Question

Consider $f:R^{+}\to R$ such that $f(3)=1$ for $a\in R^{+}$ and $f(x)\cdot f(y)+f\left(\frac{3}{x}\right)f\left(\frac{3}{y}\right)=2f(xy)\forall x,y\in R^{+}$ then $f(97)$ can be

• A. $1$
• B. $-1$
• C. $2$
• D. $97$

Answer 1

Answer: (A)

Put $x=y=1$
$f^2(1)+f^2(3)=2f(1)$
$\Rightarrow (f(1)-1{)}^2=0\Rightarrow f(1)=1$
$f(x)f(1)+f\left(\frac{3}{x}\right)f(3)=2f(x)$
$\Rightarrow f(x)=f\left(\frac{3}{x}\right)\forall x>0$ $...(1)$
$\Rightarrow f(x)f\left(\frac{3}{x}\right)+f\left(\frac{3}{x}\right)f(x)=2f(3)$
$\Rightarrow f(x)f\left(\frac{3}{x}\right)=1$ $....(2)$
$\Rightarrow$ from $(1)$ and $(2)$
$f^2(x)=1\forall x>0$
Put $x=y=\sqrt{t}$
$f^2(\sqrt{t})+f^2\left(\frac{3}{\sqrt{t}}\right)=2f(t)$ or $f(t)>0$
$\Rightarrow f(x)=1\forall x>0$