Given `f(x)={(x^2 xx e^(2(x-1)), 0 le x le 1), (acos(2x-2)+bx^2, 1 lt x le2):}` is differentiable at `x=1`, then the value of `|a - b|` is

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is differentiable at \( x = 1 \). This requires that both the function values and the derivatives from the left and right sides at \( x = 1 \) are equal.

### Step 1: Find \( f(1^-) \)
The function for \( 0 \leq x \leq 1 \) is given by:
\[ f(x) = x^2 e^{2(x-1)} \]
Calculating \( f(1) \):
\[ f(1^-) = 1^2 e^{2(1-1)} = 1 \cdot e^0 = 1 \]

### Step 2: Find \( f(1^+) \)
The function for \( 1 < x \leq 2 \) is given by:
\[ f(x) = a \cos(2x - 2) + b x^2 \]
Calculating \( f(1) \):
\[ f(1^+) = a \cos(2(1) - 2) + b(1^2) = a \cos(0) + b = a + b \]

### Step 3: Set the function values equal
Since the function is differentiable at \( x = 1 \), we have:
\[ f(1^-) = f(1^+) \]
Thus,
\[ 1 = a + b \quad (1) \]

### Step 4: Find \( f'(1^-) \)
Now we differentiate \( f(x) = x^2 e^{2(x-1)} \):
Using the product rule:
\[ f'(x) = \frac{d}{dx}(x^2) \cdot e^{2(x-1)} + x^2 \cdot \frac{d}{dx}(e^{2(x-1)}) \]
Calculating the derivatives:
\[ \frac{d}{dx}(x^2) = 2x \]
\[ \frac{d}{dx}(e^{2(x-1)}) = e^{2(x-1)} \cdot 2 \]
Thus,
\[ f'(x) = 2x e^{2(x-1)} + x^2 \cdot 2 e^{2(x-1)} = e^{2(x-1)}(2x + 2x^2) \]
Now, substituting \( x = 1 \):
\[ f'(1^-) = e^{2(1-1)}(2(1) + 2(1^2)) = 1(2 + 2) = 4 \]

### Step 5: Find \( f'(1^+) \)
Now we differentiate \( f(x) = a \cos(2x - 2) + b x^2 \):
\[ f'(x) = -2a \sin(2x - 2) + 2bx \]
Substituting \( x = 1 \):
\[ f'(1^+) = -2a \sin(0) + 2b(1) = 0 + 2b = 2b \]

### Step 6: Set the derivatives equal
Since the function is differentiable at \( x = 1 \), we have:
\[ f'(1^-) = f'(1^+) \]
Thus,
\[ 4 = 2b \quad (2) \]
From this, we can solve for \( b \):
\[ b = 2 \]

### Step 7: Substitute \( b \) back into equation (1)
Substituting \( b = 2 \) into equation (1):
\[ 1 = a + 2 \implies a = 1 - 2 = -1 \]

### Step 8: Find \( |a - b| \)
Now we calculate \( |a - b| \):
\[ |a - b| = |-1 - 2| = |-3| = 3 \]

### Final Answer
Thus, the value of \( |a - b| \) is:
\[ \boxed{3} \]