The question in the image asks which of the following statements is false, where $x$ is a real number lying in the interval $(0, 1)$, and it provides five options.

Let's evaluate each one:

1. $e^x < e$:

   • Since $x \in (0, 1)$, $e^x$ will always be less than $e$ because exponentiating a number less than 1 results in a value less than $e$. So, this statement is true.

2. $e^{-x} < \frac{1}{e}$:

   • Here $e^{-x} = \frac{1}{e^x}$. Since $x \in (0, 1)$, $e^x$ is between $1$ and $e$, so $\frac{1}{e^x}$ (which is $e^{-x}$) is between $\frac{1}{e}$ and $1$, meaning $e^{-x}$ is greater than $\frac{1}{e}$. Thus, the statement $e^{-x} < \frac{1}{e}$ is false.

3. $\ln(x) < 0$:

   • Since $x \in (0, 1)$, $\ln(x)$ is negative because the logarithm of any number less than 1 is negative. So, this statement is true.

4. $e^{-x} < 1$:

   • Since $e^{-x} = \frac{1}{e^x}$, and $e^x > 1$ for $x \in (0, 1)$, this implies $e^{-x}$ is indeed less than 1. So, this statement is true.

5. $\ln\left( \frac{1}{x} \right) > 0$:

   • Since $x \in (0, 1)$, $\frac{1}{x} > 1$, and the logarithm of a number greater than 1 is positive. So, this statement is true.

Conclusion:

The false statement is $e^{-x} < \frac{1}{e}$, which corresponds to the second option.

Would you like more details or explanations? Here are 5 related questions to expand the topic:

1. What is the behavior of the exponential function $e^x$ when $x$ is negative?
2. How does the logarithmic function behave for values in the interval $(0, 1)$?
3. Why is $e^{-x}$ always greater than $\frac{1}{e}$ for $x \in (0, 1)$?
4. How do exponential and logarithmic functions relate to each other in terms of inverse functions?
5. What are the properties of $\ln(x)$ when $x \to 0^+$?

Tip:

When working with logarithmic inequalities, always keep in mind the domain restrictions to avoid undefined results (e.g., $\ln(x)$ is only defined for $x > 0$).