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2014 Paper 2 Q7
D: 1600.0 B: 1486.9
modulus function curve sketching graphical solution piecewise linear number of solutions inequality quadrilateral absolute value equations

  1. The function \(\f\) is defined by \(\f(x)= |x-a| + |x-b| \), where \(a < b\). Sketch the graph of \(\f(x)\), giving the gradient in each of the regions \(x < a\), \(a < x < b\) and \(x > b\). Sketch on the same diagram the graph of \(\g(x)\), where \(\g(x)= |2x-a-b|\). What shape is the quadrilateral with vertices \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\)?
  2. Show graphically that the equation \[ |x-a| + |x-b| = |x-c|\,, \] where \(a < b\), has \(0\), \(1\) or \(2\) solutions, stating the relationship of \(c\) to \(a\) and \(b\) in each case.
  3. For the equation \[ |x-a| + |x-b| = |x-c|+|x-d|\,, \] where \(a < b\), \(c < d\) and \(d-c < b-a\), determine the number of solutions in the various cases that arise, stating the relationship between \(a\), \(b\), \(c\) and \(d\) in each case.

Solution:

  1. \(\,\)
    TikZ diagram
    \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\) forms a rectangle.
  2. There are no solutions if \(a < c < b\):
    TikZ diagram
    There is one solution if \(a=c\) or \(a = b\)
    TikZ diagram
    And there are two solution if \(c \not \in [a,b]\)
    TikZ diagram
    There is exactly one solution unless....
    TikZ diagram
    ... there are infinitely many solutions when the gradients line up perfectly, ie when \(a+b=c+d\)
    TikZ diagram

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2011 Paper 2 Q1
D: 1600.0 B: 1500.0
curve sketching square root functions graphical solution simultaneous equations algebraic manipulation domain restrictions intersection of curves

  1. Sketch the curve \(y=\sqrt{1-x} + \sqrt{3+x}\;\). Use your sketch to show that only one real value of \(x\) satisfies \[ \sqrt{1-x} + \sqrt{3+x} = x+1\,, \] and give this value.
  2. Determine graphically the number of real values of \(x\) that satisfy \[ 2\sqrt{1-x} = \sqrt{3+x} + \sqrt{3-x}\;. \] Solve this equation.

Solution:

  1. TikZ diagram
    Clearly the only solution is \(x = 1\)
  2. TikZ diagram
    There is clearly only one solution, with \(x \approx -2\) \begin{align*} && 4(1-x) &= 6+2\sqrt{9-x^2} \\ && -2x-1 &=\sqrt{9-x^2} \\ \Rightarrow && 4x^2+4x+1 &= 9-x^2 \\ \Rightarrow && 0 &= 5x^2+4x-8 \\ &&x&= \frac{-2\pm 2\sqrt{11}}{5} \\ \Rightarrow && x &= -\left ( \frac{2+2\sqrt{11}}{5} \right) \end{align*}

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