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The solution set of the inequality $|3 x| ≥ |6-3 x|$ is: |
$(-\infty, 1]$ $[1, \infty)$ $(-\infty, 1) \cup(1, \infty)$ $(-\infty,-1) \cup(-1, \infty)$ |
$[1, \infty)$ |
The correct answer is Option (2) → $[1, \infty)$ Given inequality $|3x|\ge|6-3x|$ Square both sides $9x^2\ge(6-3x)^2$ $9x^2\ge36-36x+9x^2$ $0\ge36-36x$ $36x\ge36$ $x\ge1$ The solution set is $x\ge1$. |
