Solution set of the inequations $x^{2} - 4 x \leq 12$ and $x^{2} - 2 x \geq 15$ taken together is

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$(5, 6)$

$[2, 5] .$

$[5, 6]$

$[- 3, 5]$

answer is B.

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Detailed Solution

The given inequations are $x^{2} - 4 x \leq 12 \Rightarrow x^{2} - 4 x - 12 \leq 0$

$\Rightarrow (x - 6) (x + 2) \leq 0 \Rightarrow x \in [- 2, 6]$ ….. ( 1 )

And $x^{2} - 2 x \geq 15 \Rightarrow x^{2} - 2 x - 15 \geq 0$

$\Rightarrow (x - 5) (x + 3) \geq 0$

$\Rightarrow x \leq - 3 o r x \geq 5$

$\Rightarrow x \in (- \infty, - 3] \cup [5, \infty)$ …… ( 2 )

$∴$ Solution set of the inequations is $[- 2, 6] \cap (- \infty, - 3] \cup [5, \infty) = [5, 6]$

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