MAT 1330 Final: A1_solutions

• School: University of Ottawa
• Department: Mathematics
• Course: MAT 1330
• Professor: Frithjof Lutscher

Document Summary

Case 1: x 2 > 0, which is the case when x > 2. 5 > 3(x 2) which leads to. 3 ) is the solution to this case. Case 2: x 2 < 0, which is the case when x < 2. Then 5 < 3(x 2) leads to the contradiction. 3 < x but we had assumed x < 2. So we have no solution in this case. The solution hence comes from case 1 and is given by x (2, 11. 3 ). (b) |x2 + 2x 12| = 1. Evaluating the absolute value function hence gives x2 + 2x 12 = 1 which leads to two cases: Case 1: x2 + 2x 12 = 1, which gives solutions x1,2 = 1 . Case 2: x2 + 2x 12 = 1 which gives solutions x3,4 = 1 .