2001 Test 2 solution

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10 Apr 2012
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MAT 137Y, Solutions to Term Test 2, 2001-2002
1. Evaluate the following expressions.
(5%) (i) d
dx sin2
1

x
cosx
x
. (Do NOT simplify your answer.)
2sin
1

x
cosx
x
cos
1

x
cosx
x
x
1
sinx
2
x
cosx

1
x
cosx
x2.
(5%) (ii) lim
x
π
1
cosx
π
x
2.
Let Lbe the limit. Applying L’Hˆopital’s Rule twice, L
lim
x
π
sinx
2
π
x
lim
x
π
cosx
2
1
2
(5%) (iii) lim
x
x2
x2
sinx.
Dividing both sides by x2, we have lim
x
1
1
sinx
x2
. Since sinx
x2
0 by the squeeze theorem, it follows that
the limit is 1.
Note that if you apply L’Hˆopital’s Rule twice, we get lim
x
2
2
sinx, which does not exist. However, you
cannot apply L’Hˆopital’s Rule when the resulting limit does not exist.
(5%) (iv) lim
x
0
sinx
x2
cosx
x
.
Let Lbe the limit. Obtaining a common denominator and applying L’Hˆopital’s Rule once,
L
lim
x
0
sinx
xcosx
x2
lim
x
0
cosx
xsinx
cosx
2x
lim
x
0
sinx
2
0
For the rest of this question, suppose gand hare functions such that
h
0
2
h
0
5
g
0
3
g
0
2
g
2
4
g
2
1
h
1
2
h
1
3
h
2
6
h
2
3
(3%) (v)
g
h
0
g
h
0

h
0
g
2

5
4
5
20
(3%) (vi)
h
g
2
h
g
2

g
2
h
1

4
3
4
12
(15%) 2. Find the equation of the line through the point
3
5
which cuts off the least area from the first quadrant.
Let
x
0
and
0
y
be the points on the line which intersect the point
3
5
. Then the slope mcan be calculated
in two ways, giving us
m
5
y
3
5
3
x

5
y
15
3
x
1

y
5
15
3
x
1
Then the area of the triangle is A
x
1
2xy
1
2x
5
15
3
x
1
5
2x
15
2x
3
x, for x
3
. Differentiating,
A
x
5
2
15
2
3
3
x
2
5
3
x
2
15
2
3
x
2
0

3
x
2
3

x
0 or x
6
1
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