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FM June 2024 p11 q06
4162
The curve C has equation \(y = \frac{x^2 + ax + 1}{x + 2}\), where \(a > \frac{5}{2}\).
Find the equations of the asymptotes of C.
Show that C has no stationary points.
Sketch C, stating the coordinates of the point of intersection with the y-axis and labelling the asymptotes.
Sketch the curve with equation \(y = \left| \frac{x^2 + ax + 1}{x + 2} \right|\).
On your sketch in part (i), draw the line \(y = a\).
It is given that \(\left| \frac{x^2 + ax + 1}{x + 2} \right| < a\) for \(-5 - \sqrt{14} < x < -3\) and \(-5 + \sqrt{14} < x < 3\). Find the value of \(a\).
Solution
(a) The vertical asymptote is at \(x = -2\). The oblique asymptote is found by dividing \(x^2 + ax + 1\) by \(x + 2\), giving \(y = x + a - 2\).
(b) Differentiate \(y = \frac{x^2 + ax + 1}{x + 2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(x+2)(2x+a) - (x^2+ax+1)}{(x+2)^2}\). Simplifying gives \(\frac{dy}{dx} = \frac{x^2 + 4x + 2a - 1}{(x+2)^2}\). Setting the numerator to zero gives \(x^2 + 4x + 2a - 1 = 0\). The discriminant is \(16 - 4(2a-1) = 20 - 8a\), which is less than zero for \(a > \frac{5}{2}\), so no stationary points exist.
(c) The curve intersects the y-axis at \((0, \frac{1}{2})\). The asymptotes are \(x = -2\) and \(y = x + a - 2\).
(d)(i) Sketch the curve \(y = \left| \frac{x^2 + ax + 1}{x + 2} \right|\) with the same asymptotes and a reflection in the x-axis where the original curve is negative.
(d)(ii) Draw the line \(y = a\) on the sketch.
(d)(iii) Solve \(\frac{x^2 + ax + 1}{x + 2} = a\) and \(\frac{x^2 + ax + 1}{x + 2} = -a\). This gives the quadratic \(x^2 + 1 - 2a = 0\) or \(x^2 + 2ax + 1 = 0\). The roots are \(-a - \sqrt{a^2 - 2a - 1} < x < -a + \sqrt{a^2 - 2a - 1}\). Solving for \(a\) gives \(a = 5\).
