Let f: R rightarrow R satisfy f(x+y)=2^{x} f( | vedclass

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(C) Given $f(x+y)=2^{x} f(y)+4^{y} f(x)$.Setting $y=2$,we get $f(x+2)=2^{x} f(2)+4^{2} f(x) = 3 \cdot 2^{x} + 16 f(x)$.Differentiating with respect to

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$248$

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(C) Given $f(x+y)=2^{x} f(y)+4^{y} f(x)$.Setting $y=2$,we get $f(x+2)=2^{x} f(2)+4^{2} f(x) = 3 \cdot 2^{x} + 16 f(x)$.Differentiating with respect to $x$,we get $f^{\prime}(x+2) = 3 \cdot 2^{x} \ln 2 + 16 f^{\prime}(x)$.For $x=2$,$f^{\prime}(4) = 3 \cdot 2^{2} \ln 2 + 16 f^{\prime}(2) = 12 \ln 2 + 16 f^{\prime}(2)$ ... $(i)$.Alternatively,setting $x=2$ in the original equation,$f(2+y)=2^{2} f(y)+4^{y} f(2) = 4 f(y) + 3 \cdot 4^{y}$.Differentiating with respect to $y$,we get $f^{\prime}(2+y) = 4 f^{\prime}(y) + 3 \cdot 4^{y} \ln 4 = 4 f^{\prime}(y) + 6 \cdot 4^{y} \ln 2$.For $y=2$,$f^{\prime}(4) = 4 f^{\prime}(2) + 6 \cdot 4^{2} \ln 2 = 4 f^{\prime}(2) + 96 \ln 2$ ... $(ii)$.Equating $(i)$ and $(ii)$,$12 \ln 2 + 16 f^{\prime}(2) = 4 f^{\prime}(2) + 96 \ln 2$.$12 f^{\prime}(2) = 84 \ln 2 \implies f^{\prime}(2) = 7 \ln 2$.Substituting into $(ii)$,$f^{\prime}(4) = 4(7 \ln 2) + 96 \ln 2 = 28 \ln 2 + 96 \ln 2 = 124 \ln 2$.Finally,$14 \cdot \frac{f^{\prime}(4)}{f^{\prime}(2)} = 14 \cdot \frac{124 \ln 2}{7 \ln 2} = 2 \cdot 124 = 248$.

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

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