Question

Let $f\left(x\right)=x+\sin x$. If $g$ denotes the inverse function of $f$, the value of $g^{{}^{\prime }}\left(\frac{\pi }{4}+\frac{1}{\sqrt{2}}\right)$ is equal to

• A. $\sqrt{2}-1$
• B. $\frac{\sqrt{2}+1}{\sqrt{2}}$
• C. $2-\sqrt{2}$
• D. $\sqrt{2}+1$

Answer 1

Answer: (C)

$2-\sqrt{2}$

Given that
$f\left(x\right)=x+\sin x$
On differentiating we get,
$f^{{}^{\prime }}\left(x\right)=1+\cos x=\frac{dy}{dx}$
But we know $g$ is inverse function of $f$,
$\therefore g^{{}^{\prime }}\left(y\right)=\frac{dx}{dy}=\frac{1}{1+\cos x}$
$y=\frac{\pi }{4}+\frac{1}{\sqrt{2}}=x+\sin x$
$\therefore x=\frac{\pi }{4}$
$\therefore g^{{}^{\prime }}\left(\frac{\pi }{4}+\frac{1}{\sqrt{2}}\right)=\frac{1}{1+\frac{1}{\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{2}+1}=2-\sqrt{2}$