Q72 / 120 · VITEEE 2013 · Mathematics
In the expansion of $\frac{a+b x}{e^{x}}$, the coefficient of $x^{r}$ is
- A$\frac{a-b}{r !}$
- B$\frac{a-b r}{r !}$
- Correct answerC$(-1)^{r} \frac{a-b r}{r !}$
- DNone of these
Solution
$\begin{array}{l}
(a+b x) e^{-x} \\
=(a+b x) \\
\quad\left\{1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots+(-1)^{n} \frac{x^{n}}{n !}+\ldots\right\}
\end{array}$
$\therefore$ The coefficient of $x^{\mathrm{r}}=\mathrm{a}$.
$\frac{(-1)^{\mathrm{r}}}{\mathrm{r} !}+\mathrm{b} \frac{(-1)^{\mathrm{r}-1}}{(\mathrm{r}-1) !}=\frac{(-1)^{\mathrm{r}}}{\mathrm{r} !}(\mathrm{a}-\mathrm{br})$
Hence, the correct answer is option: 'C'.