Invertibility and Derivative Properties of Rational Function

To determine the correct statement, let us analyze the properties of the function $f : (0,1) \rightarrow \mathbb{R}$ defined by:
$f(x) = \frac{b-x}{1-bx}, \quad \text{where } 0 < b < 1.$

1. Injectivity (One-One)

We find the derivative of $f(x)$ with respect to $x$:
$f'(x) = \frac{\frac{d}{dx}(b-x) \cdot (1-bx) - (b-x) \cdot \frac{d}{dx}(1-bx)}{(1-bx)^2}$
$f'(x) = \frac{(-1)(1-bx) - (b-x)(-b)}{(1-bx)^2} = \frac{-1 + bx + b^2 - bx}{(1-bx)^2} = \frac{b^2 - 1}{(1-bx)^2}$

Since $0 < b < 1$, we have $b^2 - 1 < 0$.
Also, for any $x \in (0,1)$, $(1-bx)^2 > 0$.
Thus,
$f'(x) < 0 \quad \forall x \in (0,1).$

Since $f'(x) < 0$ everywhere on $(0,1)$, $f(x)$ is strictly decreasing on $(0,1)$. This means $f$ is an injective (one-one) function.

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2. Surjectivity (Onto)

Since $f(x)$ is continuous and strictly decreasing on $(0,1)$, we evaluate its limits at the boundaries of the domain $(0,1)$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{b-x}{1-bx} = b$
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{b-x}{1-bx} = \frac{b-1}{1-b} = -1$

Thus, the range of $f$ is:
$\text{Range}(f) = (-1, b)$

The given codomain of the function $f : (0,1) \rightarrow \mathbb{R}$ is $\mathbb{R}$.
Since $\text{Range}(f) = (-1, b) \neq \mathbb{R}$, $f$ is not surjective (onto).

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3. Conclusion on Invertibility

A function $f: A \rightarrow B$ is invertible if and only if it is a bijection (both one-one and onto).

Since $f$ is not surjective onto the codomain $\mathbb{R}$, $f$ is not invertible on $(0,1)$.

Consequently:

• Option (A) is correct.
• Options (B), (C), and (D) are incorrect because the inverse function $f^{-1}$ does not exist for $f : (0,1) \rightarrow \mathbb{R}$.

Correct Answer: A