗ ∗ ∗ ∗
Define
6
of the following
8
terms in two sentences or less.
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
This Section of the exam is worth 10%.
∗ ∗ ∗ ∗ ∗∗
1. Continuous function at some point
x
2. Strictly concave function
3. Homogeneous function of degree
k
4. Mean Value Theorem
5. Complement of a set
X
6. Contrapositive of
P
⇒
R
7. Compact set
8. Diameter of a set
X
Solution
:
1. A function
f
is said to be continuous at some point
x
if
∀
δ
>
0,
∃
✏
>
0 so that if
d
(
x,y
)
<
✏
then
d
(
f
(
x
)
, f
(
y
))
<
δ
.
2. A function
f
is said to be strictly concave if for all
x
≠
y
,
f
(
↵
x
+
(
1
−
↵
)
y
)
>
↵
f
(
x
)
+
(
1
−
↵
)
f
(
y
)
for all
↵
∈
(
0
,
1
)
.
3. A function
f
is said to be homogeneous of degree
k
if for all
λ
>
0 and for all
x
>
0,
f
(
λ
x
)
=
λ
k
f
(
x
)
.
4. If
f
∈
C
1
over
[
a,b
]
then there exists some
c
∈
(
a,b
)
so that:
f
′
(
c
)(
b
−
a
)
=
f
(
b
)
−
f
(
a
)
5. The complement of a set
X
is
X
c
=
{
x
∶
x
∉
X
}
.
1
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6. The contrapositive of
P
⇒
R
is
∼
R
⇒
∼
P
.
7. A set is said to be compact if it is both closed and bounded.
8. The diameter of a set
X
is diam
(
X
)
=
sup
{
d
(
x,y
)
∶
x,y
∈
X
}
.
Section B: Proofs
∗ ∗ ∗ ∗ ∗ ∗ ∗
Choose
6
of the following
8
questions.
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗∗
This section of the exam is worth 40%.
∗ ∗ ∗ ∗ ∗∗
For each of the following, answer true or false. If true, prove. If false, derive a counterexample
1. If
f
∈
C
1
is homogeneous of degree
k
then
f
′
is homogeneous of degree
k
−
1.
Solution
: True. If
f
is homogeneous of degree
k
then:
f
(
λ
x
)
=
λ
k
f
(
x
)
Taking the derivative of the above identity with respect to
x
yields:
f
′
(
λ
x
)
λ
=
λ
k
f
′
(
x
)
⇒
f
′
(
λ
x
)
=
λ
k
−
1
f
′
(
x
)
and hence
f
′
is homogeneous of degree
k
−
1.
2. The function
f
(
x
)
=
20
+
x
is weakly concave and weakly convex over
R
.
Solution
: True.
f
′′
(
x
)
=
0 and so
f
′′
≥
0 and
f
′′
≤
0 for all
x
∈
R
. Hence,
f
is weakly
concave and weakly convex over
R
.
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