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2006 Paper 3 Q1
Sketch the curve with cartesian equation \[ y = \frac{2x(x^2-5)}{x^2-4} \] and give the equations of the asymptotes and of the tangent to the curve at the origin. Hence determine the number of real roots of the following equations:
- \(3x(x^2-5)= (x^2-4)(x+3)\,\);
- \(4x(x^2-5)= (x^2-4)(5x-2)\,\);
- \(4x^2(x^2-5)^2= (x^2-4)^2(x^2+1)\,\).
Solution: \begin{align*} && y &= \frac{2x(x^2-5)}{x^2-4} \\ &&&= 2x(x^2-5)(-\tfrac14)(1-\tfrac14x^2)^{-1} \\ &&&= \tfrac52x + \cdots \\ &&&= \frac{2x(x^2-4)-2x}{x^2-4} \\ &&&= 2x - \frac{2x}{x^2-4} \end{align*}
- We are looking for the intersections of \(y = \frac23(x+3)\) and \(y = f(x)\)
Therefore 3 real roots.
- We are looking for intersections of \(y = \frac12(5x-2)\) and \(y = f(x)\)
so one solution.
- We are looking for intersections of \(y = f(x)^2\) and \(y = x^2+1\), or \(y = \sqrt{x^2+1}\) and \(y = f(x)\) where \(f(x) \geq 0\)
So \(3\) solutions.
2004 Paper 3 Q2
The equation of a curve is \(y=\f ( x )\) where \[ \f ( x ) = x-4-\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r} \;. \]
- Write down the equations of the vertical and oblique asymptotes to the curve and show that the oblique asymptote is a tangent to the curve.
- Show that the equation \(\f ( x ) =0\) has a double root.
- Sketch the curve.
Solution:
- Clearly \(x = 0\) and \(x = 4\) are vertical asymptotes. Notice that \(\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r}\) tends to \(0\) as \(x \to \infty\). Therefore the oblique asymptote is \(y = x-4\).
- \begin{align*} && 0 &= \frac{x^2(x-4)^2-4^2(2x+1)^2}{x^2(x-4)} \\ &&&= \frac{(x(x-4)-4(2x+1))(x(x-4)+4(2x+1))}{x^2(x-4)} \\ &&&= \frac{(x^2-12x-4)(x^2+4x+4)}{x^2(x-4)}\\ &&&= \frac{(x^2-12x-4)(x+2)^2}{x^2(x-4)} \end{align*} Therefore \(f(x) = 0\) has a double root at \(x = -2\). Notice it also has roots at \(6 \pm 2\sqrt{10}\)
2000 Paper 3 Q4
The function \(\f(x)\) is defined by $$ \f(x) = \frac{x( x - 2 )(x-a)}{ x^2 - 1}. $$ Prove algebraically that the line \(y = x + c\) intersects the curve \(y = \f ( x )\) if \(\vert a \vert \ge1\), but there are values of \(c\) for which there are no points of intersection if \(\vert a \vert <1\). Find the equation of the oblique asymptote of the curve \(y=\f(x)\). Sketch the graph in the two cases
- \(a<-1\)
- \(-1 < a < -\frac12\)
Solution: \begin{align*} && x+ c &= f(x) \\ \Rightarrow && (x+c)(x^2-1) &= x(x-2)(x-a) \\ \Rightarrow && x^3 + cx^2-x-c &= x^3-(2+a)x^2+2ax \\ \Rightarrow && 0 &= (c+2+a)x^2-(1+2a)x-c \\ && 0 &\leq \Delta = (1+2a)^2 + 4(2+c+a)c \\ &&&= 4c^2+(4a+8)c + (1+2a)^2 \\ && \Delta_c &= 16(a+2)^2-16(1+2a)^2 \\ &&&= 16(1-a)(3a+3) \\ &&&= 48(1-a^2) \end{align*} Therefore if \(|a| \geq 1\) we must have \(\Delta_c \leq 0\) which means \(\Delta \geq 0\) and so there are always solutions. If \(|a| < 1\) there are values for \(c\) where \(\Delta < 0\) and there would be no solutions. \begin{align*} && y &= \frac{x^3-(2+a)x^2+2ax}{x^2-1} \\ &&&= \frac{(x^2-1)(x-(2+a))+(2a+1)x-(2+a)}{x^2-1} \\ &&&= x - (2+a) + \frac{(2a+1)x-(2+a)}{x^2-1} \end{align*} therefore the oblique asymptote has equation \(y = x - (2+a)\)
