Q32 / 60 · JEE Advanced 2011 (Paper 2) · Mathematics
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,
- Correct answerA$f$ if not invertible on $(0,1)$
- B$f \neq f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$
- C$f=f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$
- D$f^{-1}$ is differentiable on $(0,1)$
Solution
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.
Hence, the correct answer is option: 'A'.