If `f(x)={(5x+1,xlt2),( int_0^x ((5+abs(1-t)))dt,xge2):}`

Options

A. `f(x)` is differentiable `forall x in R`
B. `f(x)` continuous at x=2 but not differentiable at x=2
C. `f(x)` continuous at x=2 but not differentiable at x=1
D. None of these

Answer

To solve the given problem, we need to analyze the function \( f(x) \) defined in two parts:

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

2. For \( x \geq 2 \):
\[ f(x) = \int_0^x (5 + |1 - t|) dt \]

### Step 1: Check Continuity at \( x = 2 \)

To check the continuity of \( f(x) \) at \( x = 2 \), we need to evaluate the left-hand limit \( f(2^-) \) and the right-hand limit \( f(2^+) \).

**Left-hand limit \( f(2^-) \)**:
\[ f(2^-) = 5(2) + 1 = 10 + 1 = 11 \]

**Right-hand limit \( f(2^+) \)**:
\[ f(2^+) = \int_0^2 (5 + |1 - t|) dt \]
To evaluate this integral, we need to consider the expression \( |1 - t| \):
- For \( t < 1 \), \( |1 - t| = 1 - t \)
- For \( t \geq 1 \), \( |1 - t| = t - 1 \)

Thus, we split the integral into two parts:
\[ f(2^+) = \int_0^1 (5 + (1 - t)) dt + \int_1^2 (5 + (t - 1)) dt \]

Calculating the first integral:
\[ \int_0^1 (5 + 1 - t) dt = \int_0^1 (6 - t) dt = \left[ 6t - \frac{t^2}{2} \right]_0^1 = 6(1) - \frac{1^2}{2} = 6 - 0.5 = 5.5 \]

Calculating the second integral:
\[ \int_1^2 (5 + t - 1) dt = \int_1^2 (4 + t) dt = \left[ 4t + \frac{t^2}{2} \right]_1^2 = \left(4(2) + \frac{2^2}{2}\right) - \left(4(1) + \frac{1^2}{2}\right) = (8 + 2) - (4 + 0.5) = 10 - 4.5 = 5.5 \]

Thus,
\[ f(2^+) = 5.5 + 5.5 = 11 \]

Since both limits are equal:
\[ f(2^-) = f(2^+) = 11 \]
Thus, \( f(x) \) is continuous at \( x = 2 \).

### Step 2: Check Differentiability at \( x = 2 \)

Next, we need to check the differentiability of \( f(x) \) at \( x = 2 \) by finding the derivatives from the left and right.

**Left-hand derivative \( f'(2^-) \)**:
\[ f'(x) = 5 \quad \text{for } x < 2 \]
Thus,
\[ f'(2^-) = 5 \]

**Right-hand derivative \( f'(2^+) \)**:
To find \( f'(2^+) \), we differentiate the integral:
\[ f'(x) = \frac{d}{dx} \int_0^x (5 + |1 - t|) dt = 5 + |1 - x| \]
For \( x = 2 \):
\[ f'(2^+) = 5 + |1 - 2| = 5 + 1 = 6 \]

### Step 3: Conclusion

Since:
\[ f'(2^-) = 5 \quad \text{and} \quad f'(2^+) = 6 \]
We find that \( f'(2^-) \neq f'(2^+) \), indicating that \( f(x) \) is not differentiable at \( x = 2 \).

### Final Answer

- \( f(x) \) is continuous at \( x = 2 \).
- \( f(x) \) is not differentiable at \( x = 2 \).