Question

JEE mains question is in the image

Answer 1

Given to evaluate,

$\displaystyle\longrightarrow I=\int\limits_0^2||x-1|-x|\ dx$

Replace $x$ by $x+1$ so that the limits get reduced by 1.

Then the integral becomes,

$\displaystyle\longrightarrow I=\int\limits_{-1}^1||x+1-1|-(x+1)|\ dx$

$\displaystyle\longrightarrow I=\int\limits_{-1}^1||x|-x-1|\ dx$

Let us define a function $f:\mathbb{R}\to\mathbb{R}$ such that,

$\longrightarrow f(x)=||x|-x-1|$

For $x\geq0,$

$\longrightarrow f(x)=|x-x-1|$

$\longrightarrow f(x)=1$

Then the integral can be split as,

$\displaystyle\longrightarrow I=\int\limits_{-1}^0||x|-x-1|\ dx+\int\limits_0^1||x|-x-1|\ dx$

$\displaystyle\longrightarrow I=\int\limits_{-1}^0||x|-x-1|\ dx+\int\limits_0^11\ dx$

$\displaystyle\longrightarrow I=1+\int\limits_{-1}^0||x|-x-1|\ dx$

Replace $x$ by $x-\dfrac{1}{2}$ in the integral, so that the limits get added by $\dfrac{1}{2}.$

$\displaystyle\longrightarrow I=1+\int\limits_{-\frac{1}{2}}^{\frac{1}{2}}\left|\left|x-\dfrac{1}{2}\right|-\left(x-\dfrac{1}{2}\right)-1\right|\ dx$

$\displaystyle\longrightarrow I=1+\int\limits_{-\frac{1}{2}}^{\frac{1}{2}}\left|\left|x-\dfrac{1}{2}\right|-x-\dfrac{1}{2}\right|\ dx$

Let us define a function $g:\left[-\dfrac{1}{2},\ \dfrac{1}{2}\right]\to\mathbb{R}$ such that,

$\displaystyle\longrightarrow g(x)=\left|\left|x-\dfrac{1}{2}\right|-x-\dfrac{1}{2}\right|$

For $x\in\left[0,\ \dfrac{1}{2}\right]$ only,

$\displaystyle\longrightarrow g(x)=\left|-x+\dfrac{1}{2}-x-\dfrac{1}{2}\right|$

$\displaystyle\longrightarrow g(x)=2x$

And we see that,

$\displaystyle\longrightarrow g(-x)=\left|\left|-x-\dfrac{1}{2}\right|+x-\dfrac{1}{2}\right|$

$\displaystyle\longrightarrow g(-x)=\left|x+\dfrac{1}{2}+x-\dfrac{1}{2}\right|$

$\displaystyle\longrightarrow g(-x)=2x$

Hence $g$ is an even function in the domain.

Therefore,

$\displaystyle\longrightarrow\int\limits_{-\frac{1}{2}}^{0}\left|\left|x-\dfrac{1}{2}\right|-x-\dfrac{1}{2}\right|\ dx=\int\limits_0^{\frac{1}{2}}\left|\left|x-\dfrac{1}{2}\right|-x-\dfrac{1}{2}\right|\ dx$

Or,

$\displaystyle\longrightarrow\int\limits_{-\frac{1}{2}}^{\frac{1}{2}}\left|\left|x-\dfrac{1}{2}\right|-x-\dfrac{1}{2}\right|\ dx=2\int\limits_0^{\frac{1}{2}}\left|\left|x-\dfrac{1}{2}\right|-x-\dfrac{1}{2}\right|\ dx$

$\displaystyle\longrightarrow\int\limits_{-\frac{1}{2}}^{\frac{1}{2}}\left|\left|x-\dfrac{1}{2}\right|-x-\dfrac{1}{2}\right|\ dx=2\int\limits_0^{\frac{1}{2}}2x\ dx$

$\displaystyle\longrightarrow\int\limits_{-\frac{1}{2}}^{\frac{1}{2}}\left|\left|x-\dfrac{1}{2}\right|-x-\dfrac{1}{2}\right|\ dx=2\Big[x^2\Big]_0^{\frac{1}{2}}$

$\displaystyle\longrightarrow\int\limits_{-\frac{1}{2}}^{\frac{1}{2}}\left|\left|x-\dfrac{1}{2}\right|-x-\dfrac{1}{2}\right|\ dx=\dfrac{1}{2}$

Then our integral becomes nothing but,

$\displaystyle\longrightarrow I=1+\dfrac{1}{2}$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{3}{2}}}$

Hence (D) is the answer.