Question

A function $f\left(x\right)$ satisfying ${\int }_0^{1}f\left(tx\right)dt=nf\left(x\right)$, where $x>0$ is

• A. $f\left(x\right)=c.x^{\frac{1-n}{n}}$
• B. $f\left(x\right)=c.x^{\frac{n}{n-1}}$
• C. $f\left(x\right)=c.x^{\frac{1}{n}}$
• D. $f\left(x\right)=c.x^{(1-n)}$

Answer 1

The correct option is A $f\left(x\right)=c.x^{\frac{1-n}{n}}$

${\int }_0^{1}f\left(tx\right)dt={\int }_0^1\frac{f\left(tx\right)d\left(tx\right)}{x}=\frac{1}{x}{\int }_0^{x}f\left(k\right)dk=nf\left(x\right)$

$\Rightarrow {\int }_0^{x}f\left(k\right)dk=nxf\left(x\right)$

On differentiating the above equation,

$f\left(x\right)=nf\left(x\right)+nx{f}^{{}^{\prime }}\left(x\right)$

Let $y=f\left(x\right)$

$\Rightarrow y-ny=nx\frac{dy}{dx}$

$\Rightarrow \frac{(1-n)dx}{nx}=\frac{dy}{y}$

On Integrating Both sides,

$\ln y=\frac{1-n}{n}\ln x+\ln c\Rightarrow y=f\left(x\right)=c{x}^{\frac{1-n}{n}}$

So, Option A is correct answer.