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2004 Paper 2 Q2
Prove that, if \(\vert \alpha\vert < 2\sqrt{2},\) then there is no value of \(x\) for which \begin{equation} x^2 -{\alpha}\vert x \vert + 2 < 0\;. \tag{\(*\)} \end{equation} Find the solution set of \((*)\) for \({\alpha}=3\,\). For \({\alpha} > 2\sqrt{2}\,\), the sum of the lengths of the intervals in which \(x\) satisfies \((*)\) is denoted by \(S\,\). Find \(S\) in terms of \({\alpha}\) and deduce that \(S < 2{\alpha}\,\). Sketch the graph of \(S\,\) against \(\alpha \,\).
Solution: There are two cases to consider by they are equivalent to \(x^2 \pm \alpha x + 2 < 0\), which has no solution solutions if \(\Delta < 0\), ie if \(\alpha^2 - 4\cdot1\cdot2 < 0 \Leftrightarrow |\alpha| < 2\sqrt{2}\). If \(\alpha = 3\), we have \begin{align*} && 0 & > x^2-3x+2 \\ &&&= (x-2)(x-1) \\ \Rightarrow && x & \in (1,2) \\ \\ && 0 &> x^2+3x+2 \\ &&& = (x+2)(x+1) \\ \Rightarrow && x &\in (-2,-1) \end{align*} Both cases work here, so \(x \in (-2, -1) \cup (1,2)\). \begin{align*} && 0 &> x^2 \pm \alpha x + 2 \\ &&&= (x \pm \tfrac{\alpha}{2})^2 -\frac{\alpha^2-8}{4} \end{align*} The potential intervals therefore are \((\frac{\alpha -\sqrt{\alpha^2-8}}{2}, \frac{\alpha +\sqrt{\alpha^2-8}}{2})\) and \((\frac{-\alpha -\sqrt{\alpha^2-8}}{2}, \frac{-\alpha +\sqrt{\alpha^2-8}}{2})\). Neither of these intervals overlap with \(0\), since \(\alpha^2 > \alpha^2-8\), and their lengths are both \(\sqrt{\alpha^2-8}\), therefore \(S = 2\sqrt{\alpha^2-8} < 2\alpha\)
