Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. $97$

Answer 1

Answer: (A)

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