Question

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

• A. 248
• B. 248.0
• C. 248.00

Answer 1

Answer: (A), (B), (C)

$\because f(x+y)=2^{x}f\left(y\right)+4^{y}f\left(x\right)\dots \left(i\right)$
$\therefore 2^{x}f\left(y\right)+4^{y}f\left(x\right)=2^{y}f\left(x\right)+4^{x}f\left(y\right)$
$(4^y-2^y)f\left(x\right)=(4^x-2^x)f\left(y\right)\dots \left(ii\right)$
$\frac{f\left(x\right)}{4^x-2^x}=\frac{f\left(y\right)}{4^y-2^y}=k\left(\text{Say}\right)$
$\therefore f\left(x\right)=k(4^x-2^x)$
$\because f\left(2\right)=3\text{then}k=\frac{1}{4}$
$\therefore f\left(x\right)=\frac{4^x-2^x}{4}$
$\therefore f^{\prime }\left(x\right)=\frac{4^x\ln 4-2^x\ln 2}{4}$
$f^{\prime }\left(x\right)=\frac{({2.4}^x-2^x)\ln 2}{4}$
$\therefore \frac{f^{\prime }\left(4\right)}{f^{\prime }\left(2\right)}=\frac{2\times 256-16}{2\times 16-4}$
$\therefore 14\times \frac{f^{\prime }\left(4\right)}{f^{\prime }\left(2\right)}=248$