Question

If $f\left(x\right)=\tan (\sqrt{x-2}+\sqrt{4-x})$, then the range of $f\left(x\right)$ is

• A. $[\tan \sqrt{2},\tan 2]$
• B. $[\tan 2,\infty )$
• C. $(-\infty ,\tan \sqrt{2}]$
• D. $(-\infty ,\tan 2]\cup [\tan \sqrt{2},\infty )$

Answer 1

The correct option is D $(-\infty ,\tan 2]\cup [\tan \sqrt{2},\infty )$

Clearly the domain of $f$ is: $x\in [2,4]$
Let's find the range of $y=\sqrt{x-2}+\sqrt{4-x}$
$\frac{dy}{dx}=\frac{1}{2\sqrt{x-2}}-\frac{1}{2\sqrt{4-x}}=0$
$\Rightarrow x=3$ (point of maxima)
at $x=2,y=\sqrt{2}$
at $x=3,y=2$
at $x=4,y=\sqrt{2}$
So $y\in [\sqrt{2},2]$
but $f\left(x\right)=\tan \left(y\right),y\in [\sqrt{2},2]$ - $\left\{\pi /2\right\}$

So, the range of $f\left(x\right)$ is $(-\infty ,\tan 2]\cup [\tan \sqrt{2},\infty )$