A) 0
B) 1
C) 2
D) None of these
Correct Answer: C
Solution :
\[\frac{{{e}^{7x}}+{{e}^{3x}}}{{{e}^{5x}}}={{e}^{2x}}+{{e}^{-2x}}\] We know \[{{e}^{2x}}=1+\frac{2x}{1\ !}+\frac{{{(2x)}^{2}}}{2\ !}+\frac{{{(2x)}^{3}}}{3\ !}+........\infty \]......(i) \[{{e}^{-2x}}=1-\frac{2x}{1\ !}+\frac{{{(2x)}^{2}}}{2\ !}-\frac{{{(2x)}^{3}}}{3\ !}+\frac{{{(2x)}^{4}}}{4\ !}-.......\infty \] ......(ii) Adding (i) and (ii), we get \[{{e}^{2x}}+{{e}^{-2x}}=2\left[ 1+\frac{{{(2x)}^{2}}}{2\ !}+\frac{{{(2x)}^{4}}}{4\ !}+......\infty \right]\] Hence the constant term is 2.
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