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

$f(x)=|(2 x-1)(x+2)|+\frac{\sin 2 x}{2}$ <br/><br/> $0 \leq x<\frac{1}{2} \quad f(x)=(1-2 x)(x+2)+\frac{\sin 2 x}{2}$ <br/><br/> $f^{\prime}(x)=-4 x-3+\cos 2 x<0$ <br/><br/> For $x \geq \frac{1}{2}: \quad f^{\prime}(x)=4 x+3+\cos 2 x>0$ <br/><br/> So, minima occurs at $x=\frac{1}{2}$ <br/><br/> $$ \begin{aligned} \left.f(x)\right|_{\min } &=\left|2\left(\frac{1}{2}\right)^{2}+\frac{3}{2}-2\right|+\sin \left(\frac{1}{2}\right) \cdot \cos \left(\frac{1}{2}\right) \\\\ &=\frac{1}{2} \sin 1 \end{aligned} $$ <br/><br/> So, maxima is possible at $x=0$ or $x=1$ <br/><br/> Now checking for $x=0$ and $x=1$, we can see it attains its maximum value at $x=1$ <br/><br/> $$ \begin{aligned} \left.f(x)\right|_{\max }=&|2+3-2|+\frac{\sin 2}{2} \\\\ &=3+\frac{1}{2} \sin 2 \end{aligned} $$ <br/><br/> Sum of absolute maximum and minimum value <br/><br/>$=3+\frac{1}{2}(\sin 1+\sin 2)$ <br/><br/>$=3+\frac{1}{2}(\sin 1+2\sin 1\cos 1)$ <br/><br/>= $3 + {1 \over 2}(1 + 2\cos (1))\sin (1)$