Question

In the expansion of $\dfrac{{{e^{7x}} + {e^{3x}}}}{{{e^{5x}}}}$ , the constant term is
A) $0$
B) $1$
C) $2$
D) None of these

Answer 1

Hint: The exponential function $f(x) = {e^x}$ is the unique function which is equal to its own derivative , with the initial value $f(0) = {e^0} = 1$ . Let $a,b,c$ are real numbers and the function , $\dfrac{{{e^{ax}} + {e^{bx}}}}{{{e^{cx}}}}$ is expressed as ${e^{(a - c)x}} + {e^{(b - c)x}}$ , then use the expansion of ${e^x}$ and find the required answer . The expansion of $f(x) = {e^x}$ is an infinite polynomial or power series.

Complete step by step answer:
From the given data we get $\dfrac{{{e^{7x}} + {e^{3x}}}}{{{e^{5x}}}}$
Separate the denominator and we get
$ = \dfrac{{{e^{7x}}}}{{{e^{5x}}}} + \dfrac{{{e^{3x}}}}{{{e^{5x}}}}$
We know that $\dfrac{1}{{{e^x}}} = {e^{ - x}}$ , we use this in the above equation and we get
$ = {e^{7x}} \times {e^{ - 5x}} + {e^{3x}} \times {e^{ - 5x}}$
$ = {e^{(7x - 5x)}} + {e^{(3x - 5x)}}$
Calculate and we get
$ = {e^{2x}} + {e^{ - 2x}}$
Now we expand the above exponential function and we get
$ = 1 + \dfrac{{2x}}{{1!}} + \dfrac{{4{x^2}}}{{2!}} + ...... + 1 - \dfrac{{2x}}{{1!}} + \dfrac{{4{x^2}}}{{2!}} + ......$
Now arrange and we get
$ = 2 + 2\left( {\dfrac{{4{x^2}}}{{2!}} + \dfrac{{16{x^4}}}{{4!}} + ......} \right)$
Now the above equation , we will see that $2$ is the constant term .
Therefore option (C) is correct.

Note:
We know the rule of sum of power of a function . If there are two exponential functions ${e^{3a}}$ and ${e^{3b}}$ then if we multiply them we add the power of the exponential function . i.e., ${e^{3a}} \times {e^{3b}} = {e^{(3a + 3b)}}$ . If we have two exponential function like ${e^{ax}}$ and ${e^{ - bx}}$ then we get when we multiply them ${e^{ax}} \times {e^{ - bx}} = {e^{(ax - bx)}}$ . We have to know about the expansion of the exponential function of ${e^x}$ is $1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ......$