Question

Given $f ( x )$ is a polynomial function of $x$, satisfying $f ( x ) \cdot f ( y )= f ( x )+ f ( y )+ f ( xy )-2$ and that $f (2)=5$. Then $f (3)$ is equal to

• A. 10
• B. 24
• C. 15
• D. none

Answer 1

Answer: (A)

$\text { Given } f ( x ) \cdot f ( y )= f ( x )+ f ( y )+ f ( xy )-2 $
$\text { put } y =1 / x$
$ f ( x ) \cdot f \left(\frac{1}{ x }\right)= f ( x )+ f \left(\frac{1}{ x }\right)+ f (1)-2 $
$\text { put } x =1, y =1 $
$ f (1) \cdot f (1)=2 f (1)+ f (1)-2 $
$f ^2(1)-3 f (1)+2=0 $
$( f (1)-1)( f (1)-2)=0 \Rightarrow f (1)=1 \text { or } f (1)=2 $
$\text { but } f (1) \neq 1 \Rightarrow f (1)=2 (\text { as in this case } f ( x )=1 \text { for all } x )$
Hence $f(x) \cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$
Let $f(x)=x^n+1 $
$f (2)=2^{ n }+1=5 \Rightarrow n =2 $
$\therefore f ( x )= x ^2+1 \Rightarrow f (3)=10$