QJEEM22Ma24f8: JEE Main 2022 Mathematics - Class 12 JEEM22Ma24f8 Solution: Let . Consider Then,

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

$\therefore$ $$f(x)=\{\begin{array}{ccc}-x& ,& x<-1\\ x+2& ,& -1\le x<0\\ 3x+2& ,& 0\le x<1\\ x+4& ,& x\ge 1\end{array}$$

$\therefore$ $f'\left( { - {3 \over 2}} \right) + f'\left( { - {1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) + f'\left( {{3 \over 2}} \right) = - 1 + 1 + 3 + 1 = 4$

and $\int\limits_{ - 2}^2 {f(x)dx = \int\limits_{ - 2}^{ - 1} {f(x)dx + \int\limits_{ - 1}^0 {f(x)dx + \int\limits_0^1 {f(x)dx + \int\limits_1^2 {f(x)dx} } } } }$

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

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

$\therefore$ Only (S2) is correct