JEE Main 2020 (Online) 9th January Evening ShiftQuestion 1
Given : $f(x)=\begin{cases}x, & 0\le x<\frac12\\ \frac12, & x=\frac12\\ 1-x, & \frac12<x\le1\end{cases}$ and $g(x)=\left(x-\frac12\right)^2$, $x\in\mathbf{R}$. Then the area (in sq. units) of the region bounded by the curves, $y=f(x)$ and $y=g(x)$ between the lines, $2x=1$ and $2x=\sqrt3$, is :
Application of Integrals
Solution
For $x>\frac12$, $f(x)=1-x$. Put $u=x-\frac12\in\left[0,\frac{\sqrt3-1}2\right]$. $1-x=g\Rightarrow\frac12-u=u^2\Rightarrow u=\frac{\sqrt3-1}2$, i.e. $x=\frac{\sqrt3}2$, so $f\ge g$ on the whole interval. $A=\int_0^{u_1}\left(\frac12-u-u^2\right)du=\frac{u_1}2-\frac{u_1^2}2-\frac{u_1^3}3$, with $u_1^2=1-\frac{\sqrt3}2$, $u_1^3=\frac{3\sqrt3}4-\frac54$. $A=\frac{\sqrt3-1}4-\frac12+\frac{\sqrt3}4-\frac{\sqrt3}4+\frac5{12}=\frac{\sqrt3}4-\frac13$. Option (D).