Let f : ( 0 , 1 ) R be defined by f ( x ) = b - x 1 - b x , where b is a constant such that 0 b 1 . Then [2011]

Question

Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x},$ where $b$ is a constant such that $0 < b < 1$ . Then [2011]

Smart Achievers · Accepted Answer

(1, 2)

$Given: f (x) = \frac{b - x}{1 - b x}, 0 < b < 1$

Let $f (x_{1}) = f (x_{2})$ $\Rightarrow \frac{b - x_{1}}{1 - b x_{1}} = \frac{b - x_{2}}{1 - b x_{2}}$

$\Rightarrow b - b^{2} x_{2} - x_{1} + b x_{1} x_{2} = b - x_{2} - b^{2} x_{1} + b x_{1} x_{2}$

$\Rightarrow x_{2} (1 - b^{2}) = x_{1} (1 - b^{2})$ $\Rightarrow x_{1} = x_{2}$ as $1 - b^{2} \neq 0$

$∴$ $f is one-one.$

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

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

For $y = \frac{1}{b},$ $x$ is not defined.

$∴$ $f is not onto and hence not invertible.$

Also, $f' (x) = \frac{- 1 (1 - b x) - (- b) (b - x)}{{(1 - b x)}^{2}} = \frac{b^{2} - 1}{{(1 - b x)}^{2}}$

$∴$ $f' (b) = \frac{1}{b^{2} - 1}$ and $f' (0) = b^{2} - 1$ $\Rightarrow f' (b) = \frac{1}{f' (0)}$

$∴$ $(1) and (2) are the correct options.$

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