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

Option: 1

\mathrm{f}  is not invertible on  \mathrm{(0,1)}


Option: 2

\mathrm{ f \neq f^{-1} on (0,1) and f^{\prime}(b)=\frac{1}{f^{\prime}(0)} }
 


Option: 3

\mathrm{f=f^{-1} on (0,1) and f^{\prime}(b)=\frac{1}{f^{\prime}(0)} }


Option: 4

\mathrm{f^{-1}} is differentiable on (0,1)


Answers (1)

best_answer

\mathrm{f(x)=\frac{b-x}{1-b x} ; \quad f(0,1) \rightarrow R}

Here,  \mathrm{ 0<b<1}

\mathrm{f(x) =\frac{x-b}{b x-1} \\ }

\mathrm{\lim _{x \rightarrow \pm \infty} f(x) =\frac{1}{b} }

From graph, we can conclude that in \mathrm{x \in(0,1) }

\mathrm{ f(x)_{\max }=b \text { at } x=0 }
\mathrm{ f(x)_{\operatorname{man}}=\frac{1-b}{b-1}=-1 \text { at } x=1 }

Hence, Range\mathrm{ (-1, b) \neq } codomain \mathrm{(R) } Hence, \mathrm{ f(x)}  is into, not invertible.
 

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Sayak

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