Question

The greatest and least values of \(f(x)=\tan ^{-1} x-\frac{1}{2} \ln x\) on \(\left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]\) are
(A) \(f_{\min }=\sqrt{3}-1\)
(B) \(f_{\max }=\pi / 6+\frac{1}{4} \ell \ln 3\)
(C) \(f_{\min }=\pi / 3-\frac{1}{4} \ell \ln 3\)
(D) \(f_{\max }=\pi / 12+\ln 5\)

Answer 1

Ans: \((B, C)\)
Hint \(: f(x)=\tan ^{-1} x-1 / 2 \ln x, x \in[1 / \sqrt{3}, \sqrt{3}]\)
$$
f^{\prime}(x)=\frac{1}{1+x^{2}}-\frac{1}{2 x}=\frac{2 x-\left(1+x^{2}\right)}{2 x\left(1+x^{2}\right)}=\frac{-(x-1)^{2}}{2 x\left(1+x^{2}\right)}
$$
when \(x>0, f^{\prime}(x)<0\)
\(f(x)\) is decreasing function.
$$
\begin{aligned}
&f_{\min }=f(\sqrt{3})=\tan ^{-1}(\sqrt{3})-1 / 2 \ln (\sqrt{3})=\pi / 3-\frac{1}{4} \ln 3 \\
&f_{\max }=f(1 / \sqrt{3})=\pi / 6+1 / 4 \ln 3
\end{aligned}
$$