To solve the problem, we need to analyze the function \( f(x) \) and the conditions given regarding the vectors. Let's break down the steps:
### Step 1: Define the function \( f(x) \)
We start with the function defined as:
\[
f(x) = \int_{0}^{x} (|t-1| - |t+2| + t - 2) \, dt
\]
### Step 2: Simplify the integrand
To evaluate the integral, we need to consider the absolute values in the integrand. We will analyze the expression \( |t-1| - |t+2| + t - 2 \) based on the critical points \( t = -2 \) and \( t = 1 \).
1. **For \( t < -2 \)**:
\[
|t-1| = -(t-1) = -t + 1, \quad |t+2| = -(t+2) = -t - 2
\]
Thus,
\[
|t-1| - |t+2| + t - 2 = (-t + 1) - (-t - 2) + t - 2 = 1 + 2 - 2 = 1
\]
2. **For \( -2 \leq t < 1 \)**:
\[
|t-1| = -(t-1) = -t + 1, \quad |t+2| = t + 2
\]
Thus,
\[
|t-1| - |t+2| + t - 2 = (-t + 1) - (t + 2) + t - 2 = -2t - 3 + t = -t - 3
\]
3. **For \( t \geq 1 \)**:
\[
|t-1| = t - 1, \quad |t+2| = t + 2
\]
Thus,
\[
|t-1| - |t+2| + t - 2 = (t - 1) - (t + 2) + t - 2 = -1 - 2 = -3
\]
### Step 3: Set up the piecewise function for \( f(x) \)
Now we can express \( f(x) \) as a piecewise function based on the intervals derived:
1. **For \( x < -2 \)**:
\[
f(x) = \int_{0}^{x} 1 \, dt = x
\]
2. **For \( -2 \leq x < 1 \)**:
\[
f(x) = \int_{0}^{-2} 1 \, dt + \int_{-2}^{x} (-t - 3) \, dt = -2 + \left[-\frac{t^2}{2} - 3t\right]_{-2}^{x}
\]
Evaluating this gives:
\[
= -2 + \left(-\frac{x^2}{2} - 3x + 4 + 6\right) = -\frac{x^2}{2} - 3x + 8
\]
3. **For \( x \geq 1 \)**:
\[
f(x) = \int_{0}^{-2} 1 \, dt + \int_{-2}^{1} (-t - 3) \, dt + \int_{1}^{x} (-3) \, dt
\]
Evaluating this gives:
\[
= -2 + 4 + 6 - 3(x - 1) = 8 - 3x
\]
### Step 4: Find the second derivative \( f''(x) \)
Now we calculate \( f''(x) \) for each piece:
1. **For \( x < -2 \)**: \( f'(x) = 1 \) and \( f''(x) = 0 \)
2. **For \( -2 \leq x < 1 \)**: \( f'(x) = -x - 3 \) and \( f''(x) = -1 \)
3. **For \( x \geq 1 \)**: \( f'(x) = -3 \) and \( f''(x) = 0 \)
### Step 5: Analyze the condition \( f''(a) \neq 1 \)
From our analysis, \( f''(a) \) can only be \( 0 \) or \( -1 \) based on the intervals. Thus, the condition \( f''(a) \neq 1 \) is satisfied for all \( a \).
### Step 6: Determine the condition for parallel vectors
The vectors \( a \hat{i} - b^2 \hat{j} \) and \( \hat{i} + 3b \hat{j} \) are parallel if:
\[
\frac{a}{1} = \frac{-b^2}{3b}
\]
This simplifies to:
\[
a = -\frac{b^2}{3}
\]
### Step 7: Find the range of \( b \)
Given \( a \) must be in the range \( -2 < a < 1 \):
\[
-2 < -\frac{b^2}{3} < 1
\]
This leads to:
\[
6 > b^2 > -3 \quad \Rightarrow \quad b^2 < 6 \quad \Rightarrow \quad -\sqrt{6} < b < \sqrt{6}
\]
### Step 8: Count integral values of \( b \)
The integral values of \( b \) in the range \( -\sqrt{6} < b < \sqrt{6} \) are:
\[
-2, -1, 0, 1, 2
\]
Thus, there are **5 integral values** for \( b \).
### Final Answer
The number of integral values of \( b \) can be **5**.
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