JEE Main 2022 — Mathematics Calculus

Given $f(x)=2+|x|-|x-1|+|x+1|,x\in R$f(x)=2+∣x∣−∣x−1∣+∣x+1∣,x∈R \Rightarrow f(x)={\begin{matrix}-x & , & x<-1 \\ x+2 & , & -1\leq x<0 \\ 3x+2 & , & 0\leq x<1 \\ x+4 & , & x\geq 1\end{matrix}

\Rightarrow {f}^{'}(x)={\begin{matrix}-1 & , & x<-1 \\ 1 & , & -1<x<0 \\ 3 & , & 0<x<1 \\ 1 & , & x>1\end{matrix}

$$\therefore {f}^{'}(-\frac{3}{2})+{f}^{'}(-\frac{1}{2})+{f}^{'}(\frac{1}{2})+{f}^{'}(\frac{3}{2})$$∴f′(−23​)+f′(−21​)+f′(21​)+f′(23​)

$$=-1+1+3+1=4$$=−1+1+3+1=4

i.e. $(S1)$(S1) is incorrect.

Now $\displaystyle {\int }_{-2}^{2}f(x)dx={\int }_{-2}^{-1}f(x)dx+{\int }_{-1}^{0}f(x)dx+{\int }_{0}^{1}f(x)dx+{\int }_{1}^{2}f(x)dx$∫−22​f(x)dx=∫−2−1​f(x)dx+∫−10​f(x)dx+∫01​f(x)dx+∫12​f(x)dx

$$={[-\frac{{x}^{2}}{2}]}_{-2}^{-1}+{[\frac{{(x+2)}^{2}}{2}]}_{-1}^{0}+{[\frac{{(3x+2)}^{2}}{6}]}_{0}^{1}+{[\frac{{(x+4)}^{2}}{2}]}_{1}^{2}$$=[−2x2​]−2−1​+[2(x+2)2​]−10​+[6(3x+2)2​]01​+[2(x+4)2​]12​

$$(-\frac{1}{2}+2)+(2-\frac{1}{2})+(\frac{25}{6}-\frac{4}{6})+(18-\frac{25}{2})$$(−21​+2)+(2−21​)+(625​−64​)+(18−225​)

$$=\frac{3}{2}+\frac{3}{2}+\frac{7}{2}+\frac{11}{2}=12$$=23​+23​+27​+211​=12

i.e. $(S2)$(S2) is correct.