Problems

Explain why, if \(\mathrm{A, B}\) and \(\mathrm{C}\) are three events, \[ \mathrm{P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) +P(A \cap B \cap C)}, \] where \(\mathrm{P(X)}\) denotes the probability of event \(\mathrm{X}\). A cook makes three plum puddings for Christmas. He stirs \(r\) silver sixpences thoroughly into the pudding mixture before dividing it into three equal portions. Find an expression for the probability that each pudding contains at least one sixpence. Show that the cook must stir 6 or more sixpences into the mixture if there is to be less than \({1 \over 3}\) chance that at least one of the puddings contains no sixpence. Given that the cook stirs 6 sixpences into the mixture and that each pudding contains at least one sixpence, find the probability that there are two sixpences in each pudding.

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Solution:

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When we add everything in \(A\),\(B\), \(C\) we overcount the overlaps. When we remove the overlaps we remove the centre section too many times, so we have to add it back on in the end. Let \(X_i\) be the probability that the \(i\)th pudding contains a sixpence. \begin{align*} && \mathbb{P}(X_1^c \cup X_2^c \cup X_3^c) &=\mathbb{P}(X_1^c \cap X_2^c \cap X_3^c) + \mathbb{P}(X_1^c)+\mathbb{P}(X_2^c)-\mathbb{P}(X_3^c)+\\ &&&\quad\quad-\mathbb{P}(X_1^c \cap X_2^c )-\mathbb{P}( X_2^c \cap X_3^c)-\mathbb{P}(X_1^c \cap X_3^c) \\ &&&= 0 + (\tfrac23)^r+ (\tfrac23)^r+ (\tfrac23)^r + \\ &&&\quad\quad - (\tfrac13)^r- (\tfrac13)^r- (\tfrac13)^r \\ &&&= \frac{3\cdot2^r-3}{3^{r}} \\ \Rightarrow && \mathbb{P}(\text{all contain a sixpence}) &= 1 - \frac{3\cdot2^r-3}{3^{r}} \\ &&&= \frac{3^r-3\cdot2^r+3}{3^r} \end{align*} When \(r = 5\) we have \(\frac{3 \cdot 32-3}{3^5} = \frac{31}{81} > \frac13\) When \(r = 6\) we have \(\frac{3 \cdot 64-3}{3^6} = \frac{7}{27} < \frac13\) Therefore, the chef must stir in at least \(6\). \begin{align*} && \mathbb{P}(\text{two in each}|\text{at least 1 in each}) &= \frac{ \mathbb{P}(\text{two in each} \cap \text{at least 1 in each}) }{ \mathbb{P}(\text{at least 1 in each}) } \\ &&&= \frac{ \mathbb{P}(\text{two in each}) }{ \mathbb{P}(\text{at least 1 in each}) } \\ &&&= \frac{90/3^6}{20/27} \\ &&&= \frac{1/3}{2} = \frac16 \end{align*}

Show that \[ \int_0^a \frac{\sinh x}{2\cosh^2 x -1} \, \mathrm{d} x = \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}\cosh a -1}{\sqrt{2}\cosh a +1}\r + \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \] and find \[ \int_0^a \frac{\cosh x}{1+2\sinh^2 x} \, \mathrm{d} x \, . \] Hence show that \[ \int_0^\infty \frac{\cosh x - \sinh x}{1+2\sinh^2 x} \, \mathrm{d} x = \frac{\pi}{2\sqrt{2}} - \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \, . \] By substituting \(u = \e^x\) in this result, or otherwise, find \[ \int_1^\infty \frac{1}{1+u^4} \, \mathrm{d} u \, . \]

The equation of a curve is \(y=\f ( x )\) where \[ \f ( x ) = x-4-\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r} \;. \]

1. 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.
2. Show that the equation \(\f ( x ) =0\) has a double root.
3. Sketch the curve.

