If \(2 f(\sin x)+f(\cos x)=x\), then \(f^{\prime}(x)=\)
If \(2 f(\sin x)+f(\cos x)=x\), then \(f^{\prime}(x)=\)
\(\frac{1}{\sqrt{1-x^2}}\)
\(\frac{-1}{\sqrt{1-x^2}}\)
\(\frac{x}{\sqrt{1-x^2}}\)
\(\frac{-x}{\sqrt{1-x^2}}\)
Solution
It is given that,
\(2 f(\sin x)+f(\cos x)=x\) ...(i)
by replacing \(x\) by \(\frac{\pi}{2}-x\), we get
\(2 f(\cos x)+f(\sin x)=\frac{\pi}{2}-x\) ...(ii)
from Eqs. (i) and (ii), we get
\(3 f(\sin x)=3 x-\frac{\pi}{2}\)
\(\Rightarrow 3 f(x)=3 \sin ^{-1} x-\frac{\pi}{2}\left(\right.\) on replacing \(x\) by \(\left.\sin ^{-1} x\right)\)
Now, on differentiating both sides w.r.t. ' \(x\) ', we get
\(3 f^{\prime}(x)=3 \frac{1}{\sqrt{1-x^2}} \Rightarrow f^{\prime}(x)=\frac{1}{\sqrt{1-x^2}}\)
Hence, option (a) is correct.