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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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2004 Paper 3 Q2
D: 1700.0 B: 1516.0
curve sketching asymptotes oblique asymptote rational function double root tangent vertical asymptote

The equation of a curve is \(y=\f ( x )\) where \[ \f ( x ) = x-4-\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r} \;. \]

  1. Write down the equations of the vertical and oblique asymptotes to the curve and show that the oblique asymptote is a tangent to the curve.
  2. Show that the equation \(\f ( x ) =0\) has a double root.
  3. Sketch the curve.

Solution:

  1. Clearly \(x = 0\) and \(x = 4\) are vertical asymptotes. Notice that \(\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r}\) tends to \(0\) as \(x \to \infty\). Therefore the oblique asymptote is \(y = x-4\).
  2. \begin{align*} && 0 &= \frac{x^2(x-4)^2-4^2(2x+1)^2}{x^2(x-4)} \\ &&&= \frac{(x(x-4)-4(2x+1))(x(x-4)+4(2x+1))}{x^2(x-4)} \\ &&&= \frac{(x^2-12x-4)(x^2+4x+4)}{x^2(x-4)}\\ &&&= \frac{(x^2-12x-4)(x+2)^2}{x^2(x-4)} \end{align*} Therefore \(f(x) = 0\) has a double root at \(x = -2\). Notice it also has roots at \(6 \pm 2\sqrt{10}\)
  3. TikZ diagram

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