Let `f(x)={{:(int_(0)^(x)|1-t|dt",",x gt1),(x-(1)/(2)",",xle1):}` Then

Options

A. f(x) is continuous at x = 1
B. f(x) is not continuous at x = 1
C. f(x) is differentiable at x = 1
D. f(x) is not differentiable at x = 1

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} \int_{0}^{x} |1-t| dt & \text{if } x > 1 \\ x - \frac{1}{2} & \text{if } x \leq 1 \end{cases} \]

### Step 1: Evaluate the integral for \( x > 1 \)

For \( x > 1 \), we need to evaluate the integral \( \int_{0}^{x} |1-t| dt \). The absolute value \( |1-t| \) changes its expression depending on whether \( t \) is less than or greater than 1.

- For \( t \in [0, 1] \), \( |1-t| = 1-t \)
- For \( t \in [1, x] \), \( |1-t| = t-1 \)

Thus, we can split the integral:

\[ \int_{0}^{x} |1-t| dt = \int_{0}^{1} (1-t) dt + \int_{1}^{x} (t-1) dt \]

### Step 2: Calculate the first integral

Calculate \( \int_{0}^{1} (1-t) dt \):

\[ \int_{0}^{1} (1-t) dt = \left[ t - \frac{t^2}{2} \right]_{0}^{1} = \left( 1 - \frac{1}{2} \right) - (0) = \frac{1}{2} \]

### Step 3: Calculate the second integral

Now calculate \( \int_{1}^{x} (t-1) dt \):

\[ \int_{1}^{x} (t-1) dt = \left[ \frac{t^2}{2} - t \right]_{1}^{x} = \left( \frac{x^2}{2} - x \right) - \left( \frac{1^2}{2} - 1 \right) = \left( \frac{x^2}{2} - x \right) - \left( \frac{1}{2} - 1 \right) \]

This simplifies to:

\[ \frac{x^2}{2} - x + \frac{1}{2} \]

### Step 4: Combine the results

Combining both integrals, we have:

\[ f(x) = \frac{1}{2} + \left( \frac{x^2}{2} - x + \frac{1}{2} \right) = \frac{x^2}{2} - x + 1 \]

Thus, for \( x > 1 \):

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

### Step 5: Define \( f(x) \) for \( x \leq 1 \)

For \( x \leq 1 \), we have:

\[ f(x) = x - \frac{1}{2} \]

### Step 6: Check continuity at \( x = 1 \)

To check continuity at \( x = 1 \), we need to find:

- \( f(1) \)
- Left-hand limit \( \lim_{x \to 1^-} f(x) \)
- Right-hand limit \( \lim_{x \to 1^+} f(x) \)

Calculating these:

1. \( f(1) = 1 - \frac{1}{2} = \frac{1}{2} \)
2. Left-hand limit:
\[ \lim_{x \to 1^-} f(x) = 1 - \frac{1}{2} = \frac{1}{2} \]
3. Right-hand limit:
\[ \lim_{x \to 1^+} f(x) = \frac{1^2}{2} - 1 + 1 = \frac{1}{2} \]

Since all three values are equal, \( f(x) \) is continuous at \( x = 1 \).

### Step 7: Check differentiability at \( x = 1 \)

To check differentiability, we need to find the left-hand derivative and right-hand derivative at \( x = 1 \):

1. Left-hand derivative:
\[ f'(x) = 1 \quad \text{for } x \leq 1 \]
So, \( \lim_{x \to 1^-} f'(x) = 1 \).

2. Right-hand derivative:
\[ f'(x) = x \quad \text{for } x > 1 \]
So, \( \lim_{x \to 1^+} f'(x) = 1 \).

Since the left-hand and right-hand derivatives are equal, \( f(x) \) is differentiable at \( x = 1 \).

### Conclusion

- \( f(x) \) is continuous at \( x = 1 \).
- \( f(x) \) is differentiable at \( x = 1 \).