Problems

Solution: \begin{align*} S &= \sin^{2}\theta+\sin^{2}\phi+\sin^{2}\psi+2\sin\theta\sin\phi\sin\psi \\ &= \sin^{2}\theta+\sin^{2}\phi+\sin^{2}(\tfrac\pi2-\theta-\phi)+2\sin\theta\sin\phi\sin(\tfrac\pi2-\theta-\phi) \\ &= \sin^{2}\theta+\sin^{2}\phi+\cos^{2}(\theta+\phi)+2\sin\theta\sin\phi\cos(\theta+\phi) \\ &= \sin^{2}\theta+\sin^{2}\phi+\left ( \cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)\right)^2+2\sin\theta\sin\phi\left ( \cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)\right) \\ &= \sin^{2}\theta+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi-\sin^2 \theta \sin^2 \phi \\ &= \sin^{2}\theta(1-\sin^2 \phi)+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi \\ &= \sin^{2}\theta\cos^2 \phi+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi \\ &= \sin^{2}\phi+\cos^2 \phi \\ &= 1 \end{align*} Suppose \(\theta = \phi = \tfrac15 \pi, \psi = \tfrac1{10}\pi\). Also let \(s = \sin \tfrac1{10}\) \begin{align*} 1 &= 2\sin^2 \tfrac15 \pi + \sin^2 \tfrac1{10} \pi + 2 \sin^2\tfrac15 \pi \sin \tfrac1{10} \pi \\ &= 8\sin^2 \tfrac1{10} \pi \cos^2 \tfrac1{10} \pi + \sin^2 \tfrac1{10} \pi + 8 \sin^2 \tfrac1{10} \pi \cos^2 \tfrac1{10} \pi \sin \tfrac1{10} \pi \\ &= 8\sin^2 \tfrac1{10} \pi(1- \sin^2 \tfrac1{10} \pi) + \sin^2 \tfrac1{10} \pi + 8 \sin^2 \tfrac1{10} \pi (1-\sin^2 \tfrac1{10} \pi) \sin \tfrac1{10} \pi \\ &= 8s^2(1-s^2)+s^2 + 8s^2(1-s^2)s \\ &= -8 s^5 - 8 s^4 + 8 s^3 + 9 s^2 \end{align*} Therefore \(s\) is a root of \(8s^5+8s^4-8s^3-9s^2+1 = 0\), but notice that \begin{align*} 8s^5+8s^4-8s^3-9s^2+1 &= (s-1)(8 s^4 + 16 s^3 + 8 s^2 - s - 1 ) \\ &= (s-1)(s+1)(8s^3+8s^2-1) \end{align*} Therefore since \(\sin \tfrac{1}{10} \pi \neq \pm 1\) it must be a root of \(8x^3+8x^2-1=0\)

Solution: \(V_h = \frac43 \pi r_h^3, S_h = 4 \pi r_h^2\) \(\frac{\d V_h}{\d t} = -k4\pi r_h^2 \Rightarrow 4\pi r_h^2 \frac{\d r_h}{\d t} = -k 4\pi r_h^2 \Rightarrow \frac{\d r_h}{\d t} = -k\) Therefore \(r_h = 2R - kt, r_b = 3R - kt\). The height will halve when \(2kt = \frac{5}{2}R \Rightarrow kt = \frac{5}{4}R\) and the two sections will have radii \(\frac{3}{4}R\) and \(\frac{7}{4}R\) and the ratio of the volumes will be: \begin{align*} \frac{\frac{3^3}{4^3}+\frac{7^3}{4^3}}{2^3+3^3} = \frac{37}{224} \end{align*} \begin{align*} && \frac{\d V}{\d t} &= -4\pi k(r_h^2+r_b^2) \\ && \frac{\d S}{\d t} &= -8\pi k (r_h+r_b) \\ \Rightarrow && \frac{\d V}{\d S} &= \frac{r_h^2 + r_b^2}{2(r_h+r_b)} \\ &&&= \frac{(2R-kt)^2+(3R-kt)^2}{2(5R-2kt)} \\ &&&= \frac{13R^2-10Rkt+2k^2t^2}{2(5R-2kt)} \\ &&&= \frac{13R^2-10Rs + 2s^2}{2(5R-2s)} \end{align*} Where \(s = kt\) and \(0 \leq s \leq 2R\). We can maximise this but differentiating wrt to \(s\). \begin{align*} \Rightarrow && &= \frac{(-10R+4s)(10R-4s)+4(13R^2-10Rs+2s^2)}{4(5R-2s)^2} \\ &&&= \frac{-48R^2+40Rs-8s^2}{4(5R-2s)^2} \\ &&&= \frac{-8(s-2R)(s-3R)}{4(5R-2s)^2} \\ &&&<0 \end{align*} Therefore it is largest when \(s = 0\), ie \(\frac{13R^2}{10R} = \frac{13}{10}R\)

Solution: \begin{align*} && | \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} | &= d \\ && \arg \left ( \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} \right) &= \arg (z_{n+1}-z_{n}) - \arg(z_{n}-z_{n-1}) \\ &&&= \theta \\ \Rightarrow && \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} &= d e^{i \theta} \end{align*} \begin{align*} && z_1 &= 0 \\ && z_2 &= 1 \\ && \frac{z_3-z_2}{z_2-z_1} &= de^{i \theta} \\ \Rightarrow && z_3 &= de^{i \theta} + 1 \\ && \frac{z_4-z_3}{z_3-z_2} &= de^{i \theta} \\ \Rightarrow && z_4 &= (d e^{i \theta})^2 + d e^{i \theta} + 1\\ \Rightarrow && z_{n+1} &= \frac{(de^{i \theta})^{n}-1}{de^{i \theta}-1} \end{align*} If \(d