JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 21

Swap $x$ and $y$. $2^xf(y)+4^yf(x)=2^yf(x)+4^xf(y)$, so $\dfrac{f(x)}{4^x-2^x}$ is constant: $f(x)=c\left(4^x-2^x\right)$, and $f(2)=12c=3$.

Derivative. $f'(x)=c\ln2\left(2\cdot4^x-2^x\right)$, so $\dfrac{f'(4)}{f'(2)}=\dfrac{512-16}{32-4}=\dfrac{124}7$.

Value $14\times\frac{124}7=248$.

Answer: 248