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2017 Paper 3 Q4
D: 1700.0 B: 1484.0
integration geometric mean logarithms exponential function change of base proof functional equation properties of integrals

For any function \(\f\) satisfying \(\f(x) > 0\), we define the geometric mean, F, by \[ F(y) = e^{\frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \quad (y > 0). \]

  1. The function f satisfies \(\f(x) > 0\) and \(a\) is a positive number with \(a\ne1\). Prove that \[ F(y) = a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx}. \]
  2. The functions f and g satisfy \(\f(x) > 0\) and \(\g(x) > 0\), and the function \(\h\) is defined by \(\h(x) = \f(x)\g(x)\). Their geometric means are F, G and H, respectively. Show that \(H(y)= \F(y) \G(y)\,\).
  3. Prove that, for any positive number \(b\), the geometric mean of \(b^x\) is \(\sqrt{b^y}\,\).
  4. Prove that, if \(\f(x)>0\) and the geometric mean of \(\f(x)\) is \(\sqrt{\f(y)}\,\), then \(\f(x) = b^x\) for some positive number \(b\).

Solution:

  1. \begin{align*} && a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} &= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} \\ &&&= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \frac{\ln f(x)}{\ln a} \, dx} \\ &&&= e^{ \frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \\ &&&= F(y) \end{align*}
  2. \(\,\) \begin{align*} && H(y) &= e^{\frac1y \int_0^y \ln h(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln (f(x)g(x))\d x} \\ &&&= e^{\frac1y \int_0^y \left ( \ln f(x)+\ln g(x) \right)\d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x +\frac1y \int_0^y \ln g(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x }e^{\frac1y \int_0^y \ln g(x) \d x} \\ &&&= F(y)G(y) \end{align*}
  3. Suppose \(f(x) = b^x\), then \begin{align*} && F(y) &= b^{\frac1y \int_0^y \log_b f(x) \d x} \\ &&&= b^{\frac1y \int_0^y x \d x}\\ &&&= b^{\frac1y \frac{y^2}{2}} \\ &&&= b^{\frac{y}2} = \sqrt{b^y} \end{align*}
  4. Suppose the geometric mean of \(f(x)\) is \(\sqrt{f(y)}\) then the geometric mean of \(f(x)^2\) is \(f(y)\) by the the second part.

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