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FM June 2025 p12 q07
4113
The curve \(C\) has equation \(y = \frac{2x^2 - 5x}{2x^2 - 7x - 4}\).
Find the equations of the asymptotes of \(C\).
Find the coordinates of any stationary points on \(C\).
Sketch \(C\), stating the coordinates of the intersections with the axes.
Sketch the curve with equation \(y = \left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right|\).
Find in exact form the set of values of \(x\) for which \(\left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right| < \frac{1}{9}\).
Solution
(a) Vertical asymptotes occur where the denominator is zero: \(2x^2 - 7x - 4 = 0\). Solving gives \(x = -\frac{1}{2}\) and \(x = 4\). The horizontal asymptote is found by comparing the degrees of the numerator and denominator, giving \(y = 1\).
(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{(2x^2 - 7x - 4)(4x - 5) - (2x^2 - 5x)(4x - 7)}{(2x^2 - 7x - 4)^2}\). Simplify and solve \(4x^2 + 16x - 20 = 0\) to find \(x = -5\) and \(x = 1\). Substitute back to find \(y\)-coordinates: \((-5, \frac{25}{7})\) and \((1, \frac{1}{3})\).
(c) The curve intersects the axes at \((0,0)\) and \((\frac{5}{2}, 0)\).
(d) Sketch the curve \(y = \left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right|\) with correct shape and position.
(e) Solve \(\left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right| < \frac{1}{9}\). This leads to solving \(16x^2 - 38x + 4 = 0\) and \(20x^2 - 52x - 4 = 0\). The critical points are \(x = \frac{19}{16} - \frac{3}{16} \sqrt{33}, \ x = \frac{19}{16} + \frac{3}{16} \sqrt{33}, \ x = \frac{13}{10} - \frac{3}{10} \sqrt{21}, \ x = \frac{13}{10} + \frac{3}{10} \sqrt{21}\). The solution is \(\frac{13}{10} - \frac{3}{10} \sqrt{21} < x < \frac{19}{16} - \frac{3}{16} \sqrt{33}\) or \(\frac{19}{16} + \frac{3}{16} \sqrt{33} < x < \frac{13}{10} + \frac{3}{10} \sqrt{21}\).
