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(A) Given that $f: R \rightarrow R$ is an injective function satisfying $f(x) f(y) = f(x+y)$ for all $x, y \in R$. This functional equation is satisfi

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$AP$ always

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(A) Given that $f: R ightarrow R$ is an injective function satisfying $f(x) f(y) = f(x+y)$ for all $x, y \in R$. This functional equation is satisfied by the exponential function $f(x) = a^x$ for some $a > 0, a 
eq 1$. Since $f(x), f(y), f(z)$ are in $G$.$P$.,we have $(f(y))^2 = f(x) \cdot f(z)$. Substituting $f(x) = a^x$, we get $(a^y)^2 = a^x \cdot a^z$. This simplifies to $a^{2y} = a^{x+z}$. Equating the exponents, we get $2y = x+z$. This condition implies that $x, y, z$ are in $A$.$P$.

Let $f: R \rightarrow R$ be such that $f$ is injective and $f(x) f(y) = f(x+y)$ for $\forall x, y \in R$. If $f(x), f(y), f(z)$ are in $G$.$P$.,then $x, y, z$ are in:

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