A) \[a>2\]
B) \[a=2\]
C) \[a<2\]
D) None of these
Correct Answer: A
Solution :
| If \[z=x+iy\]is a complex number satisfying the given conditions then |
| \[{{a}^{2}}-3a+2=\,|z+\sqrt{2}|\,=\,|z+i\sqrt{2}+\sqrt{2}-i\,\sqrt{2}|\]\[\le \,|z+i\sqrt{2}|+\sqrt{2}|\,1-i|<{{a}^{2}}+2\] |
| \[\Rightarrow \,\,-\,3a<0\Rightarrow a>0\] ?..(i) |
| Since \[|z+\sqrt{2}|={{a}^{2}}-3a+2\]represents a circle with centre at \[A(-\sqrt{2},0)\] and radius \[\sqrt{{{a}^{2}}-3a+2},\] and \[|z+i\sqrt{2}|\,<{{a}^{2}}\] represents the interior of the circle with centre at \[B\,(0,\sqrt{2})\] and radius a. |
| Therefore there will be a complex number satisfying the given condition and the given inequality if the distance AB is less than the sum or difference of the radii of the two circles, i.e., if \[\sqrt{{{(-\sqrt{2}-0)}^{2}}+{{(0+\sqrt{2})}^{2}}}<\sqrt{{{a}^{2}}-3a+2\pm }\,a\] |
| \[\Rightarrow \,2\pm a<\sqrt{{{a}^{2}}-3a+2}\Rightarrow 4+{{a}^{2}}\pm 4a<{{a}^{2}}-3a+2\]\[\Rightarrow \,-a<-2\] or \[7a<-2\,\,\Rightarrow \,a>2\] or \[a<-\frac{2}{7}\] |
| But \[a>0\] from (i), therefore\[a>2\]. |
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