Answer:
Option B
Explanation:
Here area between 0 to b is R1 and b to 1 is R2
∴ \int_{0}^{b} (1-x)^{2}dx-\int_{b}^{1} (1-x)^{2}dx=\frac{1}{4}
\Rightarrow \left(\frac{(1-x)^{3}}{-3}\right)_{0}^{b}-\left(\frac{(1-x)^{3}}{-3}\right)^{1}_{b}=\frac{1}{4}
\Rightarrow -\frac{1}{3}{(1-b)^{3}-1})+\frac{1}{3}(0-(1-b)^{3})=\frac{1}{4}
\Rightarrow -\frac{2}{3}(1-b)^{3}=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12}
\Rightarrow (1-b)^{3}=\frac{1}{8}
\Rightarrow (1-)=\frac{1}{2} \Rightarrow b= \frac{1}{2}