f(x)=2tan−1⁡x+sin−1⁡2x1+x2,x>1 then f(5) equal to

Given f(x)=2tan−1⁡x+sin−1⁡2x1+x2 ,x>1We know that If x>1 then 2tan-1x=π-sin−1⁡2x1+x2Now f(x)=2tan−1⁡x+π−2tan−1⁡x⇒f(x)=π for all x>1⇒f(5)=π

f(x)=2tan−1⁡x+sin−1⁡2x1+x2,x>1 then f(5) equal to

Given f(x)=2tan−1⁡x+sin−1⁡2x1+x2 ,x>1We know that If x>1 then 2tan-1x=π-sin−1⁡2x1+x2Now f(x)=2tan−1⁡x+π−2tan−1⁡x⇒f(x)=π for all x>1⇒f(5)=π

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$f (x) = 2 \tan^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}}), x > 1 then f (5)$ equal to

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$\tan^{- 1} (\frac{65}{156})$

$4 \tan^{- 1} (5)$

$\frac{π}{2}$

$π$

answer is B.

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Detailed Solution

$\begin{array}{l} Given f (x) = 2 \tan^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}}), x > 1 \\ W e k n o w t h a t I f x > 1 t h e n 2 \tan^{- 1} x = π - \sin^{- 1} (\frac{2 x}{1 + x^{2}}) \\ N o w f (x) = 2 \tan^{- 1} x + π - 2 \tan^{- 1} x \\ \Rightarrow f (x) = π for all x > 1 \\ \Rightarrow f (5) = π \end{array}$