Question

Let $f:R^{+}\to R$ be a differentiable function satisfying $f\left(x\right)=e+(1-x)\ln \left(\frac{x}{e}\right)+\underset{1}{\overset{x}{\int }}f\left(t\right)dt\forall x\in R^{+}$. If the area enclosed by the curve $g\left(x\right)=x(f\left(x\right)-e^x)$ lying in the fourth quadrant is $A$, then the value of $A^{-2}$ is

Answer 1

We have
$f\left(x\right)=e+(1-x)\ln \left(\frac{x}{e}\right)+\underset{1}{\overset{x}{\int }}f\left(t\right)d\left(t\right)\cdots \left(1\right)$
$\Rightarrow f\left(x\right)=e+(1-x)(\ln x-1)+\underset{1}{\overset{x}{\int }}f\left(t\right)d\left(t\right)$
Differentiating both sides with respect to $x$, we get
$f^{\prime }\left(x\right)=1-\ln x+(1-x)\frac{1}{x}+f\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)-f\left(x\right)=(\frac{1}{x}-\ln x)$
Multiplying both sides with $e^{-x},$ we get
$e^{-x}{f}^{\prime }\left(x\right)-e^{-x}f\left(x\right)=e^{-x}(\frac{1}{x}-\ln x)$
$\Rightarrow \frac{d}{dx}(e^{-x}\cdot f(x\left)\right)=e^{-x}(\frac{1}{x}-\ln x)$
Integrating both sides with respect to $x$, we get
$e^{-x}f\left(x\right)=\int e^{-x}(\frac{1}{x}-\ln x)dx$
$\Rightarrow e^{-x}f\left(x\right)=e^{-x}\ln x+\lambda$
$\Rightarrow f\left(x\right)=\ln x+\lambda e^x\cdots \left(2\right)$

Now, putting $x=1$ in equation $\left(1\right)$, we get
$f\left(1\right)=e\cdots \left(3\right)$
$\therefore$ Using $\left(2\right)$ and $\left(3\right)$, we get $\lambda =1$
Thus, $f\left(x\right)=e^x+\ln x$


Hence, $g\left(x\right)=x\ln x$
Now, $A=\left|\underset{0}{\overset{1}{\int }}x\ln xdx\right|=\frac{1}{4}$
$\Rightarrow A^{-2}=16$