Q52 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

The maximum value of $\left\{x \in \mathbf{R} / \sqrt{x+2}>\sqrt{8-x^2}\right\}=$

  • A
    2
  • B
    $\sqrt{2}+1$
  • C
    3
  • Correct answer
    D
    $2 \sqrt{2}$

Solution

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, $x+2>8-x^2 \Rightarrow x^2+x-6>0$ $\Rightarrow(x+3)(x-2)>0 \Rightarrow x \in(-\infty,-3) \cup(2, \infty) \ldots(\text { iii })$ $\therefore \text { According to Eqs. (i), (ii) and (iii) }$ $\Rightarrow x \in[2,2 \sqrt{2}] \therefore \text { Maximum value of } x=2 \sqrt{2}$

Hence, the correct answer is option: 'D'.

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