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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 \( \cdot \frac{f^{\prime}(4)}{f^{\prime}(2)} \) is equal to |
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Answer» \(f(x + y) = 2^x f(y) + 4^y f(x)\) ......(1) \(f(y + x) = 2^y f(x) + 4^x f(y)\) .......(2) Equation (1) - Equation (2), we obtain \((2^x - 4^x) f(y) + (4y - 2y )f(x) = 0\) ⇒ \((26^x - 4^x) f(y) = (2^y - 4^y) = f(x)\) ⇒ \(\frac{f(x)}{2^x - 4^x} = \frac{f(y)}{2^y - 4^y} = k\) (Let) ⇒ \(f(x) = k(2^x - 4^x)\) \(\because f(2) = 3\) \(\therefore k(2^3 - 4^3) = 3\) ⇒ \(k = \frac3{8 - 64} = \frac{3}{-56}\) \(\therefore f(x) = \frac{-3}{56} (2^x-4^x)\) \(f'(x) = \frac{-3}{56}(2^x log2 - 4^x log4)\) \(= \frac{-3}{56}(2^x - 2^{2x + 1})log 2\) \(f'(2) = \frac{-3}{56} (4 - 32) log2\) \(= \frac{-3}{56} \times - 28\, log2\) \(= 6 \,log2\) \(f'(4) = \frac{-3}{56}(16 - 512)\, log2\) \(= \frac{-3}{56}\times - 496\,log2\) \( = \frac{-3}{7}\times{-62} \,log 2\) \(= \frac{96}{7}log2\) \(\therefore \frac{14f'(4)}{f'(2)} = \frac{14 \times \frac{96}{7}\,log2}{6\, log2}\) \(= \frac{36 \times 2}{6}\) \(= \frac{96}{3}\) \(= 32\) |
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