The sum of absolute maximum and absolute minimum values of the function [f(x) = |2x^2 + 3x…

Question

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 :

Accepted Answer

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

$0 \leq x0$

So, minima occurs at $x=\frac{1}{2}$

$$
\begin{aligned}
\left.f(x)\right|_{\min } &=\left|2\left(\frac{1}{2}\right)^{2}+\frac{3}{2}-2\right|+\sin \left(\frac{1}{2}\right) \cdot \cos \left(\frac{1}{2}\right) \\
&=\frac{1}{2} \sin 1
\end{aligned}
$$

So, maxima is possible at $x=0$ or $x=1$

Now checking for $x=0$ and $x=1$, we can see it attains its maximum value at $x=1$

$$
\begin{aligned}
\left.f(x)\right|_{\max }=&|2+3-2|+\frac{\sin 2}{2} \\
&=3+\frac{1}{2} \sin 2
\end{aligned}
$$

Sum of absolute maximum and minimum value

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

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

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

Solutions · 1