Prove that, if \(c\ge a\) and \(d\ge b\), then \[ ab+cd\ge bc+ad\,. \tag{\(*\)} \]
Solution: \begin{align*} && \underbrace{(c-a)}_{\geq 0}\underbrace{(d-b)}_{\geq 0} & \geq 0 \\ \Leftrightarrow && cd -bc -ad + ab &\geq 0 \\ \Leftrightarrow && ab +cd &\geq bc+ad \\ \end{align*}
A function \(\f(x)\) is said to be convex in the interval \(a < x < b\) if \(\f''(x)\ge0\) for all \(x\) in this interval.
The polynomial \(\p(x)\) is given by \[ \ds \p(x)= x^n +\sum\limits_{r=0}^{n-1}a_rx^r\,, \] where \(a_0\), \(a_1\), \(\ldots\) , \(a_{n-1}\) are fixed real numbers and \(n\ge1\). Let \(M\) be the greatest value of \(\big\vert \p(x) \big\vert\) for $\vert x \vert\le 1\(. Then Chebyshev's theorem states that \)M\ge 2^{1-n}$.
Solution:
The function \(\f\) is defined by \[ \f(x) = \frac{\e^x-1}{\e-1}, \ \ \ \ \ x\ge0, \] and the function \(\g\) is the inverse function to \(\f\), so that \(\g(\f(x))=x\). Sketch \(\f(x)\) and \(\g(x)\) on the same axes. Verify, by evaluating each integral, that \[ \int_0^\frac12 \f(x) \,\d x + \int_0^k \g(x) \,\d x = \frac1 {2(\sqrt \e +1)}\,, \] where \(\displaystyle k= \frac 1{\sqrt\e+1}\), and explain this result by means of a diagram.
The point \(P\) has coordinates \((x,y)\) with respect to the origin \(O\). By writing \(x=r\cos\theta\) and \(y=r\sin\theta\), or otherwise, show that, if the line \(OP\) is rotated by \(60^\circ\) clockwise about \(O\), the new \(y\)-coordinate of \(P\) is \(\frac12(y-\sqrt3\,x)\). What is the new \(y\)-coordinate in the case of an anti-clockwise rotation by \(60^\circ\,\)? An equilateral triangle \(OBC\) has vertices at \(O\), \((1,0)\) and \((\frac12,\frac12 \sqrt3)\), respectively. The point \(P\) has coordinates \((x,y)\). The perpendicular distance from \(P\) to the line through \(C\) and \(O\) is \(h_1\); the perpendicular distance from \(P\) to the line through \(O\) and \(B\) is \(h_2\); and the perpendicular distance from \(P\) to the line through \(B\) and \(C\) is \(h_3\). Show that \(h_1=\frac12 \big\vert y-\sqrt3\,x\big\vert\) and find expressions for \(h_2\) and \(h_3\). Show that \(h_1+h_2+h_3=\frac12 \sqrt3\) if and only if \(P\) lies on or in the triangle \(OBC\).
Solution:
Two identical particles \(P\) and \(Q\), each of mass \(m\), are attached to the ends of a diameter of a light thin circular hoop of radius \(a\). The hoop rolls without slipping along a straight line on a horizontal table with the plane of the hoop vertical. Initially, \(P\) is in contact with the table. At time \(t\), the hoop has rotated through an angle \(\theta\). Write down the position at time \(t\) of \(P\), relative to its starting point, in cartesian coordinates, and determine its speed in terms of \(a\), \(\theta\) and \(\dot\theta\). Show that the total kinetic energy of the two particles is \(2ma^2\dot\theta^2\). Given that the only external forces on the system are gravity and the vertical reaction of the table on the hoop, show that the hoop rolls with constant speed.
Solution:
On the (flat) planet Zog, the acceleration due to gravity is \(g\) up to height \(h\) above the surface and \(g'\) at greater heights. A particle is projected from the surface at speed \(V\) and at an angle \(\alpha\) to the surface, where \(V^2 \sin^2\alpha > 2 gh\,\). Sketch, on the same axes, the trajectories in the cases \(g'=g\) and \(g' < g\). Show that the particle lands a distance \(d\) from the point of projection given by \[ d = \left(\frac {V-V'} g + \frac {V'}{ g'} \right) V\sin2\alpha\,, \] where \(V' = \sqrt{V^2-2gh\,\rm{cosec}^2\alpha\,}\,\).
