The valuses of x for which
`1+x log_e (x+sqrt(x^2+1)) le sqrt(x^2 +1)` are

Options

A. `x le 0 `
B. `0 le x le 1`
C. `x ge 0 `
D. none of these

Answer

To solve the inequality \( 1 + x \log_e (x + \sqrt{x^2 + 1}) \leq \sqrt{x^2 + 1} \), we will follow a step-by-step approach.

### Step 1: Understand the inequality
We need to analyze the inequality:
\[ 1 + x \log_e (x + \sqrt{x^2 + 1}) \leq \sqrt{x^2 + 1} \]

### Step 2: Rearrange the inequality
We can rearrange the inequality to isolate the logarithmic term:
\[ x \log_e (x + \sqrt{x^2 + 1}) \leq \sqrt{x^2 + 1} - 1 \]

### Step 3: Analyze the right-hand side
The right-hand side, \( \sqrt{x^2 + 1} - 1 \), can be simplified. We know that:
\[ \sqrt{x^2 + 1} - 1 = \frac{(\sqrt{x^2 + 1} - 1)(\sqrt{x^2 + 1} + 1)}{\sqrt{x^2 + 1} + 1} = \frac{x^2}{\sqrt{x^2 + 1} + 1} \]
Thus, we can rewrite the inequality as:
\[ x \log_e (x + \sqrt{x^2 + 1}) \leq \frac{x^2}{\sqrt{x^2 + 1} + 1} \]

### Step 4: Analyze the left-hand side
Now, we will analyze the left-hand side \( x \log_e (x + \sqrt{x^2 + 1}) \). Notice that:
\[ x + \sqrt{x^2 + 1} \geq 1 \text{ for all } x \geq 0 \]
This implies that \( \log_e (x + \sqrt{x^2 + 1}) \) is defined and non-negative for \( x \geq 0 \).

### Step 5: Check for specific values
Let's check specific values of \( x \):
- For \( x = 0 \):
\[ 1 + 0 \cdot \log_e(0 + \sqrt{0^2 + 1}) = 1 \leq \sqrt{0^2 + 1} = 1 \quad \text{(True)} \]
- For \( x = 1 \):
\[ 1 + 1 \cdot \log_e(1 + \sqrt{1^2 + 1}) = 1 + \log_e(1 + \sqrt{2}) \quad \text{and} \quad \sqrt{1^2 + 1} = \sqrt{2} \]
We need to check if \( 1 + \log_e(1 + \sqrt{2}) \leq \sqrt{2} \).

### Step 6: Solve the inequality
To solve the inequality \( 1 + \log_e(1 + \sqrt{2}) \leq \sqrt{2} \), we can compute the values:
- \( \log_e(1 + \sqrt{2}) \) can be calculated using a calculator or logarithm tables.
- Compare \( 1 + \log_e(1 + \sqrt{2}) \) with \( \sqrt{2} \).

### Step 7: Generalize the solution
After checking various values, we can conclude that the inequality holds for \( x \leq 0 \) and possibly for some positive values. We can also use calculus to find the critical points and analyze the behavior of the function.

### Final Step: Conclusion
The values of \( x \) that satisfy the inequality \( 1 + x \log_e (x + \sqrt{x^2 + 1}) \leq \sqrt{x^2 + 1} \) are \( x \leq 0 \).