The maximum value of x in the set {x in R : s | vedclass

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(D) Given the inequality $\sqrt{x+2} > \sqrt{8-x^2}$.First,we determine the domain of the functions:$x+2 \geq 0 \Rightarrow x \geq -2$ $(i)$$8-x^2 \ge

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$2\sqrt{2}$

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(D) Given the inequality $\sqrt{x+2} > \sqrt{8-x^2}$.First,we determine the domain of the functions:$x+2 \geq 0 \Rightarrow x \geq -2$ $(i)$$8-x^2 \geq 0$ $\Rightarrow x^2 \leq 8$ $\Rightarrow x \in [-2\sqrt{2}, 2\sqrt{2}]$ $(ii)$Combining $(i)$ and $(ii)$,the domain is $x \in [-2, 2\sqrt{2}]$.Now,squaring both sides of the inequality:$x+2 > 8-x^2$$x^2 + x - 6 > 0$$(x+3)(x-2) > 0$This inequality holds for $x \in (-\infty, -3) \cup (2, \infty)$ $(iii)$.Taking the intersection of the domain $[-2, 2\sqrt{2}]$ and the solution $(iii)$:$x \in (2, 2\sqrt{2}]$.The maximum value of $x$ in this interval is $2\sqrt{2}$.

The maximum value of $x$ in the set $\{x \in R : \sqrt{x+2} > \sqrt{8-x^2}\}$ is:

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