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Let $M$ and $m$ respectively be the maximum and minimum values of the function $f(x)=\tan ^{-1}(\sin x+\cos x)$ in $\left[0, \frac{\pi}{2}\right]$ . Then the value of $\tan (\mathrm{M}-\mathrm{m})$ is equal to:
Option: 1 $2+\sqrt{3}$
Option: 2 $2-\sqrt{3}$
Option: 3 $3-2 \sqrt{2}$
Option: 4 $3+2 \sqrt{2}$

Answers (1)

best_answer

For $x\in \left [ 0,\pi/2 \right ],\\$

$\sin x +\cos x \in \left [ 1,\sqrt{2} \right ] \\$

$\Rightarrow \tan^{-1}\left ( \sin x+\cos x \right )\in\left [ \tan^{-1}1,\tan^{-1}\sqrt{2} \right ]\\$

$\Rightarrow M= \tan^{-1}\sqrt{2}\:\: \: \: m= \tan^{-1}= \frac{\pi}{4}\\$

$M-m= \tan^{-1}\sqrt{2}-\tan^{-1}1= \tan^{-1}\sqrt{2}-\frac{\pi}{4}\\$

$\tan\left ( M-m \right )= \tan\left ( \tan^{-1}\sqrt{2}-\frac{\pi}{4} \right )= \frac{\sqrt{2}-1}{1+\sqrt{2}}\\$

$= \frac{\left ( \sqrt{2}-1 \right )^{2}}{2-1}$

$= 3-2\sqrt{2}\\$

Posted by

Kuldeep Maurya

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