Question

Let $\int e^x(x^3+6x^2+6x+\alpha )dx=e^{x}f\left(x\right)+C,$ where $f\left(0\right)=\alpha$ and $\int e^x\left(f\right(x)+\beta )dx=e^{x}g\left(x\right)+K$ for constants of integration $C,K.$ If $g\left(0\right)=\alpha +\beta ,$ then the number of solution(s) of the equation $e^{x^2}\left(f\right(x)-g(x\left)\right)=1$ is

• A. $2$ if $\beta >0$
• B. $1$ if $\beta =-1$
• C. $2$ if $\beta =-2$
• D. $2$ if $\beta <-1$

Answer 1

Answer: (A), (B)

$1$ if $\beta =-1$

$I=\int e^x(x^3+6x^2+6x+\alpha )dx$
$=\int e^x(\underset{f\left(x\right)}{\underset{}{x^3+3x^2}}+\underset{f^{\prime }\left(x\right)}{\underset{}{(3x^2+6x)}})dx+\alpha \int e^{x}dx$
$=e^x(x^3+3x^2+\alpha )+C$
$\therefore f\left(x\right)=x^3+3x^2+\alpha$

and $\int e^x\left(f\right(x)+\beta )dx$
$=\int e^x(x^3+3x^2+\alpha +\beta )dx$
$=e^x(x^3+\alpha +\beta )+K$
$\therefore g\left(x\right)=x^3+\alpha +\beta$

Now, $e^{x^2}\left(f\right(x)-g(x\left)\right)=1$
$\Rightarrow e^{x^2}(x^3+3x^2+\alpha -x^3-\alpha -\beta )=1$
$\Rightarrow e^{-x^2}=3x^2-\beta$

From the graph,
If $\beta \ge 0,$ then number of solutions is $2.$
If $\beta =-1,$ then number of solution is $1.$
If $\beta <-1,$ then number of solution is $0.$