Which of the following function is odd?

Options

A. In`(1+x^3)/(1-x^3)`
B. `x.(e^(ax)-1)/(e^(ax)+1)`
C. `(1+2^x)^2/2^x`
D. none of these

Answer

To determine which of the given functions is odd, we need to check each function against the definition of an odd function. A function \( f(x) \) is considered odd if it satisfies the condition:

\[ f(-x) = -f(x) \]

Let's analyze the functions step by step.

### Step 1: Analyze the first function

**Function 1:**
\[ f(x) = \frac{\ln(1 + x^3)}{\ln(1 - x^3)} \]

**Finding \( f(-x) \):**
\[ f(-x) = \frac{\ln(1 + (-x)^3)}{\ln(1 - (-x)^3)} = \frac{\ln(1 - x^3)}{\ln(1 + x^3)} \]

**Check if \( f(-x) = -f(x) \):**
To check if \( f(-x) = -f(x) \), we need to see if:
\[ \frac{\ln(1 - x^3)}{\ln(1 + x^3)} = -\frac{\ln(1 + x^3)}{\ln(1 - x^3)} \]

This simplifies to:
\[ \ln(1 - x^3) \cdot \ln(1 - x^3) = -\ln(1 + x^3) \cdot \ln(1 + x^3) \]

This is true, so **Function 1 is an odd function**.

### Step 2: Analyze the second function

**Function 2:**
\[ f(x) = \frac{x e^x - 1}{e^x + 1} \]

**Finding \( f(-x) \):**
\[ f(-x) = \frac{-x e^{-x} - 1}{e^{-x} + 1} \]

**Check if \( f(-x) = -f(x) \):**
We need to check if:
\[ \frac{-x e^{-x} - 1}{e^{-x} + 1} = -\left(\frac{x e^x - 1}{e^x + 1}\right) \]

After simplification, we find that this does not hold true, so **Function 2 is not an odd function**.

### Step 3: Analyze the third function

**Function 3:**
\[ f(x) = \frac{(1 + 2^x)^2}{2^x} \]

**Finding \( f(-x) \):**
\[ f(-x) = \frac{(1 + 2^{-x})^2}{2^{-x}} = (1 + 2^{-x})^2 \cdot 2^x \]

**Check if \( f(-x) = -f(x) \):**
We need to check if:
\[ (1 + 2^{-x})^2 \cdot 2^x = -\frac{(1 + 2^x)^2}{2^x} \]

This also does not hold true, so **Function 3 is not an odd function**.

### Conclusion

After analyzing all three functions, we find that **only Function 1 is an odd function**.

### Final Answer:
The odd function is **Function 1**.

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