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.\[OAB-\Delta OAB\]].You need to login to perform this action.
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