Question

Let $f:(0,\infty )\to (0,\infty )$ be a differentiable function satisfying, $x\underset{0}{\overset{x}{\int }}(1-t)f\left(t\right)dt=\underset{0}{\overset{x}{\int }}tf\left(t\right)dt\forall x\in R^{+}$ and $f\left(1\right)=1.$ Then $f\left(x\right)$ can be

• A. $\frac{1}{x}{e}^{(1-\frac{1}{x})}$
• B. $\frac{1}{x^2}{e}^{(1-\frac{1}{x})}$
• C. $\frac{1}{x^2}{e}^{(1+\frac{1}{x})}$
• D. $\frac{1}{x^3}{e}^{(1-\frac{1}{x})}$

Answer 1

The correct option is D $\frac{1}{x^3}{e}^{(1-\frac{1}{x})}$

We have $x\underset{0}{\overset{x}{\int }}(1-t)f\left(t\right)dt=\underset{0}{\overset{x}{\int }}tf\left(t\right)dx$
Differentiating both sides with respect to $x,$ we get
$x(1-x)f\left(x\right)+\underset{0}{\overset{x}{\int }}(1-t)f\left(t\right)dt=xf\left(x\right)$
$\Rightarrow x^{2}f\left(x\right)=\underset{0}{\overset{x}{\int }}(1-t)f\left(t\right)dt$
Differentiating both sides with respect to $x$ again, we get

$x^2{f}^{\prime }\left(x\right)+2xf\left(x\right)=(1-x)f\left(x\right)$
$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{1-3x}{x^2}$
$\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int \frac{1-3x}{x^2}dx\Rightarrow \ln f\left(x\right)=-\frac{1}{x}-3\ln x+\ln c\{\because f(x)>0\}\Rightarrow \ln \left[\frac{f\left(x\right)}{c}\right]=\frac{-1}{x}-3\ln x$
Given, $f\left(1\right)=1$
$\Rightarrow \ln \left[\frac{1}{c}\right]=-1\Rightarrow c=e\Rightarrow \ln \left(\frac{f\left(x\right)x^3}{e}\right)=-\frac{1}{x}\Rightarrow f\left(x\right)=\frac{1}{x^3}{e}^{(1-\frac{1}{x})}$