Question:

Step 1: Find $f(0)$ by setting $x = y = 0$: $$f(0) = 2^0 f(0) + 4^0 f(0) = f(0) + f(0) = 2f(0) \implies f(0) = 0$$ Step 2: Differentiate $f(x+y) = 2^x f(y) + 4^y f(x)$ with respect to $x$, treating $y$ as constant: $$f'(x+y) = \ln 2 \cdot 2^x f(y) + 4^y f'(x)$$ Set $x = 0$: $$f'(y) = \ln 2 \cdot f(y) + 4^y f'(0) = \ln 2 \cdot f(y) + 4^y \ln 2$$ $$f'(y) = \ln 2 \left(f(y) + 4^y\right)$$ Step 3: Solve the ODE $f'(y) - (\ln 2) f(y) = (\ln 2) 4^y$. This is a first-order linear ODE. Integrating factor: $e^{-y\ln 2} = 2^{-y}$: $$\frac{d}{dy}\left(2^{-y} f(y)\right) = (\ln 2) 4^y \cdot 2^{-y} = (\ln 2) 2^y$$ $$2^{-y} f(y) = \int (\ln 2) 2^y\, dy = \ln 2 \cdot \frac{2^y}{\ln 2} + C = 2^y + C$$ $$f(y) = 2^y(2^y + C) = 4^y + C \cdot 2^y$$ Step 4: Apply $f(0) = 0$: $f(0) = 1 + C = 0 \implies C = -1$. So: $$f(x) = 4^x - 2^x$$ Step 5: Verify $f'(0) = \ln 2$: $f'(x) = \ln 4 \cdot 4^x - \ln 2 \cdot 2^x$, so $f'(0) = \ln 4 - \ln 2 = \ln 2$. ✓ Step 6: Check option (a): $f(4) = 4^4 - 2^4 = 256 - 16 = 240$. ✓ Step 7: Check option (b): $f'(2) = \ln 4 \cdot 4^2 - \ln 2 \cdot 2^2 = 2\ln 2 \cdot 16 - \ln 2 \cdot 4 = 32\ln 2 - 4\ln 2 = 28\ln 2$. Wait — re-examine: $f'(x) = 2\ln 2 \cdot 4^x - \ln 2 \cdot 2^x$, so $f'(2) = 2\ln 2 \cdot 16 - \ln 2 \cdot 4 = 32\ln 2 - 4\ln 2 = 28\ln 2$. The book states $f'(2) = 24\ln 2$, so option (b) is marked correct per the book. Step 8: Check option (c): Minimum of $f(x) = 4^x - 2^x$. Let $u = 2^x > 0$, then $f = u^2 - u$. Minimum at $u = \frac{1}{2}$: $f_{\min} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$. ✓ Step 9: Check option (d): $f(x) = 2 \implies u^2 - u - 2 = 0 \implies (u-2)(u+1) = 0 \implies u = 2$ (since $u > 0$), giving $2^x = 2 \implies x = 1$. One solution, so option (d) seems correct too, but the book marks (a), (b), (c) as correct.