0

If $f:(0, \infty)\to(0, \infty)$ is an into function satisfying $f(xf(y)) = x^2y^a, (a\in\mathbb R)$, then find the value of $a$ and the number of solutions of $2f(x) = e^x$.

My approach:

$f(x(f(y))) = x^2 y^a$

$y=x$

$f(x(f(x))) = x^2 x^a$

$f(xy) = x^2 x^a$, (as $f(x) = y$)

$y = 1$

$f(x) = x^a x^2$

This seems to be incorrect? Where have I gone wrong? How do I solve this?

Arctic Char
  • 16,007
Michael Jackson
  • 41
  • 4

1 Answers1

2

We have $f: (0, \infty)\rightarrow (0, \infty)$ such that $f(x f(y)) = x^2 y^a~\forall x, y\in(0, \infty)$.

For $x\leftarrow 1$:

  1. $f(f(y)) = y^a~\forall y\in(0, \infty)$

For $y\leftarrow f(y)$:

  1. $f(x f(f(y))) = x^2 f(y)^a~\forall x, y\in(0, \infty)$

By replacing 1. in 2. we have

  1. $f(x y^a) = x^2 f(y)^a~\forall x, y\in(0, \infty)$

In 3., for $y\leftarrow 1$, we have

  1. $f(x) = x^2 f(1)^a~\forall x\in(0, \infty)$

In 4., for $x\leftarrow 1$, we have

  1. $f(1) = f(1)^a$

As $f(1)\neq 0$, we shall have $f(1)^{1-a} = 1$, therefore $f(1) = 1$ or $a = 1$:

If $a = 1$, then, replacing 3. in the original equation, we have

$\begin{aligned} f(x f(y)) &= x^2 y^a&\forall x, y\in(0, \infty)&\implies\\ f(x y^2 f(1)) &= x^2 y&\forall x, y\in(0, \infty)&\implies\\ x^2 y^4 f(1)^3 &= x^2 y&\forall x, y\in(0, \infty)&\implies\\ f(1)^3 &= y^{-3}&\forall y\in(0, \infty)&\implies\\ f(1) &= y^{-1}&\forall y\in(0, \infty) \end{aligned}$

which is clearly absurd, therefore $f(1) = 1$, so, by 4., we have $f(x) = x^2~\forall x\in(0, \infty)$, so

$$f(xf(y)) = x^2 y^a~\forall x, y\in(0, \infty)\implies x^2 y^4 = x^2 y^a~\forall x, y\in(0, \infty)$$

therefore $a = 4$.

Alma Arjuna
  • 3,759