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
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/28/2011 for the course AMS 2010 taught by Professor Andant during the Spring '03 term at SUNY Stony Brook.
 Spring '03
 Andant

Click to edit the document details