1. If (|x| – 1)/(|x| – 2) ‎≥ 0, x ∈ R, x ‎± 2 then the interval of x is

2. The solution of the -12 < (4 -3x)/(-5) < 2 is

3. If x² = -4 then the value of x is

4. Solve: |x – 3| < 5

5. The graph of the inequations x ≥ 0, y ≥ 0, 3x + 4y ≤ 12 is

6. If |x| < 5 then the value of x lies in the interval

7. Solve: f(x) = {(x - 1)×(2 - x)}/(x - 3) ≥ 0

8. If x² = 4 then the value of x is

9. The solution of the 15 < 3(x - 2)/5 < 0 is

10. Solve: 1 ≤ |x - 1| ≤ 3