Question

Let $f\left(x\right)={\cos }^{-1}\left(\frac{2x}{1+x^2}\right)+{\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then the alue of $f\left(1\right)+f\left(2\right)$, is-

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

Answer 1

The correct option is A $-\pi$

$f\left(x\right)={\cos }^{-1}\left(\frac{2x}{1+x^2}\right)+{\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$\therefore f\left(1\right)={\cos }^{-1}\left(1\right)+{\sin }^{-1}\left(0\right)$

$\therefore f\left(2\right)={\cos }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{-3}{5}\right)$

$\therefore f\left(2\right)={\left(\frac{37\pi }{180}\right)}^c+-\pi -{\left(\frac{37\pi }{180}\right)}^c$

$\therefore f\left(1\right)+f\left(2\right)=-\pi$