MCQ
$f(x) = \sqrt {2 - 2x - {x^2}} $ નો પ્રદેશ મેળવો.
- A$ - \sqrt 3 \le x \le \sqrt 3 $
- ✓$ - 1 - \sqrt 3 \le x \le - 1 + \sqrt 3 $
- C$ - 2 \le x \le 2$
- D$ - 2 + \sqrt 3 \le x \le - 2 - \sqrt 3 $
If we want the range of $f(x)$ to be real then, $0 \leq 2-2 x-x^{2}$
$\Longrightarrow x^{2}+2 x \leq 2$
$\Longrightarrow x^{2}+2 x+1 \leq 3$
$\Longrightarrow(x+1)^{2} \leq 3$
$\Longrightarrow-\sqrt{3} \leq x+1 \leq \sqrt{3}$
$\Longrightarrow-\sqrt{3}-1 \leq x \leq \sqrt{3}-1$
So the domain of $x$ is $[-1-\sqrt{3},-1+\sqrt{3}]$
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(જ્યાં $i = \sqrt { - 1} $ ).