A straight uniform rod has mass \(m\). Its ends \(P_1\) and \(P_2\) are attached to small light rings that are constrained to move on a rough rigid circular wire with centre \(O\) fixed in a vertical plane, and the angle \(P_1OP_2\) is a right angle. The rod rests with \(P_1\) lower than \(P_2\), and with both ends lower than \(O\). The coefficient of friction between each of the rings and the wire is \(\mu\). Given that the rod is in limiting equilibrium (i.e. \ on the point of slipping at both ends), show that \[ \tan \alpha = \frac{1-2\mu -\mu^2}{1+2\mu -\mu^2}\,, \] where \(\alpha\) is the angle between \(P_1O\) and the vertical (\(0<\alpha<45^\circ\)). Let \(\theta\) be the acute angle between the rod and the horizontal. Show that \(\theta =2\lambda\), where \(\lambda \) is defined by \(\tan \lambda= \mu\) and \(0<\lambda<22.5^\circ\).
In this question, you may use without proof the results: \[ \sum_{r=1}^n r = \tfrac12 n(n+1) \qquad\text{and}\qquad \sum_{r=1}^n r^2 = \tfrac1 6 n(n+1)(2n+1)\,. \] The independent random variables \(X_1\) and \(X_2\) each take values \(1\), \(2\), \(\ldots\), \(N\), each value being equally likely. The random variable \(X\) is defined by \[ X= \begin{cases} X_1 & \text { if } X_1\ge X_2\\ X_2 & \text { if } X_2\ge X_1\;. \end{cases} \]
Solution: \begin{align*} \P(X = r) &= \P(X_1 = r, X_2 \leq r) + \P(X_2 = r, X_1 < r) \\ &= \P(X_1 = r) \P(X_2 \leq r) + \P(X_2 = r)\P( X_1 < r) \\ &= \frac{1}{N} \frac{r}{N} + \frac{1}{N} \frac{r-1}{N} \\ &= \frac{2r-1}{N^2} \end{align*} \begin{align*} \E(X) &= \sum_{r=1}^N r \P(X = r) \\ &= \sum_{r=1}^N \frac{2r^2 - r}{N^2} \\ &= \frac{1}{N^2} \l \frac{N(N+1)(2N+1)}{3} - \frac{N(N+1)}{2} \r \\ &= \frac{N+1}{N} \l \frac{4N-1}{6} \r \end{align*} When \(N = 100\), this is equal to \(\frac{101 \cdot 399}{6 \cdot 100} = \frac{101 \cdot 133}{200} = 67.165\) \begin{align*} &&\frac12 &\leq \P(X \leq m) \\ &&&=\sum_{r=1}^m \P(X=r) \\ &&&=\sum_{r=1}^m \frac{2r-1}{N^2} \\ &&&= \frac{1}{N^2} \l m(m+1) - m \r \\ &&&= \frac{m^2}{N^2} \\ \Rightarrow && m^2 &\geq \frac{N^2}{2} \\ \Rightarrow && m &\geq \frac{N}{\sqrt{2}} \\ \Rightarrow && m &= \left \lceil \frac{N}{\sqrt{2}} \right \rceil \end{align*} When \(N = 100\), \(100/\sqrt{2} = \sqrt{2}50\). \(\sqrt{2} > 1.4 \Rightarrow 50\sqrt{2} > 70\) \(\sqrt{2} < 1.42 \Rightarrow 50 \sqrt{2} < 71\), therefore \(\displaystyle \left \lceil \frac{100}{\sqrt{2}} \right \rceil = 71\) \begin{align*} \lim_{N \to \infty} \frac{\frac{(N+1)(4N-1)}{6N}}{ \left \lceil\frac{N}{\sqrt{2}} \right \rceil} &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l \frac{4N^2 +3N - 1}{2N^2} \r \tag{since the floor will be irrelevant}\\ &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l 2 + \frac{3}{2N} - \frac{1}{N^2} \r \\ &= \lim_{N \to \infty} \frac{2\sqrt{2}}{3} \end{align*}