A) not invertible on (0, 1)
B) \[f\ne {{f}^{-1}}\] on (0, 1) and \[f'(b)=\frac{1}{f'(0)}\]
C) \[f={{f}^{-1}}\] on and \[f'(b)=\frac{1}{f'(0)}\]
D) \[{{f}^{-1}}\] is differentiable on (0, 1)
Correct Answer: A
Solution :
Given, \[f(x)=\frac{b-x}{1-bx}\] \[y=\frac{b-x}{1-bx}\,\,\Rightarrow \,\,x=\frac{b-y}{1-by}\] \[0<x<1\,\,\Rightarrow \,\,0<\frac{b-y}{1-by}<1\,\,\Rightarrow \,\,-1<y<b\]
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