Let `f(x)={{:(int_(0)^(x)(5+|1-t|)dt","," if "xgt 2),(5x+1","," if "x le 2):}` then the function is

Options

A. `f(x)` is continuous everywhere
B. `f(x)` is not continuous at `x=2`
C. `f(x)` is not difference at exactly three points
D. `f(x)` is continuous and differentiable at `x=2`

Explanation

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

\[ f(x) = \begin{cases} \int_{0}^{x} (5 + |1 - t|) dt & \text{if } x > 2 \\ 5x + 1 & \text{if } x \leq 2 \end{cases} \]

### Step 1: Evaluate \( f(x) \) for \( x > 2 \)

For \( x > 2 \), we need to compute the integral:

\[ f(x) = \int_{0}^{x} (5 + |1 - t|) dt \]

### Step 2: Break down the absolute value

The expression \( |1 - t| \) can be simplified based on the value of \( t \):
- For \( t < 1 \), \( |1 - t| = 1 - t \)
- For \( t \geq 1 \), \( |1 - t| = t - 1 \)

Thus, we can split the integral at \( t = 1 \):

\[ f(x) = \int_{0}^{1} (5 + (1 - t)) dt + \int_{1}^{x} (5 + (t - 1)) dt \]

### Step 3: Compute the first integral

Calculating the first integral:

\[ \int_{0}^{1} (5 + 1 - t) dt = \int_{0}^{1} (6 - t) dt \]

Calculating this gives:

\[ = \left[ 6t - \frac{t^2}{2} \right]_{0}^{1} = (6 \cdot 1 - \frac{1^2}{2}) - (6 \cdot 0 - \frac{0^2}{2}) = 6 - \frac{1}{2} = \frac{12}{2} - \frac{1}{2} = \frac{11}{2} \]

### Step 4: Compute the second integral

Now, we compute the second integral:

\[ \int_{1}^{x} (5 + (t - 1)) dt = \int_{1}^{x} (4 + t) dt \]

Calculating this gives:

\[ = \left[ 4t + \frac{t^2}{2} \right]_{1}^{x} = \left( 4x + \frac{x^2}{2} \right) - \left( 4 \cdot 1 + \frac{1^2}{2} \right) = \left( 4x + \frac{x^2}{2} \right) - (4 + \frac{1}{2}) = 4x + \frac{x^2}{2} - \frac{9}{2} \]

### Step 5: Combine the results

Now we combine both parts to find \( f(x) \) for \( x > 2 \):

\[ f(x) = \frac{11}{2} + \left( 4x + \frac{x^2}{2} - \frac{9}{2} \right) \]

Simplifying this gives:

\[ f(x) = 4x + \frac{x^2}{2} + \frac{11}{2} - \frac{9}{2} = 4x + \frac{x^2}{2} + 1 \]

### Step 6: Final expression for \( f(x) \)

Thus, for \( x > 2 \):

\[ f(x) = 4x + \frac{x^2}{2} + 1 \]

### Step 7: Evaluate \( f(x) \) for \( x \leq 2 \)

For \( x \leq 2 \):

\[ f(x) = 5x + 1 \]

### Step 8: Check continuity at \( x = 2 \)

To check continuity at \( x = 2 \):

- Calculate \( f(2) \) from both sides:
- From the left: \( f(2) = 5(2) + 1 = 10 + 1 = 11 \)
- From the right: \( f(2) = 4(2) + \frac{2^2}{2} + 1 = 8 + 2 + 1 = 11 \)

Since both limits equal \( 11 \), \( f(x) \) is continuous at \( x = 2 \).

### Step 9: Check differentiability at \( x = 2 \)

Now we check the derivatives:

- For \( x < 2 \): \( f'(x) = 5 \)
- For \( x > 2 \): \( f'(x) = 4 + x \)

At \( x = 2 \):

- From the left: \( f'(2^-) = 5 \)
- From the right: \( f'(2^+) = 4 + 2 = 6 \)

Since \( f'(2^-) \neq f'(2^+) \), \( f(x) \) is not differentiable at \( x = 2 \).

### Conclusion

Thus, the function \( f(x) \) is continuous at \( x = 2 \) but not differentiable at \( x = 2 \).