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) \[i\,\left[ x-\frac{{{x}^{3}}}{3\,!}+\frac{{{x}^{5}}}{5\,!}-.....\infty \right]\]
Correct Answer: B
Solution :
We know that \[{{e}^{x}}=1+\frac{x}{1\ !}+\frac{{{x}^{2}}}{2\ !}+\frac{{{x}^{3}}}{3\ !}+\frac{{{x}^{4}}}{4\ !}+.....\infty \] \[{{e}^{ix}}=1+\frac{(ix)}{1\ !}+\frac{{{(xi)}^{2}}}{2\ !}+\frac{{{x}^{3}}{{i}^{3}}}{3\ !}+.......\] \[=1+\frac{ix}{1\ !}-\frac{{{x}^{2}}}{2\ !}-\frac{i{{x}^{3}}}{3\ !}+\frac{{{x}^{4}}}{4\ !}+\frac{i{{x}^{5}}}{5\ !}-\frac{{{x}^{6}}}{6\ !}+......\infty \] ......(i) and \[={{\log }_{e}}\left( \frac{1+\frac{1}{x}}{1-\frac{1}{x}} \right)=2\left\{ \frac{1}{x}+\frac{1}{3{{x}^{3}}}+\frac{1}{5{{x}^{5}}}+....... \right\}\] \[=1-\frac{xi}{1\ !}-\frac{{{x}^{2}}}{2\ !}+\frac{i{{x}^{3}}}{3\ !}+\frac{{{x}^{4}}}{4\ !}-\frac{i{{x}^{5}}}{5\ !}-\frac{{{x}^{6}}}{6\ !}+......\infty \] ......(ii) Now from (i) and (ii), we have \[\frac{{{e}^{ix}}+{{e}^{-ix}}}{2}=\left( 1-\frac{{{x}^{2}}}{2\ !}+\frac{{{x}^{4}}}{4\ !}-\frac{{{x}^{6}}}{6\ !}+.... \right)\].
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