Question

Let $y=(1+x^2){\tan }^{-1}(x-x)$ and $f\left(x\right)=\frac{1}{2x}\frac{dy}{dx},$ then $f\left(x\right)+{\cot }^{-1}x$ is equal to

• A. $0$
• B. $\frac{\pi }{2}$
• C. $-\frac{\pi }{2}$
• D. $\pi$

Answer 1

The correct option is B $\frac{\pi }{2}$

$y=(1+x^2){\tan }^{-1}(x-x)$

$\Rightarrow \frac{dy}{dx}=2x{\tan }^{-1}x+(1+x^2).\frac{1}{1+x^2}-1=2x{\tan }^{-1}x$

$\Rightarrow f\left(x\right)=\frac{\frac{dy}{dx}}{2x}={\tan }^{-1}x$

$\therefore f\left(x\right)+{\cot }^{-1}x={\tan }^{-1}+{\cot }^{-1}x=\frac{\pi }{2}$