Domain

Method 1 (The Shorter way)

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Note that the function assumes non-real values upon plugging $\displaystyle x=0$ and thus, $\displaystyle {0}$ is not a point in the domain. The options thus conclude that the domain is $\displaystyle (-∞,-3] \Rightarrow \boxed{a=-3}$

Method 2 (The Longer way)

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• Note that the function is defined only for values $x \text{ such that } x \in (-∞,-3] \cup [8,∞)$ . This comes from considering that:

$\displaystyle { x^2 - 5x - 24 \geq 0 \rightarrow (x+3)(x-8) \geq 0 \rightarrow \boxed {x \in (-∞,-3] \cup [8,∞) } }$

• Moreover, by the domain of the logarithm function, it must also hold that:

$\displaystyle \sqrt{x^2 - 5x - 24} - x-2 > 0 \rightarrow x^2 - 5x - 24 > (x+2)^2$

$\text { Solving which gives } \displaystyle \boxed { x \in (-\infty,-3] }$

Hence, as the solution region is the intersection of the two cases (they must hold simultaneously), Domain is given by $\displaystyle x \in (-∞,-3] \Rightarrow \boxed{a=3}$