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1999 Paper 1 Q4
Sketch the following subsets of the \(x\)-\(y\) plane:
- \(|x|+|y|\le 1\) ;
- \(|x-1|+|y-1|\le 1 \) ;
- \(|x-1|-|y+1|\le 1 \) ;
- \(|x|\, |y-2|\le 1\) .
Solution:
1995 Paper 1 Q1
- Find the real values of \(x\) for which \[ x^{3}-4x^{2}-x+4\geqslant0. \]
- Find the three lines in the \((x,y)\) plane on which \[ x^{3}-4x^{2}y-xy^{2}+4y^{3}=0. \]
- On a sketch shade the regions of the \((x,y)\) plane for which \[ x^{3}-4x^{2}y-xy^{2}+4y^{3}\geqslant0. \]
Solution:
- \(\,\) \begin{align*} && 0 & \leq x^3 - 4x^2 - x + 4 \\ &&&= (x-1)(x^2-3x-4) \\ &&&= (x-1)(x-4)(x+1) \\ \Leftrightarrow && x &\in [-1, 1] \cup [4, \infty) \end{align*}
- \(\,\) \begin{align*} && 0 &= x^{3}-4x^{2}y-xy^{2}+4y^{3} \\ && 0 &= (x-y)(x-4y)(x+y) \end{align*} Therefore the lines are \(y = x, 4y = x, y=-x\).
- (quickest way to see this is to check the \(x\) or \(y\)-axis)
1988 Paper 2 Q8
In a crude model of population dynamics of a community of aardvarks and buffaloes, it is assumed that, if the numbers of aardvarks and buffaloes in any year are \(A\) and \(B\) respectively, then the numbers in the following year at \(\frac{1}{4}A+\frac{3}{4}B\) and \(\frac{3}{2}B-\frac{1}{2}A\) respectively. It does not matter if the model predicts fractions of animals, but a non-positive number of buffaloes means that the species has become extinct, and the model ceases to apply. Using matrices or otherwise, show that the ratio of the number of aardvarks to the number of buffaloes can remain the same each year, provided it takes one of two possible values. Let these two possible values be \(x\) and \(y\), and let the numbers of aardvarks and buffaloes in a given year be \(a\) and \(b\) respectively. By writing the vector \((a,b)\) as a linear combination of the vectors \((x,1)\) and \((y,1),\) or otherwise, show how the numbers of aardvarks and buffaloes in subsequent years may be found. On a sketch of the \(a\)-\(b\) plane, mark the regions which correspond to the following situations
- an equilibrium population is reached as time \(t\rightarrow\infty\);
- buffaloes become extinct after a finite time;
- buffaloes approach extinction as \(t\rightarrow\infty.\)
Solution: If the population in a given year is \(\mathbf{v} = \begin{pmatrix}A \\ B \end{pmatrix}\) then the population the next year is \(\mathbf{Mv}\) where \(\mathbf{M} = \begin{pmatrix} \frac14 & \frac34 \\ -\frac12 &\frac32 \end{pmatrix}\) The ratio is the same if \(\mathbf{Mv} = \lambda \mathbf{v}\) ie if \(\mathbf{v}\) is an eigenvector of \(\mathbf{M}\). The eigenvalues will be \(1\) and \(\frac38\) (by inspection) so we should be able to solve for the eigenvectors: \(\lambda = 1\) we have \(\frac14A + \frac34B = A \Rightarrow A = B\) a ratio of \(1\). \(\lambda = \frac38\) we have \(\frac14A + \frac34B = \frac38A \Rightarrow \frac34B = \frac18A \Rightarrow A = 6B\) a ratio of \(6\). If we write \(\begin{pmatrix} a \\ b \end{pmatrix}\) as \(x_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + x_6 \begin{pmatrix} 6 \\ 1 \end{pmatrix}\) we find that after \(n\) years, we have: \(x_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \l \frac38 \r^n x_6 \begin{pmatrix} 6 \\ 1 \end{pmatrix}\) for the populations. Therefore if \(x_1\) is \(< 0\) then in finite time we will end up with one population being 0. If \(x_1 > 0\) are positive we tend to a finite population and if \(x_1 = 0\) then over time the population will tend to \(0\) at infinity. In our diagram these areas correspond to (red) - die out in finite time, (green) population stable and the thick black line where the population goes extinct as \(t \to \infty\)
