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

2tan−1⁡(x)=sin−1⁡2x1+x2 if-1≤x≤1π-sin−1⁡2x1+x2 ifx>1-π-sin−1⁡2x1+x2 ifx 1⇒x=tan⁡θ⇒tan⁡θ>1⇒θ=tan−1⁡(x)⇒θ∈45∘,90∘⇒2θ∈900,180∘⇒f(x)=2θ+sin−1⁡[sin⁡2θ]=2θ+sin-1⁡[sin⁡(π−2θ)]=π⇒f(x)=π∴f(5)=π

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

2tan−1⁡(x)=sin−1⁡2x1+x2 if-1≤x≤1π-sin−1⁡2x1+x2 ifx>1-π-sin−1⁡2x1+x2 ifx 1⇒x=tan⁡θ⇒tan⁡θ>1⇒θ=tan−1⁡(x)⇒θ∈45∘,90∘⇒2θ∈900,180∘⇒f(x)=2θ+sin−1⁡[sin⁡2θ]=2θ+sin-1⁡[sin⁡(π−2θ)]=π⇒f(x)=π∴f(5)=π

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

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4tan^–1(5)

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

$\frac{π}{2}$

$π$

answer is B.

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Detailed Solution

$\begin{array}{l} 2 \tan^{- 1} (x) = \{\begin{cases} \sin^{- 1} (\frac{2 x}{1 + x^{2}}) i f - 1 \leq x \leq 1 \\ π - \sin^{- 1} (\frac{2 x}{1 + x^{2}}) i f x > 1 \\ - π - \sin^{- 1} (\frac{2 x}{1 + x^{2}}) i f x < - 1 \end{cases} \\ f (x) = 2 \tan^{- 1} (x) + \sin^{- 1} (\frac{2 x}{1 + x^{2}}), x > 1 \\ \Rightarrow x = \tan θ \Rightarrow \tan θ > 1 \Rightarrow θ = \tan^{- 1} (x) \\ \Rightarrow θ \in (45^{\circ}, 90^{\circ}) \Rightarrow 2 θ \in (90^{0}, 180^{\circ}) \\ \Rightarrow f (x) = 2 θ + \sin^{- 1} [\sin 2 θ] \\ = 2 θ + \sin^{- 1} [\sin (π - 2 θ)] = π \\ \Rightarrow f (x) = π ∴ f (5) = π \end{array}$