Question

Let $M$ and $m$ respectively be the maximum and minimum values of the function $f\left(x\right)={\tan }^{-1}(\sin x+\cos x)\text{in}[0,\frac{\pi }{2}].$ Then the value of $\tan (M-m)$ is equal to

• A. $2-\sqrt{3}$
• B. $2+\sqrt{3}$
• C. $3+2\sqrt{2}$
• D. $3-2\sqrt{2}$

Answer 1

The correct option is D $3-2\sqrt{2}$

Range of $\sin x+\cos x$ for $x\in [0,\frac{\pi }{2}]$ is $[1,\sqrt{2}]$
So, $M={\tan }^{-1}\sqrt{2}$ and $m={\tan }^{-1}1$
$\Rightarrow M-m={\tan }^{-1}\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)$
$\Rightarrow \tan (M-m)=\frac{\sqrt{2}-1}{\sqrt{2}+1}=3-2\sqrt{2}$