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)

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