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If $|x| < 1$ and $y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$, then $x$ is equal to :
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Verified Answer
The correct answer is:
$y+\frac{y^2}{2 !}+\frac{y^3}{3 !}+\ldots$
$y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$
$\Rightarrow \quad y=\log (1+x)$
$\Rightarrow \quad 1+x=e^y$
$\Rightarrow \quad 1+x=1+y+\frac{y^2}{2 !}+\ldots$
$\Rightarrow \quad x=y+\frac{y^2}{2 !}+\frac{y^3}{3 !}+\ldots$
$\Rightarrow \quad y=\log (1+x)$
$\Rightarrow \quad 1+x=e^y$
$\Rightarrow \quad 1+x=1+y+\frac{y^2}{2 !}+\ldots$
$\Rightarrow \quad x=y+\frac{y^2}{2 !}+\frac{y^3}{3 !}+\ldots$
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