Consider the function `f(x)=|x-2|+|x-5|,c inR.`
Statement 1: `f'(4)=0`
Statement 2: f is continuous in `[2,5],` differentiable in `(2,5),`

Options

A. Statement 1 is false, statement 2 is true.
B. Statement 1 is true , Statement 2 is true, statement 2 is correct explanation for Statement 1.
C. Statement 1 is true, Statement 2 is trur, Statement2 is no a correct explanation for statement 1.
D. Statement 1 is true, Statement 2 is false.

Answer

To solve the problem, we will analyze the function \( f(x) = |x - 2| + |x - 5| \) and determine the validity of the two statements provided.

### Step 1: Define the function in piecewise form
The absolute value function can be expressed in piecewise form based on the critical points at \( x = 2 \) and \( x = 5 \).

1. **For \( x < 2 \)**:
\[ f(x) = (2 - x) + (5 - x) = 7 - 2x \]

2. **For \( 2 \leq x < 5 \)**:
\[ f(x) = (x - 2) + (5 - x) = 3 \]

3. **For \( x \geq 5 \)**:
\[ f(x) = (x - 2) + (x - 5) = 2x - 7 \]

### Step 2: Analyze the function at \( x = 4 \)
Since \( 4 \) lies in the interval \( [2, 5] \), we use the piecewise definition for \( 2 \leq x < 5 \):
\[ f(4) = 3 \]

### Step 3: Find the derivative \( f'(x) \)
Now, we need to find the derivative of \( f(x) \) in the interval \( (2, 5) \):
- In the interval \( (2, 5) \), \( f(x) = 3 \), which is a constant function.
- Therefore, the derivative is:
\[ f'(x) = 0 \quad \text{for } x \in (2, 5) \]

### Step 4: Evaluate \( f'(4) \)
Since \( 4 \) is in the interval \( (2, 5) \):
\[ f'(4) = 0 \]

### Step 5: Check continuity and differentiability
- **Continuity**: The function \( f(x) \) is continuous on the closed interval \([2, 5]\) because it does not have any breaks or jumps in this interval.
- **Differentiability**: The function is differentiable in the open interval \( (2, 5) \) since it is a constant function in this interval.

### Conclusion
- **Statement 1**: \( f'(4) = 0 \) is true.
- **Statement 2**: \( f \) is continuous in \([2, 5]\) and differentiable in \( (2, 5) \) is also true.

Thus, both statements are correct.

### Final Answer
Both statements are true.

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