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The common solution set of the inequations $x^2-4x \leq 12$ and $x^2-2x \geq 15$ taken together is

If the expression $\left( mx - 1 + \frac{1}{x} \right)$ is non-negative for all positive real numbers $x$,then what must be the minimum value of $m$?

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Statement $(I)$: The set of solutions of $|x|^2 - 4|x| + 3 < 0$ is the interval $(-3, 3)$.
Statement $(II)$: If $x < 3$ or $x > 5$,then $x^2 - 8x + 15 > 0$.
Which of the above statements is (are) true?

The set of all solutions of the inequation $x^2 - 2x + 5 \leq 0$ in $R$ is

For what values of $x$ is $\frac{8x^2 + 16x - 51}{(2x - 3)(x + 4)} > 3$?

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