Question

The area (in sq. units) of the region $A=\{(x,y)$ $:|x|+|y|\le 1,2y^2\ge |x|\}$ is :

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{7}{6}$
• D. $\frac{5}{6}$

Answer 1

Answer: (D)

$|x|+|y|\le 1$
$2y^2\ge |x|$

For point of intersection
$x+y=1\Rightarrow x=1-y$
$y^2=\frac{x}{2}\Rightarrow 2y^2=x$
$2y^2=1-y\Rightarrow 2y^2+y-1=0$
$(2y-1)(y+1)=0$
$y=\frac{1}{2}$ or -1
Now Area of $\Delta OAB=\frac{1}{2}\times 1\times 1=\frac{1}{2}$
Area of Region $R_1=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$
Area of Region $R_2=\frac{1}{\sqrt{2}}\underset{0}{\overset{\frac{1}{2}}{\int }}\sqrt{x}dx=\frac{1}{6}$
Now area of shaded region in first quadrant
$=$ Area of $\Delta OAB-R_1-R_2$
$=\frac{1}{2}-\left(\frac{1}{6}\right)-\left(\frac{1}{8}\right)=\frac{5}{24}$
So required area $=4\left(\frac{5}{24}\right)=\frac{5}{6}$