JEE Advanced 2011 Paper 2Question 1
Let $f:(0,1)\to\mathbb{R}$ be defined by $$f(x)=\frac{b-x}{1-bx},$$ where $b$ is a constant such that $0<b<1$. Then
Functions
Solution
$f^{\prime}(x)=\frac{b^2-1}{(1-bx)^2}<0$, so $f$ is one-one (invertible onto its range $(-1,b)$). $f(f(x))=x$, but $f$ maps $(0,1)$ onto $(-1,b)$, not into $(0,1)$ (e.g. $f(x)<0$ for $x>b$), so $f\neq f^{-1}$ as maps on $(0,1)$ and $f^{-1}$ is not defined on all of $(0,1)$. Also $f^{\prime}(b)=\frac{-1}{1-b^2}=\frac1{f^{\prime}(0)}$ since $f^{\prime}(0)=b^2-1$. So B holds.