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. continuous at x = 2
B. differentiable at x = 2
C. discontinuous at x = 2
D. not differentiable at x = 2

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined 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 the integral for \( x > 2 \)

First, we need to evaluate the integral \( \int_{0}^{x} (5 + |1 - t|) dt \).

To do this, we need to understand how the absolute value function \( |1 - t| \) behaves:

- For \( t < 1 \), \( |1 - t| = 1 - t \)
- For \( t \geq 1 \), \( |1 - t| = t - 1 \)

Thus, we can split the integral into two parts:

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

### Step 2: Calculate the first integral

Calculating the first integral:

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

Calculating this integral:

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

### Step 3: Calculate the second integral

Now, we calculate the second integral:

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

Calculating this integral:

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

### Step 4: Combine both parts

Combining both parts, we have:

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

Thus, for \( x > 2 \):

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

### Step 5: Analyze the case for \( x \leq 2 \)

For \( x \leq 2 \):

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

### Step 6: Check continuity at \( x = 2 \)

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

1. Calculate \( f(2) \) from the left:

\[ f(2) = 5(2) + 1 = 10 + 1 = 11 \]

2. Calculate \( f(2) \) 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 7: Check differentiability at \( x = 2 \)

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

1. Calculate \( f'(x) \) for \( x < 2 \):

\[ f'(x) = 5 \]

2. Calculate \( f'(x) \) 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 the left-hand derivative does not equal the right-hand derivative, \( f(x) \) is not differentiable at \( x = 2 \).

### Conclusion

The function \( f(x) \) is continuous at \( x = 2 \) but not differentiable at \( x = 2 \).