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9231 P11 - Nov 2021 - Q07
4280
The curve \(C\) has equation \(y = \frac{4x+5}{4-4x^2}\).
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 \(y = \left| \frac{4x+5}{4-4x^2} \right|\) and find in exact form the set of values of \(x\) for which \(4|4x+5| > 5|4-4x^2|\).
Solution
Checked by expert
(a) The vertical asymptotes occur where the denominator is zero: \(4 - 4x^2 = 0\), giving \(x = 1\) and \(x = -1\). The horizontal asymptote is \(y = 0\) as \(x\) approaches infinity.
(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate: \(\frac{dy}{dx} = \frac{(4-4x^2)(4) - (4x+5)(-8x)}{(4-4x^2)^2}\). Simplify and solve \(16x^2 + 40x + 16 = 0\) to find \((-2, \frac{1}{4})\) and \(-\frac{1}{2}, 1\).
(c) The curve intersects the x-axis where \(y = 0\), giving \(x = -\frac{5}{4}\). It intersects the y-axis where \(x = 0\), giving \(y = \frac{5}{4}\).
(d) For \(y = \left| \frac{4x+5}{4-4x^2} \right|\), solve \(4|4x+5| > 5|4-4x^2|\). This gives two cases: \(\frac{4x+5}{4-4x^2} > \frac{5}{4}\) and \(\frac{4x+5}{4-4x^2} < -\frac{5}{4}\). Solving these inequalities gives the ranges \(\frac{2}{3} - \frac{1}{3}\sqrt{6} < x < -\frac{4}{3}\) and \(0 < x < \frac{2}{3} + \frac{1}{3}\sqrt{6}\), excluding \(x = \pm 1\).
