2007 Test 1 solution

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10 Apr 2012
School
Department
Course
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MAT 137Y, 2007–2008 Winter Session, Solutions to Term Test 1
1. Evaluate the following limits algebraically.
(10%) (i) lim
x0
39x2
x.
lim
x0
(39x2)(3+9x2)
x(3+9x2)=lim
x0
9(9x2)
x(3+9x2)=lim
x0
x2
x(3+9x2)
=lim
x0
x
3+9x2=0.
(10%) (ii) lim
x4|x26x+8|
x2x12 .
Factoring the numerator, x26x+8= (x4)(x2), so
lim
x4|(x4)(x2)|
(x4)(x+3)=lim
x4|x4|
x4·lim
x4|x2|
x+3
Since x<4, it follows that |x4|=4xand |x2|=x2 (for values x>2), so we have
lim
x4
4x
x4·lim
x4
x2
x+3=1·2
7=2
7.
2. Solve the following inequalities; express your answer as a union of intervals.
(10%) (i) x
x+1>3x.
Moving everything to one side, we have
x
x+13x(x+1)
x+1>0=x3x23x
x+1>0=3x22x
x+1>0=x(3x+2)
x+1<0.
We can now draw a chart of signs to determine where x(3x+2)/(x+1)is positive or negative:
x<11<x<2
32
3<x<0x>0
x +
3x+2− − + +
x+1+ + +
x(3x+2)
x+1++
Therefore the inequality holds when x(,1)(2
3,0).
(10%) (ii) sin2xcosx>0.
This is equivalent to 2sinxcosxcosx=cosx(2sinx1)>0. On the interval [0,2π], we have
cosx=0 when x=π
2,3π
2, and sinx=1
2when x=π
6,5π
6. Again we draw a chart of signs to
determine when the factors are positive or negative:
(0,π
6) (π
6,π
2) (π
2,5π
6) (5π
6,3π
2) (3π
2,2π)
cosx+ + − − +
2sinx1+ + − −
cosx(2sinx1)+ + − −
1
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