Which of the following is correct ?

Options

A. `In (a+x) lt x AA -a lt x le 0`
B. `In(1+x) lt x AA 0 lt x `
C. `In(1+x) gt x AA x gt 0`
D. `In (1+x) lt x AA x gt -1 `

Answer

To solve the problem, we will analyze each of the given options one by one to determine their validity.

### Step 1: Analyze Option 1
**Option 1:** \( \ln(a + x) < x \) for all \( x \in (-1, 0) \)

1. **Define the function:** Let \( h(x) = \ln(a + x) - x \).
2. **Differentiate:**
\[ h'(x) = \frac{1}{a + x} - 1 \]
3. **Set the inequality:**
\[ h'(x) < 0 \implies \frac{1}{a + x} < 1 \implies 1 < a + x \implies x > -1 - a \]
Since \( x \in (-1, 0) \), this condition holds true.
4. **Determine the behavior of \( h(x) \):** Since \( h'(x) < 0 \), \( h(x) \) is a decreasing function.
5. **Evaluate at \( x = 0 \):**
\[ h(0) = \ln(a) - 0 = \ln(a) \]
6. **Conclusion for Option 1:** Since \( h(x) \) is decreasing, for \( x \in (-1, 0) \), \( h(x) > h(0) \) implies \( \ln(a + x) - x > \ln(a) \) which means \( \ln(a + x) > x \). Therefore, Option 1 is **incorrect**.

### Step 2: Analyze Option 2
**Option 2:** \( \ln(1 + x) < x \) for all \( x > 0 \)

1. **Define the function:** Let \( g(x) = \ln(1 + x) - x \).
2. **Differentiate:**
\[ g'(x) = \frac{1}{1 + x} - 1 \]
3. **Set the inequality:**
\[ g'(x) < 0 \implies \frac{1}{1 + x} < 1 \implies 1 < 1 + x \implies x > 0 \]
This condition holds for \( x > 0 \).
4. **Evaluate at \( x = 0 \):**
\[ g(0) = \ln(1) - 0 = 0 \]
5. **Conclusion for Option 2:** Since \( g(x) \) is decreasing and \( g(0) = 0 \), for \( x > 0 \), \( g(x) < g(0) \) implies \( \ln(1 + x) < x \). Therefore, Option 2 is **correct**.

### Step 3: Analyze Option 3
**Option 3:** \( \ln(1 + x) > x \) for all \( x > 0 \)

From the analysis of Option 2, we found that \( \ln(1 + x) < x \) for all \( x > 0 \). Thus, Option 3 is **incorrect**.

### Step 4: Analyze Option 4
**Option 4:** \( \ln(1 + x) < x \) for all \( x > -1 \)

1. **Revisit \( g(x) \):** We already established that \( g(x) \) is decreasing for \( x > 0 \).
2. **Evaluate behavior for \( x \in (-1, 0) \):**
- For \( x = -1 \), \( g(-1) = \ln(0) - (-1) \) is undefined.
- As \( x \) approaches -1 from the right, \( g(x) \) approaches \( -\infty \).
3. **Conclusion for Option 4:** For \( x \in (-1, 0) \), \( g(x) > 0 \), which implies \( \ln(1 + x) > x \). Therefore, Option 4 is **incorrect**.

### Final Conclusion
The only correct option is **Option 2**: \( \ln(1 + x) < x \) for all \( x > 0 \).

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