The domain of definition of function f(x)=log x2-5x-24 -x-2,is

A
$(- \infty, - 3]$
B
$(- \infty, - 3] \cup [8, \infty)$
C
$(- \infty, \frac{- 28}{9})$
D
None of these

Solution:

Here,
$\sqrt{x^{2} - 5 x - 24} > x + 2$ is equivalent to the collection of two system of inequations.
Case I
$x^{2} - 5 x - 24 \geq 0 and x + 2 < 0 \Rightarrow (x - 8) (x + 3) \geq 0 and x < - 2 \Rightarrow x \leq - 3 . . . . . (i)$
Case II
$x^{2} - 5 x - 24 \geq 0, x + 2 \geq 0 and x^{2} - 5 x - 24 > {(x + 2)}^{2} \Rightarrow x \in (- \infty, - 3] \cup [8, \infty), x \in [- 2, \infty) and x^{2} - 5 x - 24 > x^{2} + 4 x + 4 \Rightarrow 9 x < - 28 \Rightarrow x < \frac{- 28}{9} ∴ No solution i . e ., x \in ϕ . . . . (ii) From Eqs . (i) and (ii), we get x \in (- \infty, - 3]$

The domain of definition of function $f (x) = \log (\sqrt{x^{2} - 5 x - 24} - x - 2),$ is

Solution:

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