JEE Main 2022 (Online) 27th July Evening ShiftQuestion 1
Let $f(x)=2+|x|-|x-1|+|x+1|$, $x\in\mathbf{R}$. Consider (S1) : $f'\left(-\frac32\right)+f'\left(-\frac12\right)+f'\left(\frac12\right)+f'\left(\frac32\right)=2$ ; (S2) : $\int_{-2}^{2}f(x)\,dx=12$. Then,
Differentiation and its Application
Solution
Concept. Open the moduli on the intervals cut by $-1,0,1$. Formulas used. (1) Area of a trapezium under a line: width $\times$ average heightStep 1 - Pieces. $$f=\begin{cases}-x, & x<-1\\ 2+x, & -1<x<0\\ 2+3x, & 0<x<1\\ 4+x, & x>1\end{cases}$$ Step 2 - (S1). $f'\left(-\frac32\right)+f'\left(-\frac12\right)+f'\left(\frac12\right)+f'\left(\frac32\right)=-1+1+3+1=4\ne2$. Wrong. Step 3 - (S2). By (1), on each unit interval: $$\int_{-2}^{-1}=\frac{2+1}{2},\quad\int_{-1}^{0}=\frac{1+2}{2},\quad\int_0^1=\frac{2+5}{2},\quad\int_1^2=\frac{5+6}{2}.$$ Sum $=\frac{3+3+7+11}{2}=12$. Correct. Answer: (D)