Question

If the range for $y=\left({\cot }^{-1}x\right)\left({\cot }^{-1}(-x)\right)$ is
$0<y\le \frac{{\pi }^a}{b}$.
Find the value of $a+b$

• A. $2$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (D)

$6$

Given, $y=\left({\cot }^{-1}x\right)\left({\cot }^{-1}(-x)\right)$
$={\cot }^{-1}\left(x\right)(\pi -{\cot }^{-1}\left(x\right))$
Now ${\cot }^{-1}\left(x\right)$ and $(\pi -{\cot }^{-1}\left(x\right))>0$
Using A.M.$\ge$G.M., we get
$\frac{{\cot }^{-1}x+(\pi -{\cot }^{-1}\left(x\right))}{2}\ge \sqrt{\left({\cot }^{-1}x\right)(\pi -{\cot }^{-1}\left(x\right))}$
$\Rightarrow$ $0<\sqrt{{\cot }^{-1}\left(x\right)(\pi -{\cot }^{-1}\left(x\right))}\le \frac{{\cot }^{-1}x+(\pi -{\cot }^{-1}\left(x\right))}{2}=\frac{\pi }{2}$
$\Rightarrow$ $0<y\le \frac{{\pi }^2}{4}$