For all `x in(0,1)` :

Options

A. `e^xlt1+x`
B. `log_e(1+x)ltx`
C. `sinxgtx`
D. `log_exgtx`

Answer

To solve the problem, we need to analyze the given inequalities for \( x \) in the interval \( (0, 1) \). We will evaluate each option step by step.

### Step 1: Evaluate the first option \( e^x < 1 + x \)

1. **Check at \( x = 0 \)**:
\[ e^0 = 1 \quad \text{and} \quad 1 + 0 = 1 \quad \Rightarrow \quad e^0 = 1 + 0 \]
So, \( e^0 = 1 + 0 \) holds true.

2. **Check at \( x = 1 \)**:
\[ e^1 = e \quad \text{and} \quad 1 + 1 = 2 \quad \Rightarrow \quad e < 2 \quad \text{(This is false since } e \approx 2.718) \]
Therefore, \( e^x < 1 + x \) does not hold for all \( x \in (0, 1) \).

**Conclusion**: The first option is incorrect.

### Step 2: Evaluate the second option \( \log(1 + x) < x \)

1. **Check at \( x = 0 \)**:
\[ \log(1 + 0) = \log(1) = 0 \quad \text{and} \quad 0 < 0 \quad \Rightarrow \quad \text{(This is false)} \]

2. **Check at \( x = 1 \)**:
\[ \log(1 + 1) = \log(2) \quad \text{and} \quad 1 \quad \Rightarrow \quad \log(2) < 1 \quad \text{(This is true since } \log(2) \approx 0.693) \]

3. **Check for \( x \in (0, 1) \)**:
The function \( \log(1 + x) \) is increasing and \( x \) is also increasing, but \( \log(1 + x) \) grows slower than \( x \). Thus, for \( x \in (0, 1) \), \( \log(1 + x) < x \) holds true.

**Conclusion**: The second option is correct.

### Step 3: Evaluate the third option \( \sin(x) > x \)

1. **Check at \( x = 0 \)**:
\[ \sin(0) = 0 \quad \text{and} \quad 0 > 0 \quad \Rightarrow \quad \text{(This is false)} \]

2. **Check at \( x = 1 \)**:
\[ \sin(1) < 1 \quad \Rightarrow \quad \text{(This is true)} \]

3. **Check for \( x \in (0, 1) \)**:
The function \( \sin(x) \) is increasing in \( (0, \frac{\pi}{2}) \) and is less than \( x \) for \( x \in (0, 1) \).

**Conclusion**: The third option is incorrect.

### Step 4: Evaluate the fourth option \( \log(x) > x \)

1. **Check at \( x = 0 \)**:
\[ \log(0) = -\infty \quad \text{and} \quad -\infty > 0 \quad \Rightarrow \quad \text{(This is false)} \]

2. **Check at \( x = 1 \)**:
\[ \log(1) = 0 \quad \text{and} \quad 0 > 1 \quad \Rightarrow \quad \text{(This is false)} \]

3. **Check for \( x \in (0, 1) \)**:
The function \( \log(x) \) is negative for \( x \in (0, 1) \) and thus cannot be greater than \( x \).

**Conclusion**: The fourth option is incorrect.

### Final Conclusion

The only correct option is the second one: \( \log(1 + x) < x \) for all \( x \in (0, 1) \).

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