Question

Find the minimum value of the function
$f\left(x\right)=\frac{{\pi }^2}{16{\cot }^{-1}(-x)}-{\cot }^{-1}x.$

• A. $\frac{-\pi }{2}$
• B. \N
• C. $\frac{\pi }{2}$
• D. $-\pi$

Answer 1

The correct option is A $\frac{-\pi }{2}$

$f\left(x\right)=\frac{{\pi }^2}{16{\cot }^{-1}(-x)}-\left(\pi -{\cot }^{-1}\left(-x\right)\right)={\cot }^{-1}\left(-x\right)+\frac{{\pi }^2}{16{\cot }^{-1}(-x)}-\pi ={\left(\sqrt{{\cot }^{-1}(-x)}-\frac{\pi }{4\sqrt{{\cot }^{-1}(-x)}}\right)}^2+\frac{\pi }{2}-\pi \ge \frac{-\pi }{2}\therefore f_{\min }=\frac{-\pi }{2}$