A) \[8\]
B) \[10\]
C) \[16\]
D) None of these
Correct Answer: A
Solution :
Let \[(\alpha ,\,\,\beta )\] be the point with integral coordinates and lying in the interior of the region common to the circle \[{{x}^{2}}+{{y}^{2}}=16\] and the parabola\[{{y}^{2}}=4x\]. Then, \[{{\alpha }^{2}}+{{\beta }^{2}}-16<0\] and \[{{\beta }^{2}}-4\alpha <0\] It is clear from the figure that \[0<\alpha <4\] \[\Rightarrow \] \[\alpha =1,\,\,2,\,\,3\] \[[\because \,\,\alpha \in Z]\] When \[\alpha =1\] \[{{\beta }^{2}}<4\alpha \] \[\Rightarrow \] \[{{\beta }^{2}}<4\] \[\Rightarrow \] \[\beta =0,\,\,1\] So, the points are \[(1,\,\,0)\] and\[(1,\,\,1)\]. When \[\alpha =2\] \[{{\beta }^{2}}<4\alpha \] \[\Rightarrow \] \[{{\beta }^{2}}<8\] \[\Rightarrow \] \[\beta =0,\,\,1,\,\,2\] So, the points are \[(2,\,\,0),\,\,(2,\,\,1)\] and\[(2,\,\,2)\]. When \[\alpha =3\] \[{{\beta }^{2}}<4\alpha \] \[\Rightarrow \] \[{{\beta }^{2}}<12\] \[\Rightarrow \] \[\beta =0,\,\,1,\,\,2,\,\,3\] So, the points are\[(3,\,\,0),\,\,(3,\,\,1),\,\,(3,\,\,2)\]and\[(3,\,\,3)\]. Out of these points, \[(3,\,\,3)\] does not satisfy\[{{\alpha }^{2}}+{{\beta }^{2}}-16<0\]. Thus, the points lying in the region are \[(1,\,\,0),\,\,(1,\,\,1),\,\,(2,\,\,0),\,\,(2,\,\,1),\,\,(2,\,\,2),\,\,(3,\,\,0),\,\,(3,\,\,1)\]and (3,2).You need to login to perform this action.
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