Many-One Function

Step 1: Simplify the function $$g(x)$$

Given:

$$f(x) = \frac{e^x}{e^x - 1}$$

We find $$f(-x)$$:

$$f(-x) = \frac{e^{-x}}{e^{-x} - 1} = \frac{\frac{1}{e^x}}{\frac{1}{e^x} - 1} = \frac{1}{1 - e^x}$$

Now, compute $$g(x) = f(x) + f(-x)$$:

$$g(x) = \frac{e^x}{e^x - 1} + \frac{1}{1 - e^x}$$ $$g(x) = \frac{e^x}{e^x - 1} - \frac{1}{e^x - 1}$$ $$g(x) = \frac{e^x - 1}{e^x - 1} = 1$$

Step 2: Evaluate Statement (I)

From the simplification above, we see that $$g(x) = 1$$ for all $$x$$ in the domain where $$e^x \neq 1$$ (which is $$x \neq 0$$).

• A constant function is not strictly decreasing. Its derivative $$g'(x) = 0$$.
• Statement (I) is False.

Step 3: Evaluate Statement (II)

A function is "one-one" (injective) if $$g(x_1) = g(x_2)$$ implies $$x_1 = x_2$$.

• Since $$g(x)$$ is a constant function ($$g(x) = 1$$), any two distinct values $$x_1, x_2$$ will result in $$g(x_1) = g(x_2) = 1$$.
• Therefore, the function is many-to-one, not one-one.
• Statement (II) is False.

Conclusion

Both statements (I) and (II) are incorrect because the function simplifies to a constant.

Correct Option: (3) Neither (I) nor (II) is true