Question

Let $f :(0,1) \rightarrow R$ 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. $f \neq f^{-1}$ on $(0,1)$ and $f'(b)=\frac{1}{f'(0)}$
• C. $f = f ^{-1}$ on $(0,1)$ and $f '( b )=\frac{1}{ f'(0)}$
• D. $f ^{-1}$ is differentiable on $(0,1)$

Answer 1

Answer: (A)

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