A) \[\frac{x}{1-{{x}^{2}}}\]
B) \[\frac{2x}{1+{{x}^{2}}}\]
C) \[\frac{1-{{x}^{2}}}{2x}\]
D) None of these
Correct Answer: B
Solution :
Given that, \[y+\frac{{{y}^{3}}}{3}+\frac{{{y}^{5}}}{5}+....+\infty =2\left[ x+\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}+...\infty \right]\] \[\Rightarrow \] \[\frac{\log (1+y)-\log (1-y)}{2}\] \[=2\left[ \frac{\log (1+x)-\log (1-x)}{2} \right]\] \[\Rightarrow \] \[\frac{1+y}{1-y}={{\left( \frac{1+x}{1-x} \right)}^{2}}\] Applying componendo and dividendo, we get \[\frac{2y}{2}=\frac{{{(1+x)}^{2}}-{{(1-x)}^{2}}}{{{(1+x)}^{2}}+{{(1-x)}^{2}}}\] \[\Rightarrow \] \[y=\frac{4x}{2+2{{x}^{2}}}\] \[\Rightarrow \] \[y=\frac{2x}{1+{{x}^{2}}}\]
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