Practicing Success
If $log_2 x ≥ 0,$ then $log_{1/\pi}\begin{Bmatrix}sin^{-1}\frac{2x}{1+x^2}+2tan^{-1}x\end{Bmatrix}$ is equal |
$log_{1/\pi}(4 tan^{-1} x)$ 0 -1 none of these |
-1 |
We have, $log_2 x ≥ 0 ⇒ x ≥ 2°= 1$, we have For x ≥ 1, we have $sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2 tan^{-1}x.$ $∴ log_{1/\pi}\begin{Bmatrix}sin^{-1}\frac{2x}{1+x^2}+2tan^{-1}x\end{Bmatrix}$ $= log_{1/\pi}\begin{Bmatrix} \pi- 2 tan^{-1} x + 2 tan^{-1} x \end{Bmatrix}$ $= log_{1/\pi} \,\,\pi = - 1. $ |