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2007 Paper 2 Q2
D: 1600.0 B: 1500.0
curve sketching cubic stationary points point of inflection symmetry integration area between curves polynomial optimisation

A curve has equation \(y=2x^3-bx^2+cx\). It has a maximum point at \((p,m)\) and a minimum point at \((q,n)\) where \(p>0\) and \(n>0\). Let \(R\) be the region enclosed by the curve, the line \(x=p\) and the line \(y=n\).

  1. Express \(b\) and \(c\) in terms of \(p\) and \(q\).
  2. Sketch the curve. Mark on your sketch the point of inflection and shade the region \(R\). Describe the symmetry of the curve.
  3. Show that \(m-n=(q-p)^3\).
  4. Show that the area of \(R\) is \(\frac12 (q-p)^4\).

Solution:

  1. \(\,\) \begin{align*} && y &= 2x^3-bx^2+cx \\ \Rightarrow && y' &= 6x^2-2bx+c \end{align*} We must have \(p, q\) are the roots of this equation, ie \(\frac13b = p+q, \frac16c = pq\)
  2. The point of inflection will be at \(\frac{b}6\) The graph will have rotational symmetry of \(180^{\circ}\) about the point of inflection.
    TikZ diagram
  3. \begin{align*} && m-n &= 2(p^3-q^3)-b(p^2-q^2)+c(p-q) \\ &&&= (p-q)(2(p^2+qp+q^2)-b(p+q)+c) \\ &&&= (p-q)(2(p^2+qp+q^2)-3(p+q)^2+6pq) \\ &&&= (p-q)(-p^2-q^2+2pq) \\ &&&= (q-p)^3 \end{align*}
  4. The area of \(R\) is \begin{align*} A &= \frac12 bh \\ &= \frac12 (q-p)(m-n) = \frac12(q-p)^4 \end{align*} as required.

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