Which of the following function is odd ?

Options

A. `ln((1+x^(3))/(1-x^(3)))`
B. `x.(e^(ax)-1)/(a^(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:

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

Let's analyze each option step by step.

### Step 1: Analyze Option 1
Given:
\[ f(x) = \frac{\ln(1 + x^3)}{\ln(1 - x^3)} \]

Now, we need to find \( f(-x) \):
\[ f(-x) = \frac{\ln(1 + (-x)^3)}{\ln(1 - (-x)^3)} = \frac{\ln(1 - x^3)}{\ln(1 + x^3)} \]

Now, we check if \( f(-x) = -f(x) \):
\[ f(-x) = \frac{\ln(1 - x^3)}{\ln(1 + x^3)} = -\frac{\ln(1 + x^3)}{\ln(1 - x^3)} = -f(x) \]

Since this condition holds, **Option 1 is an odd function**.

### Step 2: Analyze Option 2
Given:
\[ f(x) = \frac{e^x - 1}{e^x + 1} \]

Now, we find \( f(-x) \):
\[ f(-x) = \frac{e^{-x} - 1}{e^{-x} + 1} \]

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

Now, we see that \( f(-x) \) does not equal \( -f(x) \). Therefore, **Option 2 is not an odd function**.

### Step 3: Analyze Option 3
Given:
\[ f(x) = \frac{1 + 2k^{-x}}{(2k^{-x})^2} \]

Now, we find \( f(-x) \):
\[ f(-x) = \frac{1 + 2k^{x}}{(2k^{x})^2} \]

We check if \( f(-x) = -f(x) \):
Calculating \( -f(x) \):
\[ -f(x) = -\frac{1 + 2k^{-x}}{(2k^{-x})^2} = \frac{- (1 + 2k^{-x})}{(2k^{-x})^2} \]

Since \( f(-x) \neq -f(x) \), **Option 3 is not an odd function**.

### Step 4: Analyze Option 4
Since we have already determined that Options 1, 2, and 3 have been evaluated, we conclude that **Option 4 is not applicable**.

### Final Conclusion
The only function that satisfies the condition of being odd is **Option 1**.

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