Question

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