01 Test 1 sol.pdf

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10 Apr 2012
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MAT 137Y Solutions to Term Test 1, 2001-2002
1. Evaluate the following limits.
(5%) (i) lim
x
0
x
9
x
3.
Multiplying numerator and denominator by the conjugate,
lim
x
0
x
9
x
3
lim
x
0
x
9
x
3
9
x
3
9
x
3
lim
x
0
x
9
x
3
9
x
9
lim
x
0
x
9
x
3
x
lim
x
0
9
x
3
6
(5%) (ii) lim
x
1
x1
3
1
x1
2
1.
Let x
u6. Then
lim
x
1
x1
3
1
x1
2
1
lim
u
1
u2
1
u3
1
lim
u
1
u
1

u
1
u
1

u2
u
1
lim
u
1
u
1
u2
u
1
2
3
(5%) (iii) lim
x
1
cos
π
2x
x
1.
Let u
x
1. Then
lim
x
1
cos
π
2x
x
1
lim
u
0
cos
π
2
u
1

u
lim
u
0
cos
π
2u
π
2
u
lim
u
0
cos π
2ucos π
2
sin π
2usin π
2
u
lim
u
0
sin π
2u
u
π
2lim
u
0
sin π
2u
π
2u
π
2
2. Solve the following inequalities.
(8%) (i) x3
x2
1
x3
x2
x
1
1.
Isolating one side, we get
x3
x2
1
x3
x2
x
1
x3
x2
x
1
x3
x2
x
1
0

x
x3
x2
x
1
0

x
x
1
2
x
1
0
Now we consider the sign of each factor:
x

1
1
x
0 0¡x¡1 x
1
x
 
x
1
2
 
x
1
 
x

x
1
2
x
1
 
At x
0 we have equality, and at x
1 the rational function is undefined. Therefore the set of
all solutions which satisfy the inequality are x


1

0
1

1
.
(7%) (ii)
x
1

2
x
3
.
One can take cases on the values of x
1 and x
3, but it is simpler to note that both sides are
positive, so we are allowed to square both sides, giving us
x
1
 
2
x
3
!
x2
2x
1
4
x2
6x
9
"#
x2
2x
1
4x2
24x
36
#
3x2
26x
35
0
1
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