Question

Let $p$ and $q$ are positive integers, $f$ is a function defined for positive numbers & attains only positive values such that $f[xf(y)]=x^a{y}^b$ then

• A. $b^2=a$
• B. $a=b$
• C. $a^2=b$
• D. None of these

Answer 1

Answer: (C)

Given $f(x\cdot f(y)=x^a{y}^b....(i)$
Replacing $x$ by $\frac{1}{f(y)}$, we have from (i)
$f\left(x\cdot \frac{1}{x}\right)={\left(\frac{1}{f\left(y\right)}\right)}^a\cdot y^b$
$\therefore f\left(1\right)=\frac{y^b}{{\left(f\left(y\right)\right)}^a}$
$\Rightarrow {\left(f\left(y\right)\right)}^a=\frac{y^b}{f\left(1\right)}$
$\Rightarrow f\left(y\right)=\frac{y^{b/a}}{{\left(f\left(y\right)\right)}^{1/a}}$ ( Putting $y=1)$
$\Rightarrow f(1)=1$
$\therefore f(1)=\frac{y^b}{(f(y){)}^a}=1$
$\Rightarrow (f(y){)}^a=y^b$
$\Rightarrow f(1)=\frac{y^b}{(f(y){)}^a}=1$
$\Rightarrow (f(y){)}^a=y^b$
$\Rightarrow f(y)=y^{b/a}$
Replacing $y$ as $x$, we have
$f(x)=x^{b/a}....(\ast )$
$\therefore f(x\cdot y^{b/a})=x^a{y}^b$ from (i)
Let $y^{b/a}=t$
$\Rightarrow y=t^{a/b}$
$\Rightarrow f(x\cdot t)=x^a{t}^a$
$\Rightarrow f(x)=x^a...(\ast \ast )$
Now from $(\ast )$ & $(\ast \ast )$, we have
$x^{b/a}=x^a$
$\Rightarrow \frac{a}{b}=\frac{1}{a}$
$\Rightarrow b=a^2$