Let \(C\) be the set of matrices of the form \begin{equation*} \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \end{equation*} where \(a\) and \(b\) are real numbers. Show that \(C\) is closed under addition and multiplication, and that for every matrix \(Z\) in \(C\) other than the zero matrix, there is a matrix \(Z'\) in \(C\) with \(ZZ' = I\) (\(I\) being the identity \(2\times2\) matrix). Find matrices \(X\) and \(Y\) in \(C\) such that \begin{equation*} X^2 + I = 0, \quad Y^2 + Y + I = 0. \end{equation*}
Let \(\displaystyle I_n = \int_0^{\pi/2} \sin^n\theta\, d\theta, \quad n\) an integer. Show that:
Show that if \(z = x + iy\) defines a point in the \(x,y\) plane, then \begin{equation*} \left|\frac{z - z_1}{z - z_2}\right| = k \quad \text{(where \(k\) is a positive constant and \(z_1 \neq z_2\))} \end{equation*} gives the equation of a circle or straight line, depending on the value of \(k\). If \(z \to \frac{az + b}{cz + d}, ad - bc \neq 0\), \(a\), \(b\), \(c\), \(d\) complex, show that such circles or straight lines are mapped into circles or straight lines.
If \(f(x)\) is a positive function of \(x\) whose derivative is positive and \(n \geq 2\) is an integer, justify the inequality \begin{equation*} \int_1^n f(x)\, dx < \sum_{r=2}^n f(r). \end{equation*} By considering the integral of \(\ln x\), show that \(e\left(\frac{n}{e}\right)^n < n!\)
Let \(S_1\), \(S_2\) be two spheres such that the sum of the surface areas is fixed. When is the sum of the volumes a) a maximum b) a minimum? Suppose instead that the sum of the reciprocals of the areas is fixed. When (if ever) is the sum of the volumes a) a maximum b) a minimum?
Find the solution of \(\frac{dy}{dx} = xy(y-2)\) such that \(y(0) = y_0\). Sketch the forms of solution that arise for \(y_0 > 0\).
Solution: \begin{align*} && \frac{\d y}{\d x} &= x y(y-2) \\ \Rightarrow && \int x \d x &=\int \frac{1}{y(y-2)} \d y \\ \Rightarrow && \frac{x^2}{2} &=\frac12 \int \left ( \frac{1}{y-2}-\frac{1}{y} \right)\d y \\ &&&= \frac12 (\ln |y - 2| - \ln |y|)+C \\ \Rightarrow && x^2 &= \ln |1-\frac{2}{y} | +C\\ y(0) = y_0: && 0 &= \ln | 1 - \frac{2}{y_0}| + C \\ \Rightarrow && C &= - \ln |1 - \frac{2}{y_0}| \\ \Rightarrow && x^2 &= \ln \frac{1-2/y}{1-2/y_0} \\ \Rightarrow && (1-2/y_0)e^{x^2} &=1-2/y \\ \Rightarrow && y &= \frac{2}{1-(1-2/y_0)e^{x^2}} \\ &&&= \frac{2y_0}{y_0-(y_0-2)e^{x^2}} \end{align*}
The members of a family of curves in the \(x,y\) plane satisfy the differential equation \begin{equation*} y\frac{dy}{dx} - y^2 = x^2 - x. \end{equation*} By multiplying this equation by a suitable function of \(x\) and integrating, or otherwise, obtain the curve which passes through the point \((0, 1)\). Show that this curve also passes through the point \((-a, 0)\) where \(a > 0\) and \(a = -\ln a\).
Solution: \begin{align*} && y \frac{\d y}{\d x} - y^2 &= x^2 - x \\ \Rightarrow && e^{-2x}2y \frac{\d y}{\d x} - 2e^{-2 x}y^2 &= 2e^{-2x}(x^2 - x) \\ \Rightarrow && \frac{\d }{\d x} \left (y^2 e^{-2x} \right)&= 2e^{-2x}(x^2-x) \\ \Rightarrow && y^2e^{-2x} &= \int 2e^{-2x}(x^2-x) \d x \\ &&y^2e^{-2x} &= -e^{-2x}x^2 + C \\ \Rightarrow && y^2 &= -x^2 +Ce^{2x} \\ x = 0, y =1: && 1 &= C \\ \Rightarrow && y^2 &= -x^2 + e^{2x} \\ \Rightarrow && y &= \sqrt{-x^2+e^{2x}} \end{align*} This will be \(0\) when \(x^2 = e^{2x} \Rightarrow |x| = e^{x} \Rightarrow \ln |x| = x\), ie exactly the point described
A player deals cards from a pack of 52 in sets of four. The first set of four consists of cards of different suits. What is the probability that the last set of four consists of cards of different suits? Had the first set of four consisted of cards of the same suit, what would the probability have been that the last set of four were also of one suit?