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)

$\underset{0}{\overset{1}{\int }}f(tx)dt=nf(x)$
Let $tx=u\Rightarrow dt=\frac{du}{x}$
$\therefore \frac{1}{x}\underset{0}{\overset{x}{\int }}f(u)du\Rightarrow nf(x)\Rightarrow \underset{0}{\overset{x}{\int }}f(u)du=nxf(x)$
$f(x)=n\left[f(x)+x{f}^{\prime }(x)\right]\Rightarrow f(x)\left(\frac{1-n}{n}\right)=x{f}^{\prime }(x)$
$\int \frac{dx}{x}=\frac{n}{1-n}\int \frac{dy}{y}\Rightarrow \frac{1-n}{n}\ell nx=\ell ny+\ell nc$
$x^{\frac{1-n}{n}}=cy\Rightarrow y=c^{\prime }{x}^{\frac{1-n}{n}}$