A) \[{{x}^{2}}y=2x-y\]
B) \[{{x}^{2}}y=2x+y\]
C) \[x=2{{y}^{2}}-1\]
D) \[{{x}^{2}}y=2x+{{y}^{2}}\]
Correct Answer: A
Solution :
Given equation is, \[4\left[ {{x}^{2}}+\frac{{{x}^{6}}}{3}+\frac{{{x}^{10}}}{5}+... \right]={{y}^{2}}+\frac{{{y}^{4}}}{2}+\frac{{{y}^{6}}}{3}+....\] \[\Rightarrow \,\frac{4}{2}{{\log }_{e}}\left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)=-{{\log }_{e}}(1-{{y}^{2}})\] \[\Rightarrow \,{{\log }_{e}}\,{{\left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)}^{2}}={{\log }_{e}}\left( \frac{1}{1-{{y}^{2}}} \right)\]; \[\Rightarrow {{\left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)}^{2}}=\frac{1}{1-{{y}^{2}}}\] On simplification, we get\[{{x}^{2}}y=2x-y\].
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