Consider the function `f(x)={{:(x^(2)-5,xle3),(sqrt(x+13),xgt3):}`
Consider the following statements :
1. The function is discontinuous at x=3.
2. The function is not differentiable at x=0.
What of the above statements is/are correct?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To analyze the given function and the statements, let's break down the solution step by step.

### Step 1: Define the function
The function is defined as:
- \( f(x) = x^2 - 5 \) for \( x \leq 3 \)
- \( f(x) = \sqrt{x + 13} \) for \( x > 3 \)

### Step 2: Check for continuity at \( x = 3 \)
To check if the function is continuous at \( x = 3 \), we need to find the left-hand limit, right-hand limit, and the value of the function at that point.

1. **Left-hand limit as \( x \) approaches 3**:
\[ \lim_{x \to 3^-} f(x) = f(3) = 3^2 - 5 = 9 - 5 = 4 \]

2. **Right-hand limit as \( x \) approaches 3**:
\[ \lim_{x \to 3^+} f(x) = \sqrt{3 + 13} = \sqrt{16} = 4 \]

3. **Value of the function at \( x = 3 \)**:
\[ f(3) = 3^2 - 5 = 4 \]

Since the left-hand limit, right-hand limit, and the value of the function at \( x = 3 \) are all equal (4), the function is continuous at \( x = 3 \).

### Step 3: Check for differentiability at \( x = 0 \)
Next, we need to check if the function is differentiable at \( x = 0 \).

1. **Function value at \( x = 0 \)**:
Since \( 0 \leq 3 \), we use the first part of the function:
\[ f(0) = 0^2 - 5 = -5 \]

2. **Finding the derivative**:
The derivative of \( f(x) = x^2 - 5 \) for \( x < 3 \) is:
\[ f'(x) = 2x \]
Evaluating at \( x = 0 \):
\[ f'(0) = 2 \cdot 0 = 0 \]

Since the derivative exists and is finite at \( x = 0 \), the function is differentiable at that point.

### Conclusion
1. The function is **continuous** at \( x = 3 \).
2. The function is **differentiable** at \( x = 0 \).

Thus, both statements provided in the question are **incorrect**.

### Final Answer
Both statements are incorrect.

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