The function `f (x) =1+ x ln (x+ sqrt(1+ x ^(2)))-sqrt(1- x^(2))` is:

Options

A. strictly increasing `Ax in (0,1)`
B. strictly decrreasing `AA x in (-1,0)`
C. strictly decreasing for `x in (-1,0)`
D. strictly decreasing for `x in (0,1)`

Answer

To determine whether the function \( f(x) = 1 + x \ln(x + \sqrt{1 + x^2}) - \sqrt{1 - x^2} \) is strictly increasing or decreasing, we need to find its derivative \( f'(x) \) and analyze its sign.

### Step 1: Differentiate \( f(x) \)

We start by differentiating the function \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( 1 + x \ln(x + \sqrt{1 + x^2}) - \sqrt{1 - x^2} \right) \]

The derivative of a constant (1) is 0. We will differentiate the other two terms separately.

1. **Differentiate \( x \ln(x + \sqrt{1 + x^2}) \)**:
- Use the product rule: \( u = x \) and \( v = \ln(x + \sqrt{1 + x^2}) \).
- \( u' = 1 \)
- For \( v' \), we apply the chain rule:
\[ v' = \frac{1}{x + \sqrt{1 + x^2}} \cdot \left( 1 + \frac{x}{\sqrt{1 + x^2}} \right) \]
- Therefore, the derivative of this term is:
\[ \frac{d}{dx} \left( x \ln(x + \sqrt{1 + x^2}) \right) = \ln(x + \sqrt{1 + x^2}) + x \cdot \frac{1 + \frac{x}{\sqrt{1 + x^2}}}{x + \sqrt{1 + x^2}} \]

2. **Differentiate \( -\sqrt{1 - x^2} \)**:
- Using the chain rule:
\[ \frac{d}{dx}(-\sqrt{1 - x^2}) = -\frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{x}{\sqrt{1 - x^2}} \]

Putting it all together, we have:

\[ f'(x) = \ln(x + \sqrt{1 + x^2}) + \frac{x(1 + \frac{x}{\sqrt{1 + x^2}})}{x + \sqrt{1 + x^2}} + \frac{x}{\sqrt{1 - x^2}} \]

### Step 2: Analyze the sign of \( f'(x) \)

Next, we analyze the sign of \( f'(x) \):

1. **For \( x < 0 \)**:
- The term \( \ln(x + \sqrt{1 + x^2}) \) is defined and positive since \( x + \sqrt{1 + x^2} > 0 \).
- The other terms involve \( x \) which is negative, thus the entire expression \( f'(x) < 0 \) when \( x < 0 \).

2. **For \( 0 < x < 1 \)**:
- Here, \( \ln(x + \sqrt{1 + x^2}) > 0 \) and the other terms are positive.
- Therefore, \( f'(x) > 0 \) when \( 0 < x < 1 \).

3. **At the endpoints**:
- At \( x = -1 \), \( f'(x) \) approaches a negative value.
- At \( x = 1 \), \( f'(x) \) is also positive.

### Conclusion

From the analysis:
- The function \( f(x) \) is strictly decreasing on the interval \( (-1, 0) \).
- The function \( f(x) \) is strictly increasing on the interval \( (0, 1) \).

Thus, the final answer is:
- **Strictly decreasing for \( x \in (-1, 0) \)**
- **Strictly increasing for \( x \in (0, 1) \)**