If the solution set for f(x) –2 is (−∞,5). The interval in which x lies, for the equation (f(x))2≥f(x)+6 is/are

( − ∞ , 5 ] , ( − ∞ , 0 ]

f 2 ( x ) − f ( x ) − 6 ≥ 0 ; ( f ( x ) − 3 ) ( f ( x ) + 2 ) ≥ 0 hence solution is ( − ∞ , 0 ] ∪ [ 5 , ∞ ) .

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If the solution set for f(x) < 3 is $(0, \infty)$ and the solution set for f(x) > –2 is $(- \infty, 5)$ . The interval in which x lies, for the equation $(f (x))^{2} \geq f (x) + 6$ is/are

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$[5, \infty)$

$(- \infty, \infty)$

$(- \infty, 5]$

$(- \infty, 0]$

answer is C, D.

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$f^{2} (x) - f (x) - 6 \geq 0; (f (x) - 3) (f (x) + 2) \geq 0$

hence solution is $(- \infty, 0] \cup [5, \infty)$ .

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