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Let $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x}$, where $b$ is a constant such that $0 < b < 1$. Then,
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Verified Answer
The correct answers are:
$f$ if not invertible on $(0,1)$
$f$ if not invertible on $(0,1)$
Here, $f(x)=\frac{b-x}{1-b x}$ where, $0 < b < 1,0 < x < 1$ For function to be invertible it should be one-one onto.
$\therefore$ Check range :
Let $\quad f(x)=y \Rightarrow y=\frac{b-x}{1-b x}$
$$
\begin{aligned}
& \Rightarrow y-b x y=b-x \Rightarrow x(1-b y)=b-y \\
& \Rightarrow \quad x=\frac{b-y}{1-b y}
\end{aligned}
$$
where, $0 < x < 1$
$$
\begin{gathered}
\therefore \quad 0 < \frac{b-y}{1-b y} < 1 \\
\frac{b-y}{1-b y}>0 \text { and } \frac{b-y}{1-b y} < 1 \\
\quad+\quad-\quad+ \\
\Rightarrow \quad b \\
y < b \text { or } y>\frac{1}{b} \\
\frac{(b-1)(y+1)}{1-b y} < -1 < y < \frac{1}{b}
\end{gathered}
$$
From Eqs. (i) and (ii), we get $y \in\left(-1, \frac{1}{b}\right) \subset$ Codomain
Thus, $f(x)$ is not invertible.
$\therefore$ Check range :
Let $\quad f(x)=y \Rightarrow y=\frac{b-x}{1-b x}$
$$
\begin{aligned}
& \Rightarrow y-b x y=b-x \Rightarrow x(1-b y)=b-y \\
& \Rightarrow \quad x=\frac{b-y}{1-b y}
\end{aligned}
$$
where, $0 < x < 1$
$$
\begin{gathered}
\therefore \quad 0 < \frac{b-y}{1-b y} < 1 \\
\frac{b-y}{1-b y}>0 \text { and } \frac{b-y}{1-b y} < 1 \\
\quad+\quad-\quad+ \\
\Rightarrow \quad b \\
y < b \text { or } y>\frac{1}{b} \\
\frac{(b-1)(y+1)}{1-b y} < -1 < y < \frac{1}{b}
\end{gathered}
$$
From Eqs. (i) and (ii), we get $y \in\left(-1, \frac{1}{b}\right) \subset$ Codomain
Thus, $f(x)$ is not invertible.
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