If f(x) = 2 tan^{-1} x + sin^{-1} left( frac{ | vedclass

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(C) Given the function $f(x) = 2 	an^{-1} x + \sin^{-1} \left( \frac{2x}{1 + x^2} ight)$ for $x > 1$. We know that for $x > 1$,the formula for $\si

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$ \pi $

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(C) Given the function $f(x) = 2 	an^{-1} x + \sin^{-1} \left( \frac{2x}{1 + x^2} ight)$ for $x > 1$. We know that for $x > 1$,the formula for $\sin^{-1} \left( \frac{2x}{1 + x^2} ight)$ is $\pi - 2 	an^{-1} x$. Substituting this into the expression for $f(x)$: $f(x) = 2 	an^{-1} x + (\pi - 2 	an^{-1} x)$ $f(x) = \pi$. Since $f(x)$ is a constant function for $x > 1$,$f(5) = \pi$.

If $f(x) = 2 \tan^{-1} x + \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$,where $x > 1$,then $f(5)$ is equal to:

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