JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Area Under Curves

The function $y=|x-1|+|x-2|$ is piecewise linear. For $x\le 1$: $y=(1-x)+(2-x)=3-2x$. For $1<x<2$: $y=(x-1)+(2-x)=1$. For $x\ge 2$: $y=(x-1)+(x-2)=2x-3$. The minimum value is $y=1$ on $[1,2]$. The line $y=3$ intersects the curve where $3-2x=3$ (giving $x=0$) and where $2x-3=3$ (giving $x=3$). The region is a trapezoid with vertices at $(0,3)$, $(1,1)$, $(2,1)$, and $(3,3)$. The area is the area of the trapezoid: $\frac{1}{2}(b_1+b_2)h=\frac{1}{2}(1+3)\times 2=4$.