Question

Consider a function $f\left(x\right)$ where $f:R^{+}\to R$ such that $f\left(3\right)=1$
and $f\left(x\right)f\left(y\right)+f\left(\frac{3}{x}\right)f\left(\frac{3}{y}\right)=2f\left(xy\right)$ $\forall x,y\in R^{+}$, then

• A. $f\left(x\right)$ is a cubic polynomial
• B. $f\left(x\right)$ is an even function
• C. $\frac{f\left(2\right)}{f\left(1\right)+f\left(2\right)}=\frac{1}{2}$
• D. If $g\left(x\right)=x^2,$ then number of Solutions of $f\left(x\right)=g\left(x\right)$ is $1$.

Answer 1

Answer: (C), (D)

If $g\left(x\right)=x^2,$ then number of Solutions of $f\left(x\right)=g\left(x\right)$ is $1$.

$f\left(x\right)f\left(y\right)+f\left(\frac{3}{x}\right)f\left(\frac{3}{y}\right)=2f\left(xy\right),f\left(3\right)=1$
Let $x=y=1$
$\Rightarrow f(1{)}^2+1=2f(1)\Rightarrow (f\left(1\right)-1{)}^2=0\Rightarrow f\left(1\right)=1$
Let $y=1$
$\Rightarrow f\left(x\right)+f\left(\frac{3}{x}\right)=2f\left(x\right)\Rightarrow f\left(x\right)=f\left(\frac{3}{x}\right)$
Let $y=\frac{3}{x}$
$\Rightarrow f\left(x\right)\cdot f\left(\frac{3}{x}\right)=1\Rightarrow f\left(x\right)=1$
It is a constant function.
$\frac{f\left(2\right)}{f\left(1\right)+f\left(2\right)}=\frac{1}{1+1}=\frac{1}{2}$

Here $f(-x)=f\left(x\right)$ but $x\in R^{+}$. So, $f\left(x\right)$ will not be an even function.

Now,
$f\left(x\right)=g\left(x\right)\Rightarrow x^2=1\Rightarrow x=1(\because x>0)$
Hence, it has only one solution.