JEE Main 2022 — Definite Integration Question with Solution

$$f(x)=2+|x|-|x-1|+|x+1|,x\in R$$

$$\therefore$$ f(x) = \left\{ {\matrix{ { - x} & , & {x < - 1} \cr {x + 2} & , & { - 1 \le x < 0} \cr {3x + 2} & , & {0 \le x < 1} \cr {x + 4} & , & {x \ge 1} \cr } } \right.

$$\therefore$$ $$f^{\prime }\left(-\frac{3}{2}\right)+f^{\prime }\left(-\frac{1}{2}\right)+f^{\prime }\left(\frac{1}{2}\right)+f^{\prime }\left(\frac{3}{2}\right)=-1+1+3+1=4$$

and $$\underset{-2}{\overset{2}{\int }}f(x)dx=\underset{-2}{\overset{-1}{\int }}f(x)dx+\underset{-1}{\overset{0}{\int }}f(x)dx+\underset{0}{\overset{1}{\int }}f(x)dx+\underset{1}{\overset{2}{\int }}f(x)dx$$

$$={\left[-\frac{x^2}{2}\right]}_2^{-1}+{\left[\frac{{(x+2)}^2}{2}\right]}_{-1}^0+{\left[\frac{{(3x+2)}^2}{6}\right]}_0^1+{\left[\frac{{(x+4)}^2}{2}\right]}_1^2$$

$$=\frac{3}{2}+\frac{3}{2}+\frac{7}{2}+\frac{11}{2}=\frac{24}{2}=12$$

$$\therefore$$ Only (S2) is correct