A) 4
B) 3
C) 2
D) none of these
Correct Answer: B
Solution :
The equation is \[10(z\bar{z})-3i\{{{z}^{2}}-{{(\bar{z})}^{2}}\}-6=0\] | |
Or \[10({{x}^{2}}+{{y}^{2}})-3i(2x)(2iy)-6=0\] | |
\[\Rightarrow \]\[5({{x}^{2}}+{{y}^{2}})+6xy-8=0\] | ? (1) |
Let \[(r\cos \theta ,r\sin \theta )\]be a point on (1), then \[5{{r}^{2}}+6{{r}^{2}}\sin \theta \cos \theta -8=0\] |
\[\Rightarrow \]\[{{r}^{2}}=\frac{8}{5+3\sin 2\theta }\] |
Clearly, \[1\le {{r}^{2}}\le 4\Rightarrow 1\le \left| r \right|\le 2\] |
\[\therefore \]\[{{r}_{1}}={{\left| r \right|}_{\max }}=2\] and \[{{r}_{2}}={{\left| r \right|}_{\max }}=1\] |
\[\Rightarrow \]\[{{r}_{1}}+{{r}_{2}}=3\] |
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