Question

If the greatest and the least values of $f\left(x\right)={\sin }^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)-\ln x$ in $[\frac{1}{\sqrt{3}},\sqrt{3}]$ are $M$ and $m$ respectively, then

• A. $M+m=\ln 3+\frac{\pi }{6}$
• B. $M-m=\ln 3+\frac{\pi }{6}$
• C. $M+m=\frac{\pi }{2}$
• D. $M-m=\ln 3-\frac{\pi }{3}$

Answer 1

The correct option is C $M+m=\frac{\pi }{2}$

$f\left(x\right)={\sin }^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)-\ln x$
$={\tan }^{-1}x-\ln x$
$f^{\prime }\left(x\right)=\frac{1}{1+x^2}-\frac{1}{x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x-1-x^2}{x(1+x^2)}$
Clearly, denominator $>0$
And for $-x^2+x-1,$ $D=-3<0,$ thus numerator $<0$
$\therefore f^{\prime }\left(x\right)<0$

Maximum value of $f\left(x\right)$ is $f\left(\frac{1}{\sqrt{3}}\right)$
$M={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)-\ln \left(\frac{1}{\sqrt{3}}\right)$
$=\frac{\pi }{6}+\frac{1}{2}\ln 3$

Minimum value of $f\left(x\right)$ is $f\left(\sqrt{3}\right)$
$m={\tan }^{-1}\left(\sqrt{3}\right)-\ln \left(\sqrt{3}\right)$
$=\frac{\pi }{3}-\frac{1}{2}\ln 3$

Hence, $M+m=\frac{\pi }{2}$ and $M-m=\ln 3-\frac{\pi }{6}$