A) \[x=1-y+\frac{{{y}^{2}}}{2}-\frac{{{y}^{3}}}{3}+....\]
B) \[x=1+y+\frac{{{y}^{2}}}{2}+\frac{{{y}^{3}}}{3}+....\]
C) \[x=y-\frac{{{y}^{2}}}{2!}+\frac{{{y}^{3}}}{3!}-\frac{{{y}^{4}}}{4!}+....\]
D) \[x=y+\frac{{{y}^{2}}}{2!}+\frac{{{y}^{3}}}{3!}+\frac{{{y}^{4}}}{4!}+....\]
Correct Answer: D
Solution :
Given, \[y=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+....\] and \[|x|\,<1\] Then, \[y={{\log }_{e}}(1+x)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{e}^{y}}=1+x\Rightarrow \,\,{{e}^{y}}-1=x\] \[\Rightarrow \] \[\left( 1\frac{y}{11}+\frac{{{y}^{2}}}{2!}+\frac{{{y}^{3}}}{3!}+.... \right)-1=x\] \[\Rightarrow \] \[\frac{y}{1!}+\frac{{{y}^{2}}}{2!}+\frac{{{y}^{3}}}{3!}+...=x\]
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