Let `f(x) =` Degree of `(u^(x^2) + u^2 +2u + 3).` Then, at` x=sqrt2, f(x)` is

Options

A. continuous but not differentiable
B. differentiable
C. dicontinuous
D. none of these

Explanation

To solve the problem, we need to find the function \( f(x) = \text{Degree of } (u^{x^2} + u^2 + 2u + 3) \) at \( x = \sqrt{2} \).

### Step-by-step Solution:

1. **Understanding the Function**:
The function \( f(x) \) represents the degree of the polynomial \( u^{x^2} + u^2 + 2u + 3 \). The degree of a polynomial is the highest power of the variable in that polynomial.

2. **Identifying the Degree**:
- For \( x > \sqrt{2} \):
The term \( u^{x^2} \) will dominate because \( x^2 \) will be greater than 2. Therefore, the degree of the polynomial will be \( x^2 \).
- For \( x \leq \sqrt{2} \):
The term \( u^2 \) will dominate, and the degree will be 2.

3. **Finding the Value at \( x = \sqrt{2} \)**:
Since \( x = \sqrt{2} \) is the boundary point, we need to check the limits from both sides:
- For \( x \to \sqrt{2}^+ \) (approaching from the right), the degree is \( x^2 \).
- For \( x \to \sqrt{2}^- \) (approaching from the left), the degree is 2.

4. **Calculating the Limits**:
- \( \lim_{x \to \sqrt{2}^+} f(x) = \sqrt{2}^2 = 2 \)
- \( \lim_{x \to \sqrt{2}^-} f(x) = 2 \)

5. **Conclusion on Continuity**:
Since both limits are equal and equal to \( f(\sqrt{2}) = 2 \), the function is continuous at \( x = \sqrt{2} \).

6. **Finding the Derivatives**:
- **Left-Hand Derivative (LHD)** at \( x = \sqrt{2} \):
The function is constant (degree 2) for \( x < \sqrt{2} \), so:
\[ \text{LHD} = \frac{d}{dx}(2) = 0 \]
- **Right-Hand Derivative (RHD)** at \( x = \sqrt{2} \):
The degree is \( x^2 \) for \( x > \sqrt{2} \), so:
\[ \text{RHD} = \frac{d}{dx}(x^2) = 2x \quad \text{and at } x = \sqrt{2}, \text{ RHD} = 2\sqrt{2} \]

7. **Conclusion on Differentiability**:
Since LHD (0) is not equal to RHD (\( 2\sqrt{2} \)), the function is not differentiable at \( x = \sqrt{2} \).

### Final Answer:
Thus, \( f(\sqrt{2}) = 2 \).