Key Concepts and Formulas

• Extreme Value Theorem: A continuous function on a closed interval $[a, b]$ attains both an absolute maximum and an absolute minimum value within that interval.
• Critical Points: Points where the derivative of the function is either zero or undefined.
• Trigonometric Identity: $\sin(2x) = 2\sin(x)\cos(x)$

Step-by-Step Solution

Step 1: Analyze the function and rewrite it.

We are given the function $f(x) = |2x^2 + 3x - 2| + \sin x \cos x$. We can simplify the trigonometric part using the identity $\sin(2x) = 2\sin(x)\cos(x)$, so $\sin x \cos x = \frac{1}{2}\sin(2x)$. Thus, $f(x) = |2x^2 + 3x - 2| + \frac{1}{2}\sin(2x)$.

Step 2: Factor the quadratic inside the absolute value.

We factor the quadratic $2x^2 + 3x - 2$ as $(2x - 1)(x + 2)$. Therefore, $f(x) = |(2x - 1)(x + 2)| + \frac{1}{2}\sin(2x)$.

Step 3: Determine the interval where the quadratic is positive or negative.

The roots of the quadratic are $x = \frac{1}{2}$ and $x = -2$. Since we are considering the interval $[0, 1]$, the relevant root is $x = \frac{1}{2}$.

• For $0 \le x < \frac{1}{2}$, $(2x - 1)$ is negative and $(x + 2)$ is positive, so $(2x - 1)(x + 2)$ is negative.
• For $\frac{1}{2} < x \le 1$, $(2x - 1)$ is positive and $(x + 2)$ is positive, so $(2x - 1)(x + 2)$ is positive.
• At $x = \frac{1}{2}$, $(2x - 1)(x + 2) = 0$.

Step 4: Rewrite the function without the absolute value.

We can rewrite the function as a piecewise function:

$f(x) = \begin{cases} -(2x^2 + 3x - 2) + \frac{1}{2}\sin(2x), & 0 \le x < \frac{1}{2} \\ 2x^2 + 3x - 2 + \frac{1}{2}\sin(2x), & \frac{1}{2} \le x \le 1 \end{cases}$

Step 5: Evaluate the function at the endpoints and the point where the absolute value changes sign.

We need to evaluate $f(0)$, $f(\frac{1}{2})$, and $f(1)$.

• $f(0) = |2(0)^2 + 3(0) - 2| + \frac{1}{2}\sin(2(0)) = |-2| + 0 = 2$.
• $f(\frac{1}{2}) = |2(\frac{1}{2})^2 + 3(\frac{1}{2}) - 2| + \frac{1}{2}\sin(2(\frac{1}{2})) = |\frac{1}{2} + \frac{3}{2} - 2| + \frac{1}{2}\sin(1) = |0| + \frac{1}{2}\sin(1) = \frac{1}{2}\sin(1)$.
• $f(1) = |2(1)^2 + 3(1) - 2| + \frac{1}{2}\sin(2(1)) = |2 + 3 - 2| + \frac{1}{2}\sin(2) = |3| + \frac{1}{2}\sin(2) = 3 + \frac{1}{2}\sin(2)$.

Step 6: Find the derivative of each piece of the piecewise function.

For $0 \le x < \frac{1}{2}$, $f(x) = -2x^2 - 3x + 2 + \frac{1}{2}\sin(2x)$, so $f'(x) = -4x - 3 + \cos(2x)$. For $\frac{1}{2} < x \le 1$, $f(x) = 2x^2 + 3x - 2 + \frac{1}{2}\sin(2x)$, so $f'(x) = 4x + 3 + \cos(2x)$.

Step 7: Find critical points in the interval (0, 1/2) and (1/2, 1).

First, consider $f'(x) = -4x - 3 + \cos(2x)$ for $0 < x < \frac{1}{2}$. Since $-1 \le \cos(2x) \le 1$, we have $-4x - 3 - 1 \le f'(x) \le -4x - 3 + 1$, or $-4x - 4 \le f'(x) \le -4x - 2$. Since $x > 0$, $-4x < 0$, so $-4x - 2 < -2$, which means $f'(x)$ is always negative in this interval. Thus, there are no critical points in $(0, \frac{1}{2})$.

Next, consider $f'(x) = 4x + 3 + \cos(2x)$ for $\frac{1}{2} < x < 1$. Since $-1 \le \cos(2x) \le 1$, we have $4x + 3 - 1 \le f'(x) \le 4x + 3 + 1$, or $4x + 2 \le f'(x) \le 4x + 4$. Since $x > \frac{1}{2}$, $4x > 2$, so $4x + 2 > 4$, which means $f'(x)$ is always positive in this interval. Thus, there are no critical points in $(\frac{1}{2}, 1)$.

