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AMS210.01. Quiz 1
Andrei Antonenko
January 24, 2003
1. Solve the following equation:
p
x
2

4 =

y
2
Solution:
The left hand side of the equation is greater or equal to 0, and the right hand side of the
equation is less than or equal to 0. So, the equality may happen only if both sides are equal to 0, so
√
x
2

4 = 0
⇒
x
2

4 = 0
⇒
x
=
±
2, and

y
2
= 0
⇒
y
= 0. So the solution set is
{
(2
,
0);
(

2
,
0)
}
.
2. Let
f
and
g
be functions of 2 variables, and
f
(
x,y
) =
x
+ 4
y
g
(
x,y
) =
xy
Compute the following:
(a)
f
(5
,
4)
Solution:
f
(5
,
4) = 5 + 4
·
4 = 21
(b)
g
(2
,
8)
Solution:
g
(2
,
8) = 2
·
8 = 16
(c)
f
(
g
(2
,
3)
,g
(1
,
5))
Solution:
g
(2
,
3) = 2
·
3 = 6;
g
(1
,
5) = 1
·
5 = 5;
f
(6
,
5) = 6 + 4
·
5 = 26.
(d)
f
(
g
(
t,t
2
)
,g
(
t
2
+ 1
,t
))
Solution:
g
(
t,t
2
) =
t
·
t
2
=
t
3
;
g
(
t
2
+ 1
,t
) = (
t
2
+1)
t
=
t
3
+
t
;
f
(
t
3
,t
3
+
t
) =
t
3
+4(
t
3
+
t
) = 5
t
3
+ 4
t
.
3. Solve the following equation:
2001
x
3
+ 101
x
2
+ 1999
x
= 0
Solution:
2001
x
3
+ 101
x
2
+ 1999
x
= 0
⇔
x
(2001
x
2
+ 101
x
+ 1999) = 0. So, 0 is a solution for this
equation. Now, the equation 2001
x
2
+ 101
x
+ 1999 = 0 doesn’t have any solutions since its discriminant
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 Spring '03
 Andant

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