A) \[f\]is not invertible in (0, 1)
B) \[f\ne {{f}^{-1}}\]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 in (0, 1)
Correct Answer: A
Solution :
| We have, \[f\left( x \right)=\frac{b-x}{1-bx},x\in \left( 0,1 \right)\,\] and \[b\in \left( 0,1 \right)\] |
| Let \[f\left( x \right)=y\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=\frac{b-x}{1-bx}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,y-ybx=b-x\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{b-y}{1-by}\] |
| Range \[f\left( x \right)\ne R\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,f(x)\] is not invertible in (0, 1). |
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