Question

If a function $y=f\left(x\right)$ is such that $f\left(1\right)=1$ and $x{\int }_0^x(1-x)f\left(x\right)dx$,

then $f\left(x\right)$ equals?

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

Answer 1

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

$x{\int }_0^x(1-z)f\left(z\right)dz={\int }_0^{x}zf\left(z\right)dz$
Differentiating both sides with respect to $x$ we have
$\Rightarrow {\int }_0^x(1-z)f\left(z\right)dz+x(1-x)f\left(x\right)=xf\left(x\right)$
$\Rightarrow {\int }_0^x(1-z)f\left(z\right)dz-x^{2}f\left(x\right)=0$
Again differentiating with respect to $x$ we have
$(1-x)f\left(x\right)-2xf\left(x\right)-x^2{f}^{\prime }\left(x\right)=0$
$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{1-3x}{x^2}=\frac{1}{x^2}-\frac{3}{x}$
$\Rightarrow e^{\log f\left(x\right)}=e^{-1/x-3\log x+k}$ raised at the base
$\therefore k=1,asf\left(1\right)=1$ $\Rightarrow f\left(x\right)=\frac{e^{1-1/x}}{x^3}$