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FM June 2025 p11 q07
4121
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) The 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 considering the degrees of the polynomials: \(y = 1\).
(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{(2x^2 - 7x - 4)(4x - 5) - (2x^2 - 5x)(4x - 7)}{(2x^2 - 7x - 4)^2}\). Simplifying gives \(4x^2 + 16x - 20 = 0\), leading to points \((-5, \frac{25}{3})\) and \((1, \frac{1}{3})\).
(c) The curve intersects the axes at \((0,0)\) and \((\frac{5}{2},0)\).
(d) The sketch of \(y = \left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right|\) reflects the negative parts of the curve above the x-axis.
(e) Solve \(\left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right| < \frac{1}{9}\) by considering \(\frac{2x^2 - 5x}{2x^2 - 7x - 4} < \frac{1}{9}\) and \(\frac{2x^2 - 5x}{2x^2 - 7x - 4} > -\frac{1}{9}\). Solving these inequalities gives the intervals \(\frac{13}{10} < x < \frac{19}{16}\) and \(\frac{\sqrt{33}}{16} < x < \frac{13}{10} + \frac{3}{10} \sqrt{21}\).
