Discussion Forum



1)

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


A) f if 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:

Option A

Explanation:

Here , $f(x)=\frac{b-x}{1-bx}$

 where, 0<b<1,0<x<1

For function to be invertible it should
be one-one onto

 $\therefore$ Check range

 Let    f(x)=y $\Rightarrow y=\frac{b-x}{1-bx}$

 $\Rightarrow$  $y-bxy=b-x \Rightarrow x(1-by)=b-y$ 

$ \Rightarrow$   $x= \frac{b-y}{1-by}$

 where , 0<x<1

 $\therefore$ $0 < \frac{b-y}{1-by} <1$

  $\frac{b-y}{1-by} >0$ and $\frac{b-y}{1-by} <1$

 1412202116_g2.PNG

 

 $\Rightarrow$    y<b  or $y > \frac{1}{b}$ .......(i)

  $\frac{(b-1)(y+1)}{1+by}  <-1 < y < \frac{1}{b}$.......(ii)

 From eqs.(i) and (ii) , we get

  $Y \in \left(-1, \frac{1}{b}\right)\subset $ Codomain

 Thus , f(x)  is not invertible