question_answer

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)        \[\frac{\pi }{4}-\frac{1}{2}\]

Correct Answer: D

Solution :

\[{{x}^{2}}+{{y}^{2}}=1,\,\,x+y=1\]meet when                 \[{{x}^{2}}+{{(1-x)}^{2}}=1\] \[\Rightarrow \]               \[{{x}^{2}}+1+{{x}^{2}}-2x=1\] \[\Rightarrow \]               \[2{{x}^{2}}-2x=0\] \[\Rightarrow \]               \[2x(x-1)=0\] \[\Rightarrow \]               \[x=0,\,\,x=1\] \[\Rightarrow \]Meet at points\[A(1,\,\,0),\,\,B(0,\,\,1)\] \[\therefore \]Required area\[\frac{\pi }{4}-\frac{1}{2}(1).(1)\] \[=\frac{\pi }{4}-\frac{1}{2}i.e.,\]              [quad.\[OAB-\Delta OAB\]].