Step-by-step Solution

The function is given by $f(x) = \frac{b - x}{1 - bx}$, with domain $D = (0, 1)$ and codomain $R$. The constant $b$ satisfies $0 < b < 1$.

We will analyze each option systematically.

Option A: f is not invertible on (0, 1).

A function is invertible if and only if it is bijective, which means it must be both injective (one-to-one) and surjective (onto).

1. Injectivity (One-to-one): We check if the function is strictly monotonic by examining its derivative, $f'(x)$. Using the quotient rule, $(u/v)' = (u'v - uv')/v^2$: Let $u(x) = b - x$ and $v(x) = 1 - bx$. Then $u'(x) = -1$ and $v'(x) = -b$. $f'(x) = \frac{(-1)(1 - bx) - (b - x)(-b)}{(1 - bx)^2}$ $f'(x) = \frac{-1 + bx - (-b^2 + bx)}{(1 - bx)^2}$ $f'(x) = \frac{-1 + bx + b^2 - bx}{(1 - bx)^2}$ $f'(x) = \frac{b^2 - 1}{(1 - bx)^2}$ For $x \in (0, 1)$ and $b \in (0, 1)$, the term $bx < 1$, so $1 - bx > 0$. Thus, the denominator $(1 - bx)^2$ is always positive. Since $0 < b < 1$, we have $0 < b^2 < 1$, which implies $b^2 - 1 < 0$. Therefore, $f'(x) < 0$ for all $x \in (0, 1)$. This means that $f(x)$ is a strictly decreasing function on its domain. A strictly monotonic function is always injective.

2. Surjectivity (Onto): We need to find the range of $f(x)$ and check if it equals the codomain, $R$. Since $f(x)$ is continuous and strictly decreasing on $(0, 1)$, its range is the interval $(\lim_{x\to1^-} f(x), \lim_{x\to0^+} f(x))$. $\lim_{x\to0^+} f(x) = f(0) = \frac{b - 0}{1 - b(0)} = b$ $\lim_{x\to1^-} f(x) = \frac{b - 1}{1 - b(1)} = \frac{-(1 - b)}{1 - b} = -1$ So, the range of $f$ is $(-1, b)$.

   The codomain is given as $R$. Since the range $(-1, b)$ is a proper subset of $R$ (for example, $b+1 \in R$ but $b+1 \notin (-1,b)$), the function is not surjective.

3. Invertibility Conclusion: Since $f$ is not surjective, it is not bijective. Therefore, $f$ is not invertible. Option A is correct.

Options B, C, and D: Analysis of statements involving $f^{-1}$

Since we have established that $f$ is not invertible, its inverse function $f^{-1}$ does not exist. Any statement about a non-existent object is logically false or ill-defined.

• Options B and C contain statements like $f=f^{-1}$ and $f \ne f^{-1}$. As $f^{-1}$ does not exist, these statements are ill-posed and must be considered false.
• Option D states that $f^{-1}$ is differentiable. This is also false as $f^{-1}$ does not exist.

Further analysis for completeness (checking the composition $f(f(x))$): Sometimes, the notation $f=f^{-1}$ is used as a shorthand for $f(f(x))=x$ (i.e., $f$ is an involution). Let's check this interpretation. For the composition $f(f(x))$ to be defined for all $x$ in the domain of $f$, the range of $f$ must be a subset of the domain of $f$.

• Domain of $f$ is $(0, 1)$.
• Range of $f$ is $(-1, b)$. For the composition to be defined on the entire domain, we would need $(-1, b) \subseteq (0, 1)$. This is not true because the interval $(-1, 0]$ is part of the range but not part of the domain. Let's find for which $x$ values $f(f(x))$ is undefined. $f(x)$ will be in $(-1, 0]$ when: $f(x) \le 0 \implies \frac{b-x}{1-bx} \le 0$. Since $1-bx > 0$, this simplifies to $b-x \le 0$, or $x \ge b$. So for any $x \in [b, 1)$, $f(x) \in (-1, 0]$. Since these values are not in the domain of $f$, $f(f(x))$ is not defined for $x \in [b, 1)$. Therefore, the identity $f(f(x))=x$ does not hold for all $x \in (0,1)$, making the premise of options B and C false even under this alternative interpretation.

Final Conclusion

• Option A is correct because the function is not surjective from $(0,1)$ to $R$.
• Option B is incorrect because $f^{-1}$ does not exist.
• Option C is incorrect because $f^{-1}$ does not exist.
• Option D is incorrect because $f^{-1}$ does not exist.

Thus, the only correct option is A.