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2025 Paper 3 Q3
D: 1500.0 B: 1500.0
coordinate geometry line through two points x-intercept inequality AM-GM inequality harmonic mean decreasing function curve fitting cubic curve parametric

Let \(f(x)\) be defined and positive for \(x > 0\). Let \(a\) and \(b\) be real numbers with \(0 < a < b\) and define the points \(A = (a, f(a))\) and \(B = (b, -f(b))\). Let \(X = (m,0)\) be the point of intersection of line \(AB\) with the \(x\)-axis.

  1. Find an expression for \(m\) in terms of \(a\), \(b\), \(f(a)\) and \(f(b)\).
  2. Show that, if \(f(x) = \sqrt{x}\), then \(m = \sqrt{ab}\). Find, in terms of \(n\), \(a\) function \(f(x)\) such that \(m = \frac{a^{n+1} + b^{n+1}}{a^n + b^n}\).
  3. Let \(g_1(x)\) and \(g_2(x)\) be defined and positive for \(x > 0\). Let \(m = M_1\) when \(f(x) = g_1(x)\) and let \(m = M_2\) when \(f(x) = g_2(x)\). Show that if \(\frac{g_1(x)}{g_2(x)}\) is a decreasing function then \(M_1 > M_2\). Hence show that $$\frac{a+b}{2} > \sqrt{ab} > \frac{2ab}{a+b}$$
  4. Let \(p\) and \(c\) be chosen so that the curve \(y = p(c-x)^3\) passes through both \(A\) and \(B\). Show that $$\frac{c-a}{b-c} = \left(\frac{f(a)}{f(b)}\right)^{1/3}$$ and hence determine \(c\) in terms of \(a\), \(b\), \(f(a)\) and \(f(b)\). Show that if \(f\) is a decreasing function, then \(c < m\).

Solution:

  1. The line \(AB\) has equation: \begin{align*} && \frac{y+f(b)}{x-b} &= \frac{f(a)+f(b)}{a-b} \\ \Rightarrow && \frac{f(b)}{m-b} &= \frac{f(a)+f(b)}{a-b} \\ \Rightarrow && m &= \frac{a-b}{f(a)+f(b)}f(b) + b \\ &&&= \frac{af(b)+bf(a)}{f(a)+f(b)} \end{align*}
  2. Suppose \(f(x) = \sqrt{x}\) then \begin{align*} m &= \frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}} \\ &= \frac{\sqrt{ab}(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}} \\ &= \sqrt{ab} \end{align*} Suppose \(f(x) = x^{-n}\) then \begin{align*} m &= \frac{a b^{-n}+ba^{-n}}{a^{-n}+b^{-n}} \\ &= \frac{a^{n+1}+b^{n+1}}{b^n + a^n} \\ \end{align*}
  3. Without loss of generality, we can scale \(g_1(x)\) and \(g_2(x)\) so that \(g_1(a) = g_2(a)\) and \(m\) won't change for either of them. Then since \(\frac{g_1(b)}{g_2(b)} < 1\) (this function is decreasing) our line connecting \((a,g_i(a))\) and \((b,-g_i(b))\) must interect the axis first for \(g_2\), in particular \(M_1 > M_2\). Suppose \(g_1(x) =1, g_2(x) = \sqrt{x}, g_3(x) = x^{-1}\), the notice that \(\frac{g_1(x)}{g_2(x)} =\frac{g_2(x)}{g_3(x)}= x^{-1/2}\) are decreasing, therefore: \begin{align*} \frac{a+b}{1+1} &> \sqrt{ab} > \frac{1+1}{a^{-1}+b^{-1}} \\ \frac{a+b}{2} &> \sqrt{ab} > \frac{2ab}{a+b} \\ \end{align*}
  4. We must have: \begin{align*} && p(c-a)^3 &= f(a) \\ && p(c-b)^3 &= -f(b) \\ \Rightarrow &&\left ( \frac{c-a}{c-b} \right)^3 &= -\frac{f(a)}{f(b)} \\ \Rightarrow && \frac{c-a}{b-c} &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \\ \Rightarrow && c-a &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}(b-c)\\ \Rightarrow && c \left (1 + \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \right) &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}b + a \\ \Rightarrow && c &= \frac{\left (\frac{f(a)}{f(b)} \right)^{\tfrac13}b + a}{1 + \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}} \\ &&&= \frac{b[f(a)]^\tfrac13+a[f(b)]^\tfrac13}{[f(a)]^\tfrac13+[f(b)]^\tfrac13} \end{align*} We have that \(\frac{c-a}{b-c} = \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \) and \(\frac{m-a}{b-c} = \frac{f(a)}{f(b)}\). Since \(f\) is decreasing, \(\frac{f(a)}{f(b)} > 1\) and so \(\left (\frac{f(a)}{f(b)} \right)^{\tfrac13} < \frac{f(a)}{f(b)}\), therefore \(m > c\).

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