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Question: \[\log_{e}{}\lbrack(1 + x)^{1 + x}(1 - x)^{1 - x}\rbrack =\]...

log⁡e[(1+x)1+x(1−x)1−x]=\log_{e}{}\lbrack(1 + x)^{1 + x}(1 - x)^{1 - x}\rbrack =

A

x22+x44+x66+....∞\frac{x^{2}}{2} + \frac{x^{4}}{4} + \frac{x^{6}}{6} + ....\infty

B

x21.2+x43.4+x65.6+....∞\frac{x^{2}}{1.2} + \frac{x^{4}}{3.4} + \frac{x^{6}}{5.6} + ....\infty

C

2[x21.2+x43.4+x65.6+..∞]2\left\lbrack \frac{x^{2}}{1.2} + \frac{x^{4}}{3.4} + \frac{x^{6}}{5.6} + ..\infty \right\rbrack

D

None of these

Answer

2[x21.2+x43.4+x65.6+..∞]2\left\lbrack \frac{x^{2}}{1.2} + \frac{x^{4}}{3.4} + \frac{x^{6}}{5.6} + ..\infty \right\rbrack

Explanation

Solution

−1- 1

e7x+e3xe5x\frac{e^{7x} + e^{3x}}{e^{5x}}

21!+2+42!+2+4+63!+....∞=\frac{2}{1!} + \frac{2 + 4}{2!} + \frac{2 + 4 + 6}{3!} + ....\infty =

e2ee2e

3e3e

[1+12!+14!+....∞]2−[1+13!+15!+.....∞]2=\left\lbrack 1 + \frac{1}{2!} + \frac{1}{4!} + ....\infty \right\rbrack^{2} - \left\lbrack 1 + \frac{1}{3!} + \frac{1}{5!} + .....\infty \right\rbrack^{2} =.