Question

If $i = \sqrt{- 1}$, then $\frac{e^{xi} + e^{- xi}}{2} =$

• A. $1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + .....\infty$
• B. $1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - .....\infty$
• C. $x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + ....\infty$
• D. $i\left\lbrack x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - .....\infty \right\rbrack$

Answer 1

Answer: (B)

$1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - .....\infty$

Answer 2

We know that $\log_{e}\left( \frac{n - 1}{n} \right)$

$\log_{e}\left( \frac{n}{n - 1} \right)$

$\log_{4}2 - \log_{8}2 + \log_{16}2....$ ......(i)

and $e^{2}$

$\log_{e}2$ ......(ii)

Now from (i) and (ii), we have

$\log_{e}3 - 2$.