Let f(x) be a function defined as `f(x)={{:(,int_(0)^(x)(3+|t-2|),"if "x gt 4),(,2x+8,"if "x le 4):}`
Then, f(x) is

Options

A. continuous at x=4
B. neither continuous nor differentiable at x=4
C. everywhere continuous but not differentiable at x=4
D. everywhere continuous and differentiable

Explanation

To solve the problem step by step, we will analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} \int_{0}^{x} (3 + |t - 2|) \, dt & \text{if } x > 4 \\ 2x + 8 & \text{if } x \leq 4 \end{cases} \]

### Step 1: Analyze the case when \( x > 4 \)

For \( x > 4 \), we need to evaluate the integral \( \int_{0}^{x} (3 + |t - 2|) \, dt \).

The absolute value \( |t - 2| \) changes at \( t = 2 \). We need to split the integral into two parts:

\[ f(x) = \int_{0}^{2} (3 + |t - 2|) \, dt + \int_{2}^{x} (3 + |t - 2|) \, dt \]

### Step 2: Evaluate the integral from \( 0 \) to \( 2 \)

For \( t \) in the interval \( [0, 2] \), \( |t - 2| = 2 - t \). Thus,

\[ \int_{0}^{2} (3 + |t - 2|) \, dt = \int_{0}^{2} (3 + (2 - t)) \, dt = \int_{0}^{2} (5 - t) \, dt \]

Calculating this integral:

\[ \int (5 - t) \, dt = 5t - \frac{t^2}{2} \bigg|_{0}^{2} = \left(5 \cdot 2 - \frac{2^2}{2}\right) - (0) = 10 - 2 = 8 \]

### Step 3: Evaluate the integral from \( 2 \) to \( x \)

For \( t \) in the interval \( [2, x] \) (where \( x > 4 \)), \( |t - 2| = t - 2 \). Thus,

\[ \int_{2}^{x} (3 + |t - 2|) \, dt = \int_{2}^{x} (3 + (t - 2)) \, dt = \int_{2}^{x} (t + 1) \, dt \]

Calculating this integral:

\[ \int (t + 1) \, dt = \frac{t^2}{2} + t \bigg|_{2}^{x} = \left(\frac{x^2}{2} + x\right) - \left(\frac{2^2}{2} + 2\right) = \left(\frac{x^2}{2} + x\right) - (2 + 2) = \frac{x^2}{2} + x - 4 \]

### Step 4: Combine the results for \( f(x) \) when \( x > 4 \)

Putting it all together, we have:

\[ f(x) = 8 + \left(\frac{x^2}{2} + x - 4\right) = \frac{x^2}{2} + x + 4 \]

### Step 5: Analyze the case when \( x \leq 4 \)

For \( x \leq 4 \), we have:

\[ f(x) = 2x + 8 \]

### Step 6: Check continuity at \( x = 4 \)

To check continuity, we need to see if \( \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4) \).

Calculating \( f(4) \):

\[ f(4) = 2(4) + 8 = 8 + 8 = 16 \]

Now, calculate \( \lim_{x \to 4^-} f(x) \):

\[ \lim_{x \to 4^-} f(x) = 2(4) + 8 = 16 \]

Now, calculate \( \lim_{x \to 4^+} f(x) \):

\[ \lim_{x \to 4^+} f(x) = \frac{4^2}{2} + 4 + 4 = \frac{16}{2} + 4 + 4 = 8 + 4 + 4 = 16 \]

Since both limits and \( f(4) \) are equal, \( f(x) \) is continuous at \( x = 4 \).

### Step 7: Check differentiability at \( x = 4 \)

Calculate the left-hand derivative:

\[ f'(x) = \frac{d}{dx}\left(\frac{x^2}{2} + x + 4\right) = x + 1 \]
At \( x = 4 \):
\[ f'(4^-) = 4 + 1 = 5 \]

Calculate the right-hand derivative:

\[ f'(x) = \frac{d}{dx}(2x + 8) = 2 \]
At \( x = 4 \):
\[ f'(4^+) = 2 \]

Since \( f'(4^-) \neq f'(4^+) \), \( f(x) \) is not differentiable at \( x = 4 \).

### Conclusion

The function \( f(x) \) is continuous everywhere but not differentiable at \( x = 4 \).