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$\begin{array}{l} Let f (x) = {(\sin (\tan^{- 1} x) + \sin (\cot^{- 1} x))}^{2} - 1, | x | > 1 . If \frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (\sin^{- 1} (f (x))) and y (\sqrt{3}) = \frac{π}{6} then \\ y (- \sqrt{3}) is equal to: \end{array}$

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a

-π6

b

5π6

c

π3

d

2π3

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detailed solution

Correct option is A

Let tan-1x=θ θ∈-π2,-π4∪π4,π2 as |x|>1∴f(x)=(sinθ+cosθ)2-1=sin2θ=2x1+x2 Now sin-1f(x)=sin-1sin2θ=-π-2θ,θ∈-π2,π4π-2θ, θ∈π4,π2∴ddxsin-1f(x)=-11+x2,|x|>1 ∴dydx=-11+x2⇒y=C1-tan-1x,x>1C2-tan-1x, x<-1 y(3)=π6⇒C1=π2But we cannot find C2 from given information This question must be BONUS

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$Let g (x) = \{\begin{array}{cc} \frac{x^{2} + x \tan x - x \tan 2 x}{a x + \tan x - \tan 3 x}, & x \neq 0 \\ 0 & , x = 0 \end{array} If g^{1} (0) exists and is equal to non-zero value b then$ $b is equal to$

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