The area bounded by the curves $y=2 x^2$, $y=\max \{x-[x]+|x|\}$ and the lines $x=0, x=2$ (in sq units), is

• Correct answer
  A

  2
• B

  $\frac{1}{2}$
• C

  $\frac{1}{3}$
• D

  $\frac{4}{3}$

Solution

Given curves $y=2 x^2$, $$ \begin{aligned} y & =\max \{x-[x], x+|x|\} \\ & =\max \{(x), x+|x|\} \end{aligned} $$ Now, graph of $\{x\}$
$$ \text { and graph of } x+\text { mode }|x|= \begin{cases}0, & x < 0 \\ 2 x, & x \geq 0\end{cases} $$
Now, area bounded by $y=2 x^2$ and $$ y=\max \{\{x\}, x+|x|\}=x+|x| $$ and lines $x=0, x=2$ is
$\therefore$ Required area $$ \begin{aligned} & =\int_0^1\left(2 x-2 x^2\right) d x+\int_1^2\left(2 x^2-2 x\right) d x \\ & =\left[\frac{2 x^2}{2}-\frac{2 x^3}{3}\right]_0^1+\left[\frac{2 x^3}{3}-\frac{2 x^2}{2}\right]_1^2 \\ & =\left(1-\frac{2}{3}\right)+\left(\frac{16}{3}-4-\frac{2}{3}+1\right) \\ & =\frac{1}{3}+\frac{14}{3}-3=5-3=2 \end{aligned} $$

Hence, the correct answer is option: 'A'.