Question

If $\int x^3{e}^{2x}dx=\frac{e^{2x}}{8}f(x)+c$, then the sum of all the complex roots of $f(x)=1$ is

• A. $\frac{1}{2}$
• B. 3
• C. 1
• D. 2

Answer 1

Answer: (A)

Given, $\int x^3\cdot e^{2x}dx=\frac{e^{2x}}{8}f(x)+c$
By applying integration by parts, we get
$x^3\cdot \int e^{2x}dx-\int \left(\frac{d}{dx}{x}^3\int e^{2x}dx\right)dx)=\frac{e^{2x}}{8}+f(x)+c$
$x^3\cdot \frac{e^{2x}}{2}-\int 3x^2\cdot \frac{e^{2x}}{2}dx=\frac{e^{2x}}{8}f(x)+c$
$\frac{1}{2}{x}^3{e}^{2x}-\frac{3}{2}\left[x^2\cdot \frac{e^{2x}}{2}-\int 2x\cdot \frac{e^{2x}}{2}dx\right]=\frac{e^{2x}}{8}f(x)+c$
[Again applying integration by parts]
$\frac{1}{2}{x}^3{e}^{2x}-\frac{3}{4}{x}^2\cdot e^{2x}+\frac{3}{2}\int x\cdot e^{2x}dx=\frac{e^{2x}}{8}\int (x)+c$
$\frac{1}{2}{x}^3{e}^{2x}-\frac{3}{4}{x}^2\cdot e^{2x}+\frac{3}{2}\left[x\frac{e^{2x}}{2}-\int 1\cdot \frac{e^{2x}}{2}dx\right]=\frac{e^{2x}}{8}f(x)+c$
$\frac{1}{2}{x}^3{e}^{2x}-\frac{3}{4}{x}^2{e}^{2x}+\frac{3}{4}x{e}^{2x}-\frac{3}{4}\cdot \frac{e^{2x}}{2}=\frac{e^{2x}}{8}f(x)+c$
$\frac{e^{2x}}{8}\left[4x^3-6x^2+6x-3\right]+c_1=\frac{e^{2x}}{8}f(x)+c$
$\therefore$ On comparison, we get
$f(x)=4x^3-6x^2+6x-3$
But given,
$f(x)=1$
$4x^3-6x^2+6x-3=1$
$4x^3-6x^2+6x-4=0$
Divided by $2$ , we get
$2x^3-3x^2+3x-2=0$
$\Rightarrow 2\left(x^3-1\right)-3x(x-1)=0$
$\Rightarrow 2(x-1)\left(x^2+x+1\right)-3x(x-1)=0$
$\Rightarrow (x-1)\left[2x^2+2x+2-3x\right]=0$
$\Rightarrow (x-1)\left[2x^2-x+2\right]=0$
$x=1$ is real root.
So, sum of non-real complex root from the quadratic equation
$2x^2-x+2=0$ is $-\left(\frac{-1}{2}\right)=\frac{1}{2}$