Step 8: Determine absolute maximum and minimum values.

Since there are no critical points within the intervals, we only need to consider the values at the endpoints and the point $x = \frac{1}{2}$. We have $f(0) = 2$, $f(\frac{1}{2}) = \frac{1}{2}\sin(1)$, and $f(1) = 3 + \frac{1}{2}\sin(2)$.

We want to find the absolute maximum and absolute minimum values.

Since $0 < \sin(1) \le 1$ and $0 < \sin(2) \le 1$, we have $0 < \frac{1}{2}\sin(1) \le \frac{1}{2}$ and $0 < \frac{1}{2}\sin(2) \le \frac{1}{2}$.

Therefore, $f(0) = 2$, $f(\frac{1}{2}) = \frac{1}{2}\sin(1)$, and $f(1) = 3 + \frac{1}{2}\sin(2)$. The minimum value is $f(\frac{1}{2}) = \frac{1}{2}\sin(1)$. The maximum value is $f(1) = 3 + \frac{1}{2}\sin(2)$.

Step 9: Calculate the sum of the absolute maximum and minimum values.

The sum of the absolute maximum and minimum values is: $f_{max} + f_{min} = 3 + \frac{1}{2}\sin(2) + \frac{1}{2}\sin(1) = 3 + \frac{1}{2}(\sin(2) + \sin(1))$ Using the identity $\sin(2) = 2\sin(1)\cos(1)$, we have: $3 + \frac{1}{2}(2\sin(1)\cos(1) + \sin(1)) = 3 + \frac{1}{2}\sin(1)(2\cos(1) + 1) = 3 + \frac{1}{2}(1 + 2\cos(1))\sin(1)$ $3 + \sin(1) \cos(1) + \frac{1}{2}\sin(1) = 3 + 2\sin(1/2)\cos(1/2) (2\cos^2(1/2)-1) + \frac{1}{2}\sin(1)$ $= 3 + \sin(1) \cos(1) + \frac{1}{2}\sin(1)$ $= 3 + \sin(1) \cos(1) + \frac{1}{2}\sin(1) = 3 + \sin(1) (\cos(1) + \frac{1}{2})$ $= 3 + \sin(1) (2\cos^2(\frac{1}{2}) - 1 + \frac{1}{2}) = 3 + \sin(1) (2\cos^2(\frac{1}{2}) - \frac{1}{2})$ $= 3 + \sin(1) (\frac{4\cos^2(\frac{1}{2}) - 1}{2}) = 3 + \frac{\sin(1)}{2} (4\cos^2(\frac{1}{2}) - 1)$ $= 3 + \frac{\sin(1)}{2} (2(1+\cos(1)) - 1) = 3 + \frac{\sin(1)}{2} (1+2\cos(1))$ $= 3 + \frac{\sin(1) + 2\sin(1)\cos(1)}{2} = 3 + \frac{\sin(1) + \sin(2)}{2} = 3 + \frac{\sin(1)}{2} + \frac{\sin(2)}{2}$

The correct expression is $3 + \frac{1}{2}(1 + 2\cos (1))\sin (1)$. The option (A) simplifies to this expression. $3 + \frac{\sin(1)}{2} + \sin(1) \cos(1)$ $3 + \frac{\sin(1)}{2} + \sin(1) (2\cos^2(\frac{1}{2}) - 1)$ $3 + \frac{\sin(1)}{2} + 2\sin(1) \cos^2(\frac{1}{2}) - \sin(1)$ $3 + \frac{4\sin(1) \cos^2(\frac{1}{2}) - \sin(1)}{2} = 3 + \frac{\sin(1)(4\cos^2(\frac{1}{2}) - 1)}{2}$ $3 + \frac{\sin(1)(2(1 + \cos(1)) - 1)}{2} = 3 + \frac{\sin(1)(1 + 2\cos(1))}{2}$

Common Mistakes & Tips

• Remember to check the endpoints and any points where the absolute value function changes its behavior. These are potential locations for extrema.
• Be careful when differentiating piecewise functions. Ensure you are using the correct piece for the interval you are considering.
• Don't forget to use trigonometric identities to simplify expressions and make comparisons easier.

Summary

We found the absolute maximum and minimum values of the function $f(x) = |2x^2 + 3x - 2| + \sin x \cos x$ on the interval [0, 1] by analyzing the function's behavior within the interval, considering the critical points and endpoints. The sum of the absolute maximum and absolute minimum values is $3 + \frac{1}{2}(1 + 2\cos (1))\sin (1)$.

Final Answer

The final answer is \boxed{3 + \frac{1}{2}(1 + 2\cos (1))\sin (1)}, which corresponds to option (B).