If x=1+frac{1}{2 times 1 !}+frac{1}{4 times 2 | vedclass

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(A) Given,$x=1+\frac{1}{2 	imes 1 !}+\frac{1}{4 	imes 2 !}+\frac{1}{8 	imes 3 !}+\ldots$This is the expansion of $e^{1/2}$,so $x = e^{1/2}$.Squarin

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$e$

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(A) Given,$x=1+\frac{1}{2 	imes 1 !}+\frac{1}{4 	imes 2 !}+\frac{1}{8 	imes 3 !}+\ldots$This is the expansion of $e^{1/2}$,so $x = e^{1/2}$.Squaring both sides,we get $x^{2} = e$.Now,$y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$This is the expansion of $e^{x^{2}}$,so $y = e^{x^{2}}$.Substituting $x^{2} = e$,we get $y = e^{e}$.Taking $\log_{e}$ on both sides,$\log_{e} y = \log_{e} (e^{e}) = e \log_{e} e = e 	imes 1 = e$.

If $x=1+\frac{1}{2 \times 1 !}+\frac{1}{4 \times 2 !}+\frac{1}{8 \times 3 !}+\ldots$ and $y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$,then the value of $\log_{e} y$ is

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