Let `f(x)=int_(0)^(x)(|t-1|-|t+2|+t-2)dt`, such that `f''(a)ne1`. If vectors `a hati-b^(2)hatj and hati+3bhatj` are parallel for at least one `a`, then the maximum value of `(1-8b-b^(2))` is:

Options

A. (a) `4`
B. (b) `8`
C. (c) `12`
D. (d) `16`

Answer

To solve the problem step by step, we will analyze the given function and the conditions provided.

### Step 1: Define the Function
We start with the function:
\[ f(x) = \int_{0}^{x} (|t-1| - |t+2| + t - 2) dt \]

### Step 2: Find the First Derivative
Using the Fundamental Theorem of Calculus, we differentiate \(f(x)\):
\[ f'(x) = |x-1| - |x+2| + x - 2 \]

### Step 3: Analyze the Absolute Values
The expression for \(f'(x)\) will change based on the value of \(x\). We need to consider three intervals based on the points where the absolute values change:
1. \(x < -2\)
2. \(-2 \leq x < 1\)
3. \(x \geq 1\)

#### Case 1: \(x < -2\)
\[ f'(x) = -(x-1) - (-(x+2)) + x - 2 = -x + 1 + x + 2 + x - 2 = x + 1 \]

#### Case 2: \(-2 \leq x < 1\)
\[ f'(x) = -(x-1) - (x+2) + x - 2 = -x + 1 - x - 2 + x - 2 = -x - 3 \]

#### Case 3: \(x \geq 1\)
\[ f'(x) = (x-1) - (x+2) + x - 2 = x - 1 - x - 2 + x - 2 = x - 5 \]

### Step 4: Find the Second Derivative
Now we differentiate \(f'(x)\) to find \(f''(x)\):
- For \(x < -2\): \(f''(x) = 1\)
- For \(-2 < x < 1\): \(f''(x) = -1\)
- For \(x > 1\): \(f''(x) = 1\)

### Step 5: Determine Values of \(a\)
We know that \(f''(a) \neq 1\). Therefore, \(a\) must be in the interval \([-2, 1]\).

### Step 6: Analyze the Vectors
The vectors given are \(a \hat{i} - b^2 \hat{j}\) and \(\hat{i} + 3b \hat{j}\). For these vectors to be parallel, their direction ratios must be proportional:
\[ \frac{a}{1} = \frac{-b^2}{3b} \]
This simplifies to:
\[ 3a + b = 0 \quad \text{or} \quad b = -3a \]

### Step 7: Determine the Range of \(b\)
Substituting the endpoints of \(a\):
- If \(a = -2\), then \(b = -3(-2) = 6\)
- If \(a = 1\), then \(b = -3(1) = -3\)

Thus, \(b\) lies in the interval \([-3, 6]\).

### Step 8: Maximize the Expression
We need to maximize:
\[ g(b) = 1 - 8b - b^2 \]
To find the critical points, we differentiate:
\[ g'(b) = -8 - 2b \]
Setting \(g'(b) = 0\):
\[ -8 - 2b = 0 \implies b = -4 \]
Since \(-4\) is not in the interval \([-3, 6]\), we evaluate \(g(b)\) at the endpoints.

### Step 9: Evaluate at Endpoints
1. \(g(-3) = 1 - 8(-3) - (-3)^2 = 1 + 24 - 9 = 16\)
2. \(g(6) = 1 - 8(6) - 6^2 = 1 - 48 - 36 = -83\)

### Conclusion
The maximum value of \(g(b)\) occurs at \(b = -3\):
\[ \max g(b) = 16 \]

Thus, the answer is:
\[ \boxed{16} \]