If 2 f(sin x) + f(cos x) = x,then f^{prime}(x | vedclass

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(A) Given the equation: $2 f(\sin x) + f(\cos x) = x$ ...$(i)$Replacing $x$ with $\frac{\pi}{2} - x$,we get:$2 f(\cos x) + f(\sin x) = \frac{\pi}{2} -

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$\frac{1}{\sqrt{1-x^2}}$

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(A) Given the equation: $2 f(\sin x) + f(\cos x) = x$ ...$(i)$Replacing $x$ with $\frac{\pi}{2} - x$,we get:$2 f(\cos x) + f(\sin x) = \frac{\pi}{2} - x$ ...(ii)To eliminate $f(\cos x)$,multiply (ii) by $2$:$4 f(\cos x) + 2 f(\sin x) = \pi - 2x$ ...(iii)Subtracting $(i)$ from (iii):$4 f(\cos x) - f(\cos x) = \pi - 2x - x$$3 f(\cos x) = \pi - 3x$$f(\cos x) = \frac{\pi}{3} - x$Let $t = \cos x$,then $x = \cos^{-1} t$. Substituting this into the equation:$f(t) = \frac{\pi}{3} - \cos^{-1} t$Thus,$f(x) = \frac{\pi}{3} - \cos^{-1} x$.Differentiating with respect to $x$:$f^{\prime}(x) = 0 - \left( -\frac{1}{\sqrt{1-x^2}} \right) = \frac{1}{\sqrt{1-x^2}}$Therefore,the correct option is $A$.

If $2 f(\sin x) + f(\cos x) = x$,then $f^{\prime}(x) = $

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