Question

If $I_n=\int \frac{e^{(n+1)x}dx}{\left(1+e^x+\frac{e^{2x}}{2!}+\dots \dots +\frac{e^{nx}}{n!}\right)}={\lambda }_n\left(e^x-\ln \left(f_n(x)\right)\right)+C$ where $f_n(0)=1+\frac{1}{1!}+\frac{1}{2!}+\dots \dots +\frac{1}{n!}$ and $C$ is constant of integration and $g(x)=\underset{n\to \infty }{\mathrm{Lim}}\ln \left(f_n(x)\right)$, then find the number of real solutions of the equation $g(x)=4x^2$.

Answer 1

Answer: 3

Put $e^x=t$
$I_n=\int \frac{t^{n}dt}{\left(1+t+\frac{t^2}{2!}+\dots \dots +\frac{t^n}{n!}\right)}=n!\int \frac{1+t+\frac{t^2}{2!}+\dots \dots +\frac{t^n}{n!}-\left(1+t+\frac{t^2}{2!}+\dots \dots +\frac{t^{n-1}}{(n-1)!}\right)}{\left(1+t+\frac{t^2}{2!}+\dots \dots +\frac{t^n}{n!}\right)}dt$
$=n!\left(t-\int \frac{1+t+\frac{t^2}{2!}+\dots \dots +\frac{t^{n-1}}{(n-1)!}}{1+t+\frac{t^2}{2!}+\dots \dots +\frac{t^n}{n!}}dt\right)$
Let $1+t+\frac{t^2}{2!}+\dots \dots +\frac{t^n}{n!}=v;dv=1+t+\frac{t^2}{2!}+\dots ..+\frac{t^{n-1}}{(n-1)!}dt$
$\therefore I_n=n!\left(t-\ln \left(1+t+\frac{t^2}{2!}+\dots \dots +\frac{t^n}{n!}\right)\right)+c$
$=n!\left(e^x-\ln \left(1+e^x+\frac{e^{2x}}{2!}+\dots \dots +\frac{e^{nx}}{n!}\right)\right)+c$
$\therefore g(x)=\underset{n\to \infty }{\mathrm{Lim}}\ln \left(1+e^x+\frac{e^{2x}}{2!}+\dots \dots +\frac{e^{nx}}{n!}\right)=\ln \left(e^{e^x}\right)=e^x$
$\therefore$ Number of solutions of $e^x=x^2$ is 3