The sum of absolute maximum and absolute minimum values of the function $$f(x) = |2x^2 + 3x - 2| + \sin x \cos x$$ in the interval $$[0, 1]$$ is

We need the sum of the absolute maximum and minimum of $$f(x) = |2x^2 + 3x - 2| + \sin x \cos x$$ on $$[0, 1]$$.

Analyze the expression inside the absolute value.

$$2x^2 + 3x - 2 = (2x - 1)(x + 2)$$

On $$[0, 1]$$: $$(x + 2) > 0$$ always. $$(2x - 1) < 0$$ for $$x \in [0, 1/2)$$ and $$(2x - 1) > 0$$ for $$x \in (1/2, 1]$$.

Write f(x) piecewise.

For $$x \in [0, 1/2]$$: $$f(x) = -(2x^2 + 3x - 2) + \frac{\sin 2x}{2} = -2x^2 - 3x + 2 + \frac{\sin 2x}{2}$$

For $$x \in [1/2, 1]$$: $$f(x) = 2x^2 + 3x - 2 + \frac{\sin 2x}{2}$$

Find critical points on [0, 1/2].

$$f'(x) = -4x - 3 + \cos 2x$$

At $$x = 0$$: $$f'(0) = -3 + 1 = -2 < 0$$. Since $$\cos 2x \leq 1$$ and $$-4x - 3 \leq -3$$, we have $$f'(x) < 0$$ on $$(0, 1/2)$$.

So $$f$$ is strictly decreasing on $$[0, 1/2]$$.

Find critical points on [1/2, 1].

$$f'(x) = 4x + 3 + \cos 2x$$

Since $$4x + 3 \geq 5$$ for $$x \geq 1/2$$ and $$\cos 2x \geq -1$$, we have $$f'(x) \geq 4 > 0$$.

So $$f$$ is strictly increasing on $$[1/2, 1]$$.

Evaluate at key points.

$$f(0) = 2 + 0 = 2$$

$$f(1/2) = 0 + \frac{\sin 1}{2} = \frac{\sin 1}{2}$$

$$f(1) = 2 + 3 - 2 + \frac{\sin 2}{2} = 3 + \frac{\sin 2}{2}$$

Determine absolute max and min.

$$f$$ decreases from $$f(0) = 2$$ to $$f(1/2) = \frac{\sin 1}{2}$$, then increases to $$f(1) = 3 + \frac{\sin 2}{2}$$.

Absolute minimum = $$\frac{\sin 1}{2}$$ (at $$x = 1/2$$).

Absolute maximum = $$3 + \frac{\sin 2}{2}$$ (at $$x = 1$$).

Compute the sum.

$$\text{Sum} = \frac{\sin 1}{2} + 3 + \frac{\sin 2}{2} = 3 + \frac{\sin 1 + \sin 2}{2}$$

Using $$\sin 2 = 2\sin 1 \cos 1$$:

$$= 3 + \frac{\sin 1 + 2\sin 1 \cos 1}{2} = 3 + \frac{\sin 1(1 + 2\cos 1)}{2}$$

Answer: Option B