Consider the following statements:
I. `f(x)=|x-3|` is continuous at x=0.
II. `f(x)=|x-3|` is differentiable at x=0.
Which of the statements given above is/ are corrent?

Options

A. I only
B. II only
C. Both I and II
D. Neither I nor II

Explanation

To determine the correctness of the statements regarding the function \( f(x) = |x - 3| \), we will analyze its continuity and differentiability at \( x = 0 \).

### Step 1: Check Continuity at \( x = 0 \)

A function is continuous at a point if the following three conditions are satisfied:
1. \( f(c) \) is defined.
2. \( \lim_{x \to c} f(x) \) exists.
3. \( \lim_{x \to c} f(x) = f(c) \).

Here, \( c = 0 \).

**Calculate \( f(0) \):**
\[ f(0) = |0 - 3| = | -3 | = 3. \]

**Calculate the left-hand limit as \( x \) approaches 0:**
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} |x - 3| = |0 - 3| = 3. \]

**Calculate the right-hand limit as \( x \) approaches 0:**
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} |x - 3| = |0 - 3| = 3. \]

**Check if the limits equal \( f(0) \):**
\[ \lim_{x \to 0^-} f(x) = 3, \quad \lim_{x \to 0^+} f(x) = 3, \quad f(0) = 3. \]

Since all three conditions are satisfied, \( f(x) \) is continuous at \( x = 0 \).

### Step 2: Check Differentiability at \( x = 0 \)

A function is differentiable at a point if the left-hand derivative and the right-hand derivative at that point exist and are equal.

**Calculate the left-hand derivative:**
\[ f'(x) = \frac{d}{dx}(3 - x) \quad \text{for } x < 3. \]
Thus,
\[ f'(x) = -1 \quad \text{for } x < 3. \]
So,
\[ f'(0^-) = -1. \]

**Calculate the right-hand derivative:**
\[ f'(x) = \frac{d}{dx}(3 - x) \quad \text{for } x < 3. \]
Thus,
\[ f'(x) = -1 \quad \text{for } x > 3. \]
So,
\[ f'(0^+) = -1. \]

Since both the left-hand and right-hand derivatives exist and are equal:
\[ f'(0^-) = f'(0^+) = -1. \]
Thus, \( f(x) \) is differentiable at \( x = 0 \).

### Conclusion

Both statements are correct:
1. \( f(x) = |x - 3| \) is continuous at \( x = 0 \) (True).
2. \( f(x) = |x - 3| \) is differentiable at \( x = 0 \) (True).

### Final Answer
Both statements are correct.

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