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Q.
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
Relations and Functions - Part 2
Solution:
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
$f(x) = x^{b/a} \,....(*)$
$\therefore f(x \cdot y^{b/a}) = x^a \,y^b$ from (i)
Let $y^{b/a} = t $
$\Rightarrow y = t^{a/b}$
$\Rightarrow f( x \cdot t) = x^a t^a$
$\Rightarrow f(x) = x^a\,\,...(**)$
Now from $(*)$ & $(**)$, we have
$x^{b/a} = x^a $
$\Rightarrow \frac{a}{b} = \frac{1}{a}$
$\Rightarrow b = a^2$