WBJEE · Maths · Application of Derivatives
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 \(\mathrm{f}_{\min }=\sqrt{3}-1\)
- B \(\mathrm{f}_{\max }=\pi / 6+\frac{1}{4} \ell \mathrm{n} 3\)
- C \(\mathrm{f}_{\min }=\pi / 3-\frac{1}{4} \ell \mathrm{n} 3\)
- D \(\mathrm{f}_{\max }=\pi / 12+\ell \mathrm{n} 5\)
Answer & Solution
Correct Answer
(C) \(\mathrm{f}_{\min }=\pi / 3-\frac{1}{4} \ell \mathrm{n} 3\)
Step-by-step Solution
Detailed explanation
\(f(x)=\tan ^{-1} x-1 / 2 \ln x, x \in[1 / \sqrt{3}, \sqrt{3}]\) \(f^{\prime}(x)=\frac{1}{1+x^{2}}-\frac{1}{2 x}=\frac{2 x-\left(1+x^{2}\right)}{2 x\left(1+x^{2}\right)}=\frac{-(x-1)^{2}}{2 x\left(1+x^{2}\right)}\) when \(x>0, f^{\prime}(x) < 0\) \(f(x)\) is decreasing function.…
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