Question

If $f\left(x\right)={\tan }^{-1}x+\frac{1}{2}\ln x$. Then

• A. the greatest value of $f\left(x\right)$ on $[\frac{1}{\sqrt{3}},\sqrt{3}]$ is $\frac{\pi }{3}+\frac{1}{4}\ln 3$
• B. the least value of f(x) on $[\frac{1}{\sqrt{3}},\sqrt{3}]$ is $\frac{\pi }{3}-\frac{1}{4}\ln 3$
• C. $f\left(x\right)$ decreases on $(0,\infty )$
• D. $f\left(x\right)$ increases on $(-\infty ,0)$

Answer 1

The correct option is A the greatest value of $f\left(x\right)$ on $[\frac{1}{\sqrt{3}},\sqrt{3}]$ is $\frac{\pi }{3}+\frac{1}{4}\ln 3$

We know that $\ln x$ and $\tan x$ are both increasing function and Hence their sum is an increasing quantity too.

Therefore in a given domain $(a,b)$ the function will have the greatest value at $b$, therefore the greatest value shall be $f\left(b\right)$,

Using this logic in the above question,

$f\left(x\right)={\tan }^{-1}x+\frac{1}{2}\ln xf\left(\sqrt{3}\right)={\tan }^{-1}\sqrt{3}+\frac{1}{2}\ln \sqrt{3}f\left(\sqrt{3}\right)=\frac{\pi }{3}+\frac{1}{4}\ln 3$