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 \[{{a}^{2}}-3a+2=|z+\sqrt{2}|\,=\,|z+i\sqrt{2}+\sqrt{2}-i\sqrt{2}|\] \[\le |z+i\sqrt{2}|+\sqrt{2}|1-i|\] \[<{{a}^{2}}+2\] Þ \[-3a<0\,\,\Rightarrow a>0\] .....(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 \[|z+\sqrt{2}i|\]\[<{{a}^{2}}\] 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 \[\sqrt{{{(-\sqrt{2}-0)}^{2}}+{{(0+\sqrt{2})}^{2}}}<\sqrt{{{a}^{2}}-3a+2}\pm a\] Þ \[2\pm a<\sqrt{{{a}^{2}}-3a+2}\] Þ \[4+{{a}^{2}}\pm 4a<{{a}^{2}}-3a+2\] Þ\[-a<-2\]or \[7a<-2\]Þ \[a>2\]or \[a<-\frac{7}{2}\] But \[a>0\] from (i), therefore\[a>2\].
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