Question

Let $f( x )= x +\sin x$. Suppose $g$ denotes the inverse function of $f$. The value of $g^{\prime}\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)$ has the value 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)

$f(x)=y=x+\sin x $
$\frac{d y}{d x}=1+\cos x $
$g^{\prime}(y)=\frac{d x}{d y}=\frac{1}{1+\cos x} $
$\text { where } y=\frac{\pi}{4}+\frac{1}{\sqrt{2}}=x+\sin x \Rightarrow x=\frac{\pi}{4}$
$\therefore g ^{\prime}\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)=\frac{1}{1+(1 / \sqrt{2})}=\frac{\sqrt{2}}{\sqrt{2}+1}=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2}$