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TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 12 of 160

Marks: +1, -0

The maximum of {x ∊ R |

\sqrt{x+2}

\sqrt{8-x^2}

}=

2
$\sqrt{2} + 1$
3
$2\sqrt{2}$

Solution:

Given

\{x \in R: \sqrt{x+2} > \sqrt{8-x^2}\}

\therefore

Defining the function

x+2 \ge 0 \Rightarrow x \ge -2

....(i)

and

8-x^2 \ge 0 \Rightarrow x \in [-2\sqrt{2}, 2\sqrt{2}]

...(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)

....(iii)

\therefore

According to Eqs. (i), (ii) and (iii)

\Rightarrow x \in [2, 2\sqrt{2}

]

\therefore

Maximum value of

x = 2\sqrt{2}

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