Question

A function $f(x)$ satisfying $\underset{0}{\overset{1}{\int }}f(tx)dt=nf(x)$, where $x>0$, is

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

Answer 1

Answer: (A)

$\underset{0}{\overset{1}{\int }}f(tx)dt=n\cdot f(x)$
put $tx=y\Rightarrow dt=\frac{1}{x}dy$
$\therefore \frac{1}{x}\underset{0}{\overset{x}{\int }}f(y)dy=nf(x)$
$\therefore \underset{0}{\overset{x}{\int }}f(y)dy=x\cdot n\cdot f(x)$
Differentiating
$f(x)=n\left[f(x)+x{f}^{\prime }(x)\right]$
$f(x)(1-n)=nx{f}^{\prime }(x)$
$\therefore \frac{f^{\prime }(x)}{f(x)}=\frac{1-n}{nx}$
$\text{ Integrating }\ln f(x)=\left(\frac{1-n}{n}\right)\ln cx=\ln (cx{)}^{\frac{1-n}{n}}$
$\therefore f(x)=c\cdot x^{\frac{1-n}{n}}$