2009 Test 1 solution

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10 Apr 2012
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MAT137Y, 2009–2010 Winter Session, Solutions to Term Test 1
1. Evaluate the following limits. (Do not prove them using the formal definition of limit.)
(10%) (i) lim
x4
1
x1
4
x2x12.
Obtaining a common denominator and factoring the denominator, we get
lim
x4
4x
4x
(x4)(x+3)=lim
x41
4x(x+3)=1
112.
(10%) (ii) lim
x1
xx25x+5
x1.
We multiply top and bottom by both conjugates, giving us
lim
x1
xx25x+5
x1·x+x25x+5
x+x25x+5·x+1
x+1=lim
x1
[x2(x25x+5)](x+1)
(x1)(x+x25x+5)
=lim
x1
5(x1)(x+1)
(x1)(x+x25x+5)
=lim
x1
5(x+1)
x+x25x+5=5.
2.
(10%) (i) Suppose θis in the first quadrant, and cosθ=1
5. Find the exact value of sin 2θ.
We can do this by drawing a triangle. Alternatively, since sin2θ+cos2θ=1, we have
sin2θ=1cos2θ=11
25 =24
25 . Hence, sinθ=24
5=26
5, since sin θ>0 as we are
in the first quadrant.
Therefore, sin2θ=2 sinθcos θ=2·26
5·1
5=46
25 .
(10%) (ii) Consider the function f(x) = |x3|+|x|. Express f(x)as a piecewise function by
eliminating the absolute values, and (without plotting any points) sketch the graph of
f.
To express fas a piecewise function, we eliminate the absolute value first:
|x3|=(x3,x3
3x,x<3,|x|=(x,x0
x,x<0.
Therefore,
if x<0,|x3|+|x|=3x+ (x) = 32x,
if 0 x<3,|x3|+|x|=3x+x=3,
if x3,|x3|+|x|=x3+x=2x3.
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