# Given f(x) =

Question:

Given: $f(x)=\left\{\begin{array}{cc}x & , 0 \leq x<\frac{1}{2} \\ \frac{1}{2} & , \quad x=\frac{1}{2} \\ 1-x & , \quad \frac{1}{2} and$g(x)=\left(x-\frac{1}{2}\right)^{2}, x \in \mathrm{R}$. Then the area (in sq. units) of the region bounded by the curves,$y=f(x)$and$y=g(x)$between the lines,$2 x=1$and$2 x=\sqrt{3}$, is : 1. (1)$\frac{1}{3}+\frac{\sqrt{3}}{4}$2. (2)$\frac{\sqrt{3}}{4}-\frac{1}{3}$3. (3)$\frac{1}{2}-\frac{\sqrt{3}}{4}$4. (4)$\frac{1}{2}+\frac{\sqrt{3}}{4}$Correct Option: , 2 Solution: Coordinates of$P\left(\frac{1}{2}, 0\right), Q\left(\frac{\sqrt{3}}{2}, 0\right), R\left(\frac{\sqrt{3}}{2}, 1-\frac{\sqrt{3}}{2}\right)$and$S\left(\frac{1}{2}, \frac{1}{2}\right)$Required area = Area of trapezium$P Q R S-\int_{1 / 2}^{\sqrt{3} / 2}\left(x-\frac{1}{2}\right)^{2} d x=\frac{1}{2}\left(\frac{\sqrt{3}-1}{2}\right)\left(\frac{1}{2}+1-\frac{\sqrt{3}}{2}\right)-\frac{1}{3}\left(\left(x-\frac{1}{2}\right)^{3}\right)_{1 / 2}^{\sqrt{3} / 2}=\frac{\sqrt{3}}{4}-\frac{1}{3}\$