For all `x in (0,1)`

Options

A. `e^(x) lt 1 +x`
B. `log_(e) (1 +x) lt x`
C. `sin x gt x`
D. `log_(e) x gt x`

Answer

To solve the problem, we need to analyze the given functions and their relationships in the interval \( x \in (0, 1) \).

### Step-by-step Solution:

1. **Identify the Functions**: We are given two functions to compare:
- \( f(x) = e^x \)
- \( g(x) = 1 + x \)

2. **Evaluate at the Endpoints**:
- For \( x = 0 \):
\[ f(0) = e^0 = 1, \quad g(0) = 1 + 0 = 1 \]
- For \( x = 1 \):
\[ f(1) = e^1 = e \approx 2.718, \quad g(1) = 1 + 1 = 2 \]

3. **Check Intermediate Values**: We can check a midpoint, say \( x = \frac{1}{2} \):
- Calculate \( f\left(\frac{1}{2}\right) \):
\[ f\left(\frac{1}{2}\right) = e^{\frac{1}{2}} \approx 1.648 \]
- Calculate \( g\left(\frac{1}{2}\right) \):
\[ g\left(\frac{1}{2}\right) = 1 + \frac{1}{2} = 1.5 \]

4. **Comparison**:
- At \( x = 0 \): \( f(0) = g(0) \)
- At \( x = 1 \): \( f(1) > g(1) \)
- At \( x = \frac{1}{2} \): \( f\left(\frac{1}{2}\right) > g\left(\frac{1}{2}\right) \)

5. **Conclusion for the First Comparison**: Since \( f(x) \) starts at the same point as \( g(x) \) but grows faster, we can conclude that:
\[ e^x > 1 + x \text{ for all } x \in (0, 1) \]

6. **Next Comparison**: We need to analyze \( \ln(1+x) \) and \( x \).
- For \( x = 0 \):
\[ \ln(1+0) = \ln(1) = 0 \]
- For \( x = 1 \):
\[ \ln(1+1) = \ln(2) \approx 0.693 \]

7. **Check Intermediate Values**: Again, check \( x = \frac{1}{2} \):
- Calculate \( \ln(1+\frac{1}{2}) \):
\[ \ln(1.5) \approx 0.405 \]

8. **Comparison**:
- At \( x = 0 \): \( \ln(1) = 0 < 0 \)
- At \( x = 1 \): \( \ln(2) < 1 \)
- At \( x = \frac{1}{2} \): \( \ln(1.5) < \frac{1}{2} \)

9. **Conclusion for the Second Comparison**: Since \( \ln(1+x) < x \) for all \( x \in (0, 1) \).

10. **Final Conclusion**: The only valid comparison is \( \ln(1+x) < x \) and \( e^x > 1 + x \).

### Final Answer:
The correct statement is:
- \( e^x > 1 + x \) for all \( x \in (0, 1) \)
- \( \ln(1+x) < x \) for all \( x \in (0, 1) \)