Question

The greatest value of $f\left(x\right)={\tan }^{-1}x-\frac{1}{2}\log x$ on $[1/\sqrt{3},\sqrt{3}]$ is

• A. $\frac{\pi }{3}+\frac{1}{2}\log 3$
• B. $\frac{\pi }{6}+\frac{1}{4}\log 3$
• C. $\frac{\pi }{6}+\frac{1}{2}\log 3$
• D. $\frac{\pi }{2}+\frac{1}{3}\log 3$

Answer 1

The correct option is B $\frac{\pi }{6}+\frac{1}{4}\log 3$

$f\left(x\right)={\tan }^{-1}x-\frac{1}{2}\log x$

$f^{{}^{\prime }}\left(x\right)=\frac{1}{1+x^2}-\frac{1}{2x}=\frac{-{(x-1)}^2}{2x(1+x^2)}$

since, $f^{{}^{\prime }}\left(x\right)<0$ for $x\in [1/\sqrt{3},\sqrt{3}]$

Therefore, $f\left(x\right)$ is a decreasing function in $[1/\sqrt{3},\sqrt{3}]$

thus $max\left[f\left(x\right)\right]=f\left(\frac{1}{\sqrt{3}}\right)={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{1}{2}\log \left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}+\frac{1}{4}\log 3$

Ans: B