Let $f: R \rightarrow R$ satisfy $f(x+y)=2^{x} f(y)+4^{y} f(x)$, $\forall x, y \in R$. If $f(2)=3$,

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Let $f: R \rightarrow R$ satisfy $f(x+y)=2^{x} f(y)+4^{y} f(x)$, $\forall x, y \in R$. If $f(2)=3$, then $14 \cdot \frac{f^{\prime}(4)}{f^{\prime}(2)}$ is equal to

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(248) : We have, $f(x+y)=2^{x} f(y)+$ $$\begin{equation*}4^{y} f(x), \forall x, y \in R 	ag{i} \end{equation*}$$ $$ 3=2 f(1)+4 f(1) $$ $\Rightarrow \quad f(1)=\frac{1}{2}$\ $f(3)=f(2+1)=2^{2} f(1)+4^{1} f(2)$ $$ =4 	imes 3+4 	imes \frac{1}{2}=14 $$ Put $x ightarrow y$ and $y ightarrow x$ $$\begin{equation*}f(y+x)=2^{y} f(x)+4^{x} f(y) 	ag{ii} \end{equation*}$$ Subtract equation (ii) from (i)\ $0=2^{x} f(y)+4^{y} f(x)-2^{y} f(x)-4^{x} f(y)$\ $-f(x)\left(4^{y}-2^{y}ight)=f(y)\left(2^{x}-4^{x}ight)$ $$ \frac{f(x)}{4^{x}-2^{x}}=\frac{f(y)}{4^{y}-2^{y}}=K $$ $f(x)=K\left(4^{x}-2^{x}ight) ; f(y)=K\left(4^{y}-2^{y}ight)$\ Given $f(2)=3 ; 3=K\left(4^{2}-2^{2}ight) \Rightarrow K=\frac{1}{4}$\ $f(x)=\frac{1}{4}\left(4^{x}-2^{x}ight) ;$\ $f^{\prime}(x)=\frac{1}{4}\left(4^{x} \log 4-2^{x} \log 2ight)$\ $f^{\prime}(4)=\frac{1}{4}\left(4^{4} \log 4-2^{4} \log 2ight)$\ $f^{\prime}(2)=\frac{1}{4}\left(4^{2} \log 4-2^{2} \log 2ight)$\ So, $14 \cdot \frac{f^{\prime}(4)}{f^{\prime}(2)}=\frac{14 \cdot \log 2\left(4^{4} 	imes 2-2^{4}ight)}{\log 2\left(4^{2} 	imes 2-2^{2}ight)}$\ $=14 \cdot \frac{496}{28}=248$\