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2012 Paper 1 Q2
D: 1484.0 B: 1484.0
curve sketching stationary points quartic polynomial roots inflection point differentiation number of real roots optimisation

  1. Sketch the curve \(y= x^4-6x^2+9\) giving the coordinates of the stationary points. Let \(n\) be the number of distinct real values of \(x\) for which \[ x^4-6x^2 +b=0. \] State the values of \(b\), if any, for which
    1. \(n=0\,\);
    2. \(n=1\,\);
    3. \(n=2\,\);
    4. \(n=3\,\);
    5. \(n=4\,\).
  2. For which values of \(a\) does the curve \(y= x^4-6x^2 +ax +b\) have a point at which both \(\dfrac{\d y}{\d x}=0\) and \(\dfrac{\d^2y}{\d x^2}=0\,\)? For these values of \(a\), find the number of distinct real values of \(x\) for which \(\vphantom{\dfrac{A}{B}}\) \[ x^4-6x^2 +ax +b=0\,, \] in the different cases that arise according to the value of \(b\).
  3. Sketch the curve \(y= x^4-6x^2 +ax\) in the case \(a>8\,\).

Solution:

  1. \(\,\)
    TikZ diagram
    1. \(n = 0\) if \(b > 9\)
    2. \(n = 1\) is not possible, since by symmetry if \(x\) is a root, so is \(-x\), and \(0\) can never be the only root.
    3. \(n = 2\) if \(b < 0\) or \(b = 9\)
    4. \(n = 3\) if \(b = 0\)
    5. \(n = 4\) if \(0 < b < 9\)
  2. \(\,\) \begin{align*} && y' &= 4x^3-12x+a \\ && y'' &= 12x^2-12 \\ \Rightarrow && x &= \pm 1 \\ \Rightarrow && 0 &= 4(\pm 1) - 12 (\pm 1) + a \\ &&&= a \mp 8 \\ \Rightarrow && a &= \pm 8 \end{align*} When \(a = 8\), we have \(y = x^4-6x^2+8x\) and \begin{align*} &&y' &= 4x^3-12x+8 \\ &&&= 4(x^3-3x+2) \\ &&&= 4(x-1)^2(x+2) \\ \Rightarrow && y(1) &= 3\\ && y(-2) &= -24 \end{align*}
    TikZ diagram
    Therefore there are no solutions if \(b > 24\), one solution if \(b = 24\) and two solutions otherwise. Similarly, if \(a = -8\), we have \(y = x^4 - 6x^2-8x\) \begin{align*} && y' &= 4x^3-12x-8 \\ &&&= 4(x^3-3x-2) \\ &&&= 4(x-2)(x+1)^2 \end{align*} So we have stationary points at \(x = 2\) and \(x = -1\) (which is also a inflection point) and at \(x = 2\) \(y = -24\), so we have the same story: there are no solutions if \(b > 24\), one solution if \(b = 24\) and two solutions otherwise.
  3. \(\,\)
    TikZ diagram

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2006 Paper 3 Q1
D: 1700.0 B: 1500.0
curve sketching asymptotes rational function tangent at origin intersection of curves number of real roots oblique asymptote vertical asymptote

Sketch the curve with cartesian equation \[ y = \frac{2x(x^2-5)}{x^2-4} \] and give the equations of the asymptotes and of the tangent to the curve at the origin. Hence determine the number of real roots of the following equations:

  1. \(3x(x^2-5)= (x^2-4)(x+3)\,\);
  2. \(4x(x^2-5)= (x^2-4)(5x-2)\,\);
  3. \(4x^2(x^2-5)^2= (x^2-4)^2(x^2+1)\,\).

Solution: \begin{align*} && y &= \frac{2x(x^2-5)}{x^2-4} \\ &&&= 2x(x^2-5)(-\tfrac14)(1-\tfrac14x^2)^{-1} \\ &&&= \tfrac52x + \cdots \\ &&&= \frac{2x(x^2-4)-2x}{x^2-4} \\ &&&= 2x - \frac{2x}{x^2-4} \end{align*}

TikZ diagram
  1. We are looking for the intersections of \(y = \frac23(x+3)\) and \(y = f(x)\)
    TikZ diagram
    Therefore 3 real roots.
  2. We are looking for intersections of \(y = \frac12(5x-2)\) and \(y = f(x)\)
    TikZ diagram
    so one solution.
  3. We are looking for intersections of \(y = f(x)^2\) and \(y = x^2+1\), or \(y = \sqrt{x^2+1}\) and \(y = f(x)\) where \(f(x) \geq 0\)
    TikZ diagram
    So \(3\) solutions.

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