Let `f:(0,1) rarr IR` be a function defined by `f(x)=1/(1-e(-x))`,and `g(x)=f(-x)-(f(x))` Consider two statements
(I) g is an increasing function in (0, 1)
(II) g is one-one in (0, 1)Then,

Options

A. Only (I) and (II) are true
B. Both (I) and (II) are true
C. Neither (I) nor (II) is true
D. Only (II) is true.

Answer

To solve the problem, we need to analyze the function \( g(x) = f(-x) - f(x) \) where \( f(x) = \frac{1}{1 - e^{-x}} \). We will check the two statements: whether \( g \) is an increasing function in \( (0, 1) \) and whether \( g \) is one-one in \( (0, 1) \).

### Step 1: Define the function \( g(x) \)

We start with the definitions:
\[ f(x) = \frac{1}{1 - e^{-x}} \]
\[ f(-x) = \frac{1}{1 - e^{x}} \]
Thus, we can write:
\[ g(x) = f(-x) - f(x) = \frac{1}{1 - e^{x}} - \frac{1}{1 - e^{-x}} \]

### Step 2: Simplify \( g(x) \)

To simplify \( g(x) \), we find a common denominator:
\[ g(x) = \frac{(1 - e^{-x}) - (1 - e^{x})}{(1 - e^{x})(1 - e^{-x})} \]
This simplifies to:
\[ g(x) = \frac{e^{x} - e^{-x}}{(1 - e^{x})(1 - e^{-x})} \]
Recognizing that \( e^{x} - e^{-x} = 2\sinh(x) \), we can rewrite:
\[ g(x) = \frac{2\sinh(x)}{(1 - e^{x})(1 - e^{-x})} \]

### Step 3: Differentiate \( g(x) \)

To determine if \( g(x) \) is increasing, we need to find \( g'(x) \). We will use the quotient rule:
\[ g'(x) = \frac{(1 - e^{x})(1 - e^{-x}) \cdot (2\cosh(x)) - 2\sinh(x) \cdot \text{derivative of the denominator}}{(1 - e^{x})^2(1 - e^{-x})^2} \]
Calculating the derivative of the denominator involves applying the product rule.

### Step 4: Analyze \( g'(x) \)

After simplification, we find that \( g'(x) \) is positive in the interval \( (0, 1) \). This indicates that \( g(x) \) is an increasing function.

### Step 5: Check if \( g(x) \) is one-one

Since \( g(x) \) is an increasing function, it is also one-one in the interval \( (0, 1) \). A function that is strictly increasing is guaranteed to be one-to-one.

### Conclusion

Both statements are true:
1. \( g \) is an increasing function in \( (0, 1) \).
2. \( g \) is one-one in \( (0, 1) \).

Thus, the final answer is that both statements are true.