Question

The area of the region $[(x,y):x^2+y^2\le 1\le x+y]$ is;

• A. $\frac{\pi }{5}$
• B. $\frac{\pi }{4}$
• C. $\frac{{\pi }^2}{3}$
• D. $\frac{\pi }{4}-\frac{1}{2}$

Answer 1

The correct option is D $\frac{\pi }{4}-\frac{1}{2}$

$x^2+y^2=1,x+y=1$ meet when
$x^2+{(1-x)}^2=1$
$\Rightarrow x^2+1+x^2-2x=1$
$\Rightarrow 2x^2-2x=0$
$\Rightarrow 2x(x-1)=0$
$\Rightarrow x=0,x=1$
Meet at points $A(1,0),B(0,1)$
Therefore, required area $=\frac{\pi }{4}-\frac{1}{2}\left(1\right)\cdot \left(1\right)$
$=\frac{\pi }{4}-\frac{1}{2}$ i.e., [ $OAB-\Delta OAB$]