# sol1-05 - MAT 137Y 2005–2006 Term Test 1 Solutions 1(10(a...

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Unformatted text preview: MAT 137Y, 2005–2006 Term Test 1 Solutions 1. (10%) (a) Solve for all x which satisfies | 2 x | + | x + 1 | = 3 2 . By definition of absolute value, | x | = x , x ≥ ,- x , x < , | x + 1 | = x + 1 , x ≥ - 1 ,- x- 1 , x <- 1 . If x <- 1, then | 2 x | + | x + 1 | = (- 2 x )+(- x- 1 ) =- 3 x- 1 = 3 2 when x =- 5 6 , but this is not a valid solution since we are only looking at values x <- 1. If- 1 ≤ x < 0, then | 2 x | + | x + 1 | = (- 2 x )+( x + 1 ) =- x + 1 = 3 2 = ⇒ x =- 1 2 , which is a valid solution (since x ∈ [- 1 , ) ). If x ≥ 0, then | 2 x | + | x + 1 | = 2 x +( x + 1 ) = 3 x + 1 = 3 2 = ⇒ x = 1 6 , which is also a valid solution. Hence, the equation is satisfied for x =- 1 2 , 1 6 . (10%) (b) Solve the inequality x x + 1 ≤ x- 2 x + 5 and express your answer as a union of intervals. Note that the inequality does not make sense for x =- 1 ,- 5. Moving everything to one side, x x + 1 ≤ x- 2 x + 5 = ⇒ x x + 1- x- 2 x + 5 ≤ = ⇒ x ( x + 5 )- ( x- 2 )( x + 1 ) ( x + 1 )( x + 5 ) ≤ = ⇒ 6 x + 2 ( x + 1 )( x + 5 ) ≤ . We can use a chart of signs to indicate where each of the factors is positive or negative. x <- 5- 5 < x <- 1- 1 < x <- 1 3 x >- 1 3 6 x + 2--- + x + 1-- + + x + 5- + + + 6 x + 2 ( x + 1 )( x + 5 )- +- + So by the chart, we see that the inequality is satisfied for x ∈ (- ∞ ,- 5 ) ∪ (- 1 ,...
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## This note was uploaded on 04/09/2010 for the course MAT 137 taught by Professor Uppal during the Spring '08 term at University of Toronto.

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sol1-05 - MAT 137Y 2005–2006 Term Test 1 Solutions 1(10(a...

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