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If the solution set for $$\displaystyle f\left ( x \right ) < 3$$ is $$\displaystyle \left ( 0,\:\infty  \right )$$ and the solution set for $$\displaystyle f\left ( x \right ) > -2$$ is $$\displaystyle \left ( -\infty ,\: 5  \right )$$, then the true solution set for $$\displaystyle \left ( f\left ( x \right ) \right )^{2} \geq f\left ( x \right ) + 6$$, is


A
(,+)
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B
(,0]
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C
[0,5]
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D
(,0][5,)
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Solution

The correct option is C $$\displaystyle (- \infty, \:0] \cup [5, \: \infty)$$
$$f(x)^{2}-f(x)-6\geq0$$
Upon factorization we get
$$(f(x)-3)(f(x)+2)\geq 0$$ ...(i)
Now we know that $$f(x)<3 $$ includes a solution set $$(0,\infty)$$ and $$f(x)>-2$$ implies a solution set $$(-\infty,5)$$
Hence from one, we get the solution set of the above inequality as
$$(-\infty,0]\cup[5,\infty)$$

Mathematics

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