A) \[\frac{1}{2}\]
B) 1
C) 0
D) none of these
Correct Answer: C
Solution :
\[\frac{{{e}^{4x}}-1}{{{e}^{2x}}}={{e}^{2x}}-{{e}^{-2x}}\] \[=\left( 1+\frac{2x}{1!}+\frac{{{(2x)}^{2}}}{2!}+... \right)\] \[-\left( 1-\frac{2x}{1!}+\frac{{{(2x)}^{2}}}{2!}-\frac{{{(2x)}^{3}}}{3!}+... \right)\] \[=2\left( \frac{2x}{1!}+\frac{{{(2x)}^{3}}}{3!}+... \right)\] \[\therefore \]The coefficient of \[{{x}^{2}}\]in the above series is 0.
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