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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.
 Spring '08
 UPPAL

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