Question

The line x = 0 divides the area enclosed by the curves $|x-1|-y=0,|x|+y-3=0$ into two area $R_1$ and $R_2$ where $R_1<R_2$. Then the ratio of $R_1$ and $R_2$ is

• A. 1 : 2
• B. 1 : 4
• C. 1 : $\sqrt{2}$ +1
• D. 1 : 3

Answer 1

The correct option is D 1 : 3

The graph of the given curves $|x-1|-y=0,|x|+y-3=0$ is:

Area of region $R_1$ is $=\frac{1}{2}\times 2\times 1=1$ sq. unit.
Area of region $R_2$ is $={\int }_0^2[3-|x|]-[|x-1|]dx$
$={\int }_0^2[3-|x|]dx-{\int }_0^2|x-1|dx$
$={\int }_0^2(3-x)dx-{\int }_0^1-(x-1)dx-{\int }_1^2(x-1)dx$
$=[3x-\frac{x^2}{2}{]}_0^2+[\frac{x^2}{2}-x{]}_0^1-[\frac{x^2}{2}-x{]}_1^2$
$=(6-2)+\frac{1}{2}-1-(2-2)+(\frac{1}{2}-1)$
$=4-\frac{1}{2}-\frac{1}{2}$
$=4-1$
$=3$ sq. unit.
Hence, the ratio of the area's of region $R_1$ to $R_2$ is $1:3$