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

Answer 1

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

Consider $f\left(x\right)=c{x}^{\frac{1-n}{n}}$

$\therefore f\left(tx\right)=c(tx{)}^{\frac{1-n}{n}}$

$\Rightarrow {\int }_0^{1}f\left(tx\right)dt={\int }_0^{1}c(tx{)}^{\frac{1-n}{n}}dt=c{x}^{\frac{1-n}{n}}{\int }_0^1{t}^{\frac{1-n}{n}}dt$

$=c{x}^{\frac{1-n}{n}}\frac{t^{\frac{1-n}{n}}+1}{\frac{1-n}{n}+1}{|{}_0^1=c{x}^{\frac{1-n}{n}}\frac{t^{\frac{1-n+n}{n}}}{\frac{1-n+n}{n}}|}_0^1=c{x}^{\frac{1-n}{n}}\frac{t^{\frac{1}{n}}}{\frac{1}{n}}{|}_0^1$

$=nc{x}^{\frac{1-n}{n}}(1-0)=nc{x}^{\frac{1-n}{n}}=nf\left(x\right)$

$\Rightarrow {\int }_0^{1}f\left(tx\right)dt=nf\left(x\right)$

Thus, $f\left(x\right)=c{x}^{\frac{1-n}{n}}$