Question

Let $f:(0,1)\to R$ be defined by $f(x)=\frac{b-x}{1-bx}$, where be is a constant such that $0<b<1$. Then

• A. $f$ is not invertible on $(0,1)$
• B. $f\ne f^{-1}$ on $(0,1)$ and $f^{\prime }(b)=\frac{1}{f^{\prime }(0)}$
• C. $f=f^{-1}$ on $(0,1)$ and $f^{\prime }(b)=\frac{1}{f^{\prime }(0)}$
• D. $f^{-1}$ is differentiable on $(0,1)$

Answer 1

Answer: (A)

$f(x)=\frac{b-x}{1-bx}$
Let $y=\frac{b-x}{1-bx}$
$\Rightarrow x=\frac{b-y}{1-by}$
$0<x<1\Rightarrow 0<\frac{b-y}{1-by}<1$
$\frac{b-y}{1-by}>0\Rightarrow y<b$ or $y>\frac{1}{b}$
$\frac{b-y}{1-by}-1<0\Rightarrow -1<y<\frac{1}{b}$
$\therefore -1<y<b$