Given that \(\f''(x) > 0\) when \(a \le x \le b\,\), explain with the aid of a sketch why \[ (b-a) \, \f \Big( {a+b \over 2} \Big) < \int^b_a \f(x) \, \mathrm{d}x < (b-a) \, \displaystyle \frac{\f(a) + \f(b)}{2} \;. \] By choosing suitable \(a\), \(b\) and \(\f(x)\,\), show that \[ {4 \over (2n-1)^2} < {1 \over n-1} - {1 \over n} < {1 \over 2} \l {1 \over n^2} + {1 \over (n-1)^2}\r \,, \] where \(n\) is an integer greater than 1. Deduce that \[ 4 \l {1 \over 3^2} +{1 \over 5^2} + {1 \over 7^2} + \cdots \r < 1 < {1 \over 2} + \left( {1 \over 2^2} +{1 \over 3^2} + {1 \over 4^2} + \cdots \right)\,. \] Show that \[ {1 \over 2} \l {1 \over 3^2} + {1 \over 4^2} + {1 \over 5^2} + \frac 1 {6^2} + \cdots \r < {1 \over 3^2} +{1 \over 5^2} + {1 \over 7^2} + \cdots \] and hence show that \[ {3 \over 2} \displaystyle < \sum_{n=1}^\infty {1 \over n^2} <{7 \over 4}\;. \]

The triangle \(OAB\) is isosceles, with \(OA = OB\) and angle \(AOB = 2 \alpha\) where \(0< \alpha < {\pi \over 2}\,\). The semi-circle \(\mathrm{C}_0\) has its centre at the midpoint of the base \(AB\) of the triangle, and the sides \(OA\) and \(OB\) of the triangle are both tangent to the semi-circle. \(\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3, \ldots\) are circles such that \(\mathrm{C}_n\) is tangent to \(\mathrm{C}_{n-1}\) and to sides \(OA\) and \(OB\) of the triangle. Let \(r_n\) be the radius of \(\mathrm{C}_n\,\). Show that \[ \frac{r_{n+1}}{r_n} = \frac{1-\sin\alpha}{1+\sin\alpha}\;. \] Let \(S\) be the total area of the semi-circle \(\mathrm{C}_0\) and the circles \(\mathrm{C}_1\), \(\mathrm{C}_2\), \(\mathrm{C}_3\), \(\ldots\;\). Show that \[ S = {1 + \sin^2 \alpha \over 4 \sin \alpha} \, \pi r_0^2 \;. \] Show that there are values of \(\alpha\) for which \(S\) is more than four fifths of the area of triangle~\(OAB\).

Show that if \(\, \cos(x - \alpha) = \cos \beta \,\) then either \(\, \tan x = \tan ( \alpha + \beta)\,\) or \(\; \tan x = \tan ( \alpha - \beta)\,\). By choosing suitable values of \(x\), \(\alpha\) and \(\beta\,\), give an example to show that if \(\,\tan x = \tan ( \alpha + \beta)\,\), then \(\,\cos(x - \alpha) \, \) need not equal \( \cos \beta \,\). Let \(\omega\) be the acute angle such that \(\tan \omega = \frac 43\,\).

1. For \(0 \le x \le 2 \pi\), solve the equation \[ \cos x -7 \sin x = 5 \] giving both solutions in terms of \(\omega\,\).
2. For \(0 \le x \le 2 \pi\), solve the equation \[ 2\cos x + 11 \sin x = 10 \] showing that one solution is twice the other and giving both in terms of \(\omega\,\).

Given a sequence \(w_0\), \(w_1\), \(w_2\), \(\ldots\,\), the sequence \(F_1\), \(F_2\), \(\ldots\) is defined by $$F_n = w_n^2 + w_{n-1}^2 - 4w_nw_{n-1} \,.$$ Show that $\; F_{n}-F_{n-1} = \l w_n-w_{n-2} \r \l w_n+w_{n-2}-4w_{n-1} \r \; \( for \)n \ge 2\,$.

1. The sequence \(u_0\), \(u_1\), \(u_2\), \(\ldots\) has \(u_0 = 1\), and \(u_1 = 2\) and satisfies \[ u_n = 4u_{n-1} -u_{n-2} \quad (n \ge 2)\;. \] Prove that \ $ u_n^2 + u_{n-1}^2 = 4u_nu_{n-1}-3 \; $ for \(n \ge 1\,\).
2. A sequence \(v_0\), \(v_1\), \(v_2\), \(\ldots\,\) has \(v_0=1\) and satisfies \begin{equation*} v_n^2 + v_{n-1}^2 = 4v_nv_{n-1}-3 \quad (n \ge 1). \tag{\(\ast\)} \end{equation*} \makebox[7mm]{(a) \hfill}Find \(v_1\) and prove that, for each \(n\ge2\,\), either \(v_n= 4v_{n-1} -v_{n-2}\) or \(v_n=v_{n-2}\,\). \makebox[7mm]{(b) \hfill}Show that the sequence, with period 2, defined by \begin{equation*} v_n = \begin{cases} 1 & \mbox{for \(n\) even} \\ 2 & \mbox{for \(n\) odd} \end{cases} \end{equation*} \makebox[7mm]{\hfill}satisfies \((\ast)\). \makebox[7mm]{(c) \hfill}Find a sequence \(v_n\) with period 4 which has \(v_0=1\,\), and satisfies~\((\ast)\).

For \(n=1\), \(2\), \(3\), \(\ldots\,\), let \[ I_n = \int_0^1 {t^{n-1} \over \l t+1 \r^n} \, \mathrm{d} t \, . \] By considering the greatest value taken by \(\displaystyle {t \over t+1}\) for \(0 \le t \le 1\) show that \(I_{n+1} < {1 \over 2} I_{n}\,\). Show also that \(\; \displaystyle I_{n+1}= - \frac 1{\; n\, 2^n} + I_{n}\,\). Deduce that \(\; \displaystyle I_n < \frac1 {\; n \, 2^{n-1}}\,\). Prove that \[ \ln 2 = \sum_{r=1}^n {1 \over \; r\, 2^r} + I_{n+1} \] and hence show that \({2 \over 3} < \ln 2 < {17 \over 24}\,\).

Show that if \[ {\mathrm{d}y \over \mathrm{d} x}=\f(x)y + {\g(x) \over y} \] then the substitution \(u = y^2\) gives a linear differential equation for \(u(x)\,\). Hence or otherwise solve the differential equation \[ {\mathrm{d}y \over \mathrm{d} x}={y \over x} - {1 \over y}\;. \] Determine the solution curves of this equation which pass through \((1 \,, 1)\,\), \((2\, , 2)\) and \((4 \, , 4)\) and sketch graphs of all three curves on the same axes.

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Solution: \begin{align*} && \frac{\d y}{\d x} &= f(x) y + \frac{g(x)}{y} \\ && y \frac{\d y}{\d x} &= f(x) y^2 + g(x) \\ u = y^2: && \frac12 \frac{\d u}{\d x} &= f(x) u + g(x) \end{align*} Which is a linear differential equation for \(u\). \begin{align*} && \frac12 u' &= \frac1x u -1 \\ \Rightarrow && u' - \frac2xu &= -1 \\ \Rightarrow && \frac{1}{x^2} u' - \frac{2}{x^3} u &= -\frac{1}{x^2} \\ \Rightarrow && (\frac{u}{x^2})' &= - \frac{1}{x^2} \\ \Rightarrow && \frac{u}{x^2} &= \frac1x + C \\ \Rightarrow && u &= Cx^2 + x \\ \Rightarrow && y^2 &= Cx^2 + x \end{align*} If \((1,1)\) is on the curve then \(1 = C + 1 \Rightarrow C = 0 \Rightarrow y^2 = x\). If \((2,2)\) is on the curve then \(4 = 4C + 2 \Rightarrow C = \frac12 \Rightarrow y^2 = x + \frac12 x^2\). If \((3,3)\) is on the curve then \(9 = 9C + 3 \Rightarrow C = \frac23 \Rightarrow y^2 = x + \frac23 x^2\)

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A circular hoop of radius \(a\) is free to rotate about a fixed horizontal axis passing through a point \(P\) on its circumference. The plane of the hoop is perpendicular to this axis. The hoop hangs in equilibrium with its centre, \(O\), vertically below \(P\). The point \(A\) on the hoop is vertically below \(O\), so that \(POA\) is a diameter of the hoop. A mouse \(M\) runs at constant speed \(u\) round the rough inner surface of the lower part of the hoop. Show that the mouse can choose its speed so that the hoop remains in equilibrium with diameter \(POA\) vertical. Describe what happens to the hoop when the mouse passes the point at which angle \(AOM = 2 \arctan \mu\,\), where \(\mu\) is the coefficient of friction between mouse and hoop.