If f(x) is continuous such that `abs(f(x)) le 1, forall x in R " and " g(x)=(e^(f(x))-e^(-abs(f(x))))/(e^(f(x))+e^(-abs(f(x)))),` then range of g(x) is

Options

A. [0,1]
B. `[0,(e^(2)-1)/(e^(2)+1)]`
C. `[0,(e^(2)+1)/(e^(2)-1)]`
D. `[(1-e^(2))/(1+e^(2)),0]`

Answer

To solve the problem, we need to analyze the function \( g(x) \) given the constraints on \( f(x) \).

### Step-by-Step Solution:

1. **Understanding the function \( g(x) \)**:
The function is defined as:
\[ g(x) = \frac{e^{f(x)} - e^{-\lvert f(x) \rvert}}{e^{f(x)} + e^{-\lvert f(x) \rvert}} \]
Given that \( \lvert f(x) \rvert \leq 1 \), we know that \( f(x) \) can take values in the interval \([-1, 1]\).

2. **Case Analysis**:
We will analyze two cases based on the sign of \( f(x) \):
- Case 1: \( f(x) \geq 0 \)
- Case 2: \( f(x) < 0 \)

3. **Case 1: \( f(x) \geq 0 \)**:
Here, \( \lvert f(x) \rvert = f(x) \). Thus, we can rewrite \( g(x) \):
\[ g(x) = \frac{e^{f(x)} - e^{ - f(x)}}{e^{f(x)} + e^{-f(x)}} \]
This can be simplified using the identity for hyperbolic tangent:
\[ g(x) = \tanh(f(x)) \]
Since \( f(x) \) varies from \( 0 \) to \( 1 \), we find the range of \( g(x) \):
- At \( f(x) = 0 \): \( g(0) = \tanh(0) = 0 \)
- At \( f(x) = 1 \): \( g(1) = \tanh(1) \)

The value of \( \tanh(1) \) can be calculated:
\[ \tanh(1) = \frac{e^1 - e^{-1}}{e^1 + e^{-1}} = \frac{e - \frac{1}{e}}{e + \frac{1}{e}} = \frac{e^2 - 1}{e^2 + 1} \]

4. **Case 2: \( f(x) < 0 \)**:
In this case, \( \lvert f(x) \rvert = -f(x) \). Thus, we can rewrite \( g(x) \):
\[ g(x) = \frac{e^{f(x)} - e^{f(x)}}{e^{f(x)} + e^{-f(x)}} = 0 \]
This means that for any negative value of \( f(x) \), \( g(x) \) will always equal \( 0 \).

5. **Combining Results**:
From both cases, we find:
- When \( f(x) \geq 0 \), \( g(x) \) ranges from \( 0 \) to \( \frac{e^2 - 1}{e^2 + 1} \).
- When \( f(x) < 0 \), \( g(x) = 0 \).

Therefore, the overall range of \( g(x) \) is:
\[ [0, \frac{e^2 - 1}{e^2 + 1}] \]

### Final Answer:
The range of \( g(x) \) is:
\[ [0, \frac{e^2 - 1}{e^2 + 1}] \]