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FM Nov 2023 p13 q07
4198
The curve C has equation \(y = f(x)\), where \(f(x) = \frac{x^2 + 2}{x^2 - x - 2}\).
(a) Find the equations of the asymptotes of C.
(b) Find the coordinates of any stationary points on C, giving your answers correct to 1 decimal place.
(c) Sketch C, stating the coordinates of any intersections with the axes.
(d) Sketch the curve with equation \(y = \frac{1}{f(x)}\).
(e) Find the set of values for which \(\frac{1}{f(x)} < f(x)\).
Solution
(a) The vertical asymptotes occur where the denominator is zero: \(x^2 - x - 2 = 0\). Solving gives \(x = -1\) and \(x = 2\). The horizontal asymptote is found by considering the limits as \(x\) approaches infinity, giving \(y = 1\).
(b) To find stationary points, set \(\frac{dy}{dx} = 0\). The derivative is \(\frac{dy}{dx} = \frac{(x^2 - x - 2)(2x) - (x^2 + 2)(2x - 1)}{(x^2 - x - 2)^2}\). Solving \(x^2 + 8x - 2 = 0\) gives the stationary points \((-8.2, 0.9)\) and \((0.2, -0.9)\).
(c) The sketch shows the curve with asymptotes and intersection at \((0, -1)\).
(d) The curve \(y = \frac{1}{f(x)}\) is sketched by inverting the values of \(f(x)\) from part (c).
(e) Solve \(\frac{x^2 + 2}{x^2 - x - 2} = 1\) and \(\frac{x^2 + 2}{x^2 - x - 2} = -1\) to find critical points. The solution is \(-4 < x < -1, \ 0 < x < \frac{1}{2}, \ x > 2\).
