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The maximum value of $\left\{x \in \mathbf{R} / \sqrt{x+2}>\sqrt{8-x^2}\right\}=$
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Verified Answer
The correct answer is:
$2 \sqrt{2}$
Given $\left\{x \in \mathbf{R}: \sqrt{x+2}>\sqrt{8-x^2}\right\}$
$\therefore$ Defining the function
$x+2 \geq 0 \Rightarrow x \geq-2$ $\ldots$ (i)
and $8-x^2 \geq 0 \Rightarrow x \in[-2 \sqrt{2}, 2 \sqrt{2}]$ $\ldots$ (ii)
Now, $\sqrt{x+2}>\sqrt{8-x^2}$
$\therefore$ Squaring on both sides,
$x+2>8-x^2 \Rightarrow x^2+x-6>0$
$\Rightarrow(x+3)(x-2)>0 \Rightarrow x \in(-\infty,-3) \cup(2, \infty) \ldots(\text { iii })$
$\therefore \text { According to Eqs. (i), (ii) and (iii) }$
$\Rightarrow x \in[2,2 \sqrt{2}] \therefore \text { Maximum value of } x=2 \sqrt{2}$
$\therefore$ Defining the function
$x+2 \geq 0 \Rightarrow x \geq-2$ $\ldots$ (i)
and $8-x^2 \geq 0 \Rightarrow x \in[-2 \sqrt{2}, 2 \sqrt{2}]$ $\ldots$ (ii)
Now, $\sqrt{x+2}>\sqrt{8-x^2}$
$\therefore$ Squaring on both sides,
$x+2>8-x^2 \Rightarrow x^2+x-6>0$
$\Rightarrow(x+3)(x-2)>0 \Rightarrow x \in(-\infty,-3) \cup(2, \infty) \ldots(\text { iii })$
$\therefore \text { According to Eqs. (i), (ii) and (iii) }$
$\Rightarrow x \in[2,2 \sqrt{2}] \therefore \text { Maximum value of } x=2 \sqrt{2}$
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