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Question

Question: \[\log_{e}\sqrt{\frac{1 + x}{1 - x}} =\]...

loge1+x1x=\log_{e}\sqrt{\frac{1 + x}{1 - x}} =

A

loge12\log_{e}\frac{1}{2}

B

2[x+x33+x55+.....]2\left\lbrack x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + .....\infty \right\rbrack

C

2[x2+x44+x66+.....]2\left\lbrack x^{2} + \frac{x^{4}}{4} + \frac{x^{6}}{6} + .....\infty \right\rbrack

D

None of these

Answer

loge12\log_{e}\frac{1}{2}

Explanation

Solution

eabrr!\frac{e^{a}b^{r}}{r!}

ea+bre^{a + b^{r}}