log_e [(1 + x)^{1 + x} (1 - x)^{1 - x}] = | vedclass

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(C) Let $f(x) = \log_e [(1 + x)^{1 + x} (1 - x)^{1 - x}]$.Using the property $\log(ab) = \log a + \log b$,we get:$f(x) = (1 + x) \log_e(1 + x) + (1 -

Vedclass · Accepted Answer

$2 \left[ \frac{x^2}{1 \cdot 2} + \frac{x^4}{3 \cdot 4} + \frac{x^6}{5 \cdot 6} + \dots \infty \right]$

Vedclass · Answer

(C) Let $f(x) = \log_e [(1 + x)^{1 + x} (1 - x)^{1 - x}]$.Using the property $\log(ab) = \log a + \log b$,we get:$f(x) = (1 + x) \log_e(1 + x) + (1 - x) \log_e(1 - x)$.Using the logarithmic series expansions $\log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ and $\log_e(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots$:$f(x) = (1 + x)(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots) + (1 - x)(-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots)$.Expanding the terms:$f(x) = (x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots) + (x^2 - \frac{x^3}{2} + \frac{x^4}{3} - \dots) + (-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots) + (x^2 + \frac{x^3}{2} + \frac{x^4}{3} + \dots)$.Grouping like terms:$f(x) = 2(x^2(1 - \frac{1}{2}) + x^4(\frac{1}{3} - \frac{1}{4}) + x^6(\frac{1}{5} - \frac{1}{6}) + \dots)$.$f(x) = 2 [\frac{x^2}{1 \cdot 2} + \frac{x^4}{3 \cdot 4} + \frac{x^6}{5 \cdot 6} + \dots]$.Thus,the correct option is $C$.

$\log_e [(1 + x)^{1 + x} (1 - x)^{1 - x}] = $

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