- Thread starter
- #1

- Thread starter jacks
- Start date

- Thread starter
- #1

- Feb 5, 2012

- 1,621

Hi jacks,If $f$ is a Real valued function on the set of real no. such that for any real $a$ and $b$ and $f(af(b)) = ab$. Then $f(2012) = $

\(\mbox{By substituting, }a=1\mbox{ we get, }f[f(b)]=b\mbox{ for each }b\in\Re\,.\mbox{ Therefore the inverse of }f\mbox{ is itself.}\)

\(\mbox{Suppose there exist a real number }n\mbox{ such that, }f(n)=1\,.\mbox{ Then, }\)

\[f\left[mf\left(\frac{1}{m}\right)\right]=1=f(n)\mbox{ where }m\in\Re\mbox{ and }m\neq 0\]

\[\Rightarrow mf\left(\frac{1}{m}\right)=n\]

\[\Rightarrow f\left(\frac{1}{m}\right)=\frac{n}{m}~~~~~~~(1)\]

\[\Rightarrow f\left(\frac{1}{m}\right)=f\left[nf\left(\frac{1}{m}\right)\right]\]

\[\Rightarrow \frac{1}{m}=nf\left(\frac{1}{m}\right)\]

\[\Rightarrow f\left(\frac{1}{m}\right)=\frac{1}{mn}~~~~~~~~(2)\]

By (1) and (2);

\[n=\pm 1\]

Therefore \(n\) can be \(1\mbox{ or }-1\) depending on the function \(f\). Hence,

If the function \(f\) is defined such that, \(f(1)=1,\)

\[f(2012)=f(2012f(1))=2012\]

If the function \(f\) is defined such that, \(f(-1)=1,\)

\[f(2012)=f(2012f(-1))=-2012\]

- Jan 26, 2012

- 183

First I'll show that $f(1) \ne 0$

Sub $a = 0$ into the functional equation giving $f(0) = 0$

Then sub $a = b= 1$ giving $f(f(1)) =1$. If $f(1) = 0$ then$ f(0) = 1$ but $f(0) = 0$ contradiction.

Now set $b = 1$ so $f(a f(1)) = a$. If we let $f(1) = k$ then we have $f(ka) = a$ or $f(x) = \dfrac{x}{k}$

Returning back to the original function equation gives

$f(af(b)) = f\left(\dfrac{ab}{k}\right) = \dfrac{ab}{k^2} = ab$ giving $k = \pm 1$

Thus, $f(x) = \pm x$