Key Concepts and Formulas

• Function Composition and Simplification: Manipulating and simplifying algebraic expressions, especially those involving exponential functions.
• Differentiation (Quotient Rule): If $h(x) = \frac{u(x)}{v(x)}$, then $h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.
• Monotonicity and One-to-One Functions: A function is increasing if $f'(x) > 0$, decreasing if $f'(x) < 0$. A strictly monotonic function (strictly increasing or strictly decreasing) is one-to-one.

Step-by-Step Solution

Step 1: Define and Express $g(x)$

We are given $f(x) = \frac{1}{1 - e^{-x}}$ and $g(x) = f(-x) - f(x)$. The first step is to find an explicit expression for $g(x)$.

First, we find $f(-x)$ by substituting $-x$ into $f(x)$: $f(-x) = \frac{1}{1 - e^{-(-x)}} = \frac{1}{1 - e^{x}}$

Now, we substitute $f(-x)$ and $f(x)$ into the definition of $g(x)$: $g(x) = f(-x) - f(x) = \frac{1}{1 - e^x} - \frac{1}{1 - e^{-x}}$

Step 2: Simplify $g(x)$

Simplifying $g(x)$ before differentiating makes the process easier and less prone to errors.

We rewrite $e^{-x}$ as $\frac{1}{e^x}$: $g(x) = \frac{1}{1 - e^x} - \frac{1}{1 - \frac{1}{e^x}} = \frac{1}{1 - e^x} - \frac{1}{\frac{e^x - 1}{e^x}} = \frac{1}{1 - e^x} - \frac{e^x}{e^x - 1}$

We can rewrite the second term by factoring out a $-1$ from the denominator: $g(x) = \frac{1}{1 - e^x} - \frac{e^x}{-(1 - e^x)} = \frac{1}{1 - e^x} + \frac{e^x}{1 - e^x}$

Combining the fractions: $g(x) = \frac{1 + e^x}{1 - e^x}$

Step 3: Determine if $g(x)$ is increasing in $(0, 1)$

To check if $g(x)$ is increasing, we need to find its derivative $g'(x)$ and determine its sign in the interval $(0, 1)$. We will use the quotient rule.

Let $u(x) = 1 + e^x$ and $v(x) = 1 - e^x$. Then $u'(x) = e^x$ and $v'(x) = -e^x$. Applying the quotient rule: $g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} = \frac{e^x(1 - e^x) - (1 + e^x)(-e^x)}{(1 - e^x)^2}$ $g'(x) = \frac{e^x - e^{2x} + e^x + e^{2x}}{(1 - e^x)^2} = \frac{2e^x}{(1 - e^x)^2}$

Now, we analyze the sign of $g'(x)$ for $x \in (0, 1)$. Since $e^x > 0$ for all $x$, $2e^x > 0$. Also, $(1 - e^x)^2 > 0$ for $x \in (0, 1)$ because $e^x \neq 1$ in this interval. Therefore, $g'(x) = \frac{2e^x}{(1 - e^x)^2} > 0 \quad \text{for all } x \in (0, 1)$

Since $g'(x) > 0$ for all $x \in (0, 1)$, $g(x)$ is strictly increasing in $(0, 1)$. So, statement (I) is true.

Step 4: Determine if $g(x)$ is one-to-one in $(0, 1)$

A strictly monotonic function is always one-to-one. Since we found that $g(x)$ is strictly increasing in $(0, 1)$, it is also one-to-one in $(0, 1)$. So, statement (II) is true.

Common Mistakes & Tips

• Be careful with signs when applying the quotient rule.
• Always simplify the function before differentiating.
• Remember that a strictly monotonic function is always one-to-one.

Summary

We found an expression for $g(x)$, simplified it, and then found its derivative. By analyzing the sign of the derivative, we determined that $g(x)$ is strictly increasing in $(0, 1)$. Therefore, $g(x)$ is also one-to-one in $(0, 1)$. Both statements (I) and (II) are true.

Final Answer

The final answer is \boxed{A}, which corresponds to option (A).