Given: f x = x , 0 ≤ x < 1 2 1 2 , x = 1 2 1 - x , 1 2 < x ≤ 1 and g x = x - 1 2 2 , x…

1. $\frac{1}{3}+\frac{\sqrt{3}}{4}$
2. $\frac{\sqrt{3}}{4}-\frac{1}{3}$
3. $\frac{1}{2}-\frac{\sqrt{3}}{4}$
4. $\frac{1}{2}+\frac{\sqrt{3}}{4}$

Solution

Required area = Area of trapezium ABCD $-\underset{1/2}{\overset{\sqrt{3}/2}{\int }}{\left(x-\frac{1}{2}\right)}^{2}dx$
$=\frac{1}{2}\left(\frac{\sqrt{3}-1}{2}\right)\left(\frac{1}{2}+1-\frac{\sqrt{3}}{2}\right)-\frac{1}{3}{\left({\left(x-\frac{1}{2}\right)}^3\right)}_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}$
$=\frac{\sqrt{3}}{4}-\frac{1}{3}$

Asked in: JEE Main 2020 (09 Jan Shift 2)

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