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2002 Paper 3 Q3
D: 1700.0 B: 1516.0
curve sketching quadratics inequalities surds square root equations optimisation discriminant number of solutions

Let \[\f(x) = a \sqrt{x} - \sqrt{x - b}\;, \] where \(x\ge b >0\) and \(a>1\,\). Sketch the graph of \(\f(x)\,\). Hence show that the equation \(\f(x) = c\), where \(c>0\), has no solution when \(c^2 < b \l a^2 - 1 \r\,\). Find conditions on \(c^2\) in terms of \(a\) and \(b\) for the equation to have exactly one or exactly two solutions. Solve the equations

  1. \(3 \sqrt{x} - \sqrt{x - 2} = 4\, ,\)
  2. \(3 \sqrt{x} - \sqrt{x - 3} = 5\;\).

Solution: \begin{align*} && f'(x) &= \frac12 ax^{-1/2}-\frac12(x-b)^{-1/2} \\ \Rightarrow f'(x) = 0: && 0 &= \frac{a\sqrt{x-b}-\sqrt{x}}{\sqrt{x(x-b)}} \\ \Rightarrow && x &= a^2(x-b)\\ \Rightarrow && x &= \frac{a^2b}{a^2-1} \\ && f(x) &= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{a^2b}{a^2-1}-b} \\ &&&= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{b}{a^2-1}} \\ &&&= \sqrt{b(a^2-1)} \end{align*}

TikZ diagram
If \(c\) is below the turning point, ie \(c^2 < b(a^2-1)\) there is no solution. If \(c^2 = b(a^2-1)\) there is exactly one solution. If \(b(a^2-1) < c^2 < (f(b))^2 = a^2b\) then there are two solutions, otherwise there is exactly one solution.
  1. \(c^2 = 16\), \(2 \cdot (3^2-1) = 16\), so we should have exactly one solution at \(x = \frac{3^2 \cdot 2}{3^2 -1 } = \frac{9}{4}\)
  2. \(c^2 = 25\) and \(3 \cdot (3^2 - 1) = 24, 3 \cdot (3^2) = 27\), so we look for two solutions. \begin{align*} && 5 & = 3 \sqrt{x} - \sqrt{x-3} \\ \Rightarrow && 25 &= 9x+x-3-6\sqrt{x(x-3)} \\ \Rightarrow && 3\sqrt{x(x-3)} &= 5x-14 \\ \Rightarrow && 9x(x-3) &= 25x^2-140x+196 \\ \Rightarrow && 0 &= 16x^2-113x+196 \\ &&&= (x-4)(16x-49) \\ \Rightarrow && x &= 4, \frac{49}{16} \end{align*}

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