Question

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

• A. $3+\frac{\sin \left(1\right){\cos }^2\left(\frac{1}{2}\right)}{2}$
• B. $3+\frac{1}{2}(1+2\cos (1\left)\right)\sin \left(1\right)$
• C. $2+\sin \left(\frac{1}{2}\right)\cos \left(\frac{1}{2}\right)$
• D. $5+\frac{1}{2}\left(\sin \right(1)+\sin (2\left)\right)$

Answer 1

The correct option is B $3+\frac{1}{2}(1+2\cos (1\left)\right)\sin \left(1\right)$

Given, $f\left(x\right)=|2x^2+3x-2|+\sin x\cos x$
$\Rightarrow f\left(x\right)=|(2x-1)(x+2)|+\frac{\sin 2x}{2}$
$f\left(x\right)=\{\begin{array}{cc}-(2x-1)(x+2)+\frac{\sin 2x}{2},& 0\le x<\frac{1}{2}\\ (2x-1)(x+2)+\frac{\sin 2x}{2},& \frac{1}{2}\le x\le 1\end{array}$
$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{cc}-(4x+3)+\frac{\cos 2x}{2},& 0\le x<\frac{1}{2}\\ (4x+3)+\frac{\cos 2x}{2},& \frac{1}{2}<x\le 1\end{array}$

For $0\le x<\frac{1}{2}\Rightarrow f^{\prime }\left(x\right)<0$
For $\frac{1}{2}<x\le 1\Rightarrow f^{\prime }\left(x\right)>0$

Thus, $f\left(x\right)$ has local minima at $x=\frac{1}{2}.$
Therefore, absolute minimum value of $f\left(x\right):$
$f_{min}=|2{\left(\frac{1}{2}\right)}^2+\frac{3}{2}-2|+\sin \left(\frac{1}{2}\right)\cdot \cos \left(\frac{1}{2}\right)$
$\Rightarrow f_{min}=\frac{1}{2}\sin \left(1\right)$

So, maxima is possible at $x=0$ or $x=1$
Now, checking for $x=0$ and $x=1$
$f\left(0\right)=2$ and $f\left(1\right)=3+\frac{1}{2}\sin \left(2\right)$
We can see it attains its maximum value at $x=1$
$\therefore f_{max}=3+\frac{1}{2}\sin \left(2\right)$

Hence, the sum of absolute maximum and minimum values of $f\left(x\right)$ is $=3+\frac{1}{2}(\sin 1+\sin 2)$