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

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

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

$5 + \frac{1}{2} (\sin (1) + \sin (2))$

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

answer is B.

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Detailed Solution

$f (x) = | 2 x^{2} + 3 x - 2 | + \sin x \cos x f (x) = | (2 x - 1) (x + 2) | + \sin x \cos x f^{1} (x) = {\begin{matrix} 4 x + 3 + \frac{\cos 2 x}{4}, & \frac{1}{2} < x < 1 \\ - (4 x + 3) + \frac{\cos 2 x}{4}, & 0 \leq x < \frac{1}{2} \end{matrix}$
For $0 \leq x < \frac{1}{2} \Rightarrow f^{1} (x) < 0$
For $\frac{1}{2} < x \leq 1 \Rightarrow f^{1} (x) > 0$
F(x) local minima at $x = \frac{1}{2}$ and local maxima at x = 1
$f (\frac{1}{2}) + f (1) = 3 + \frac{1}{2} (1 + 2 \cos 1) \sin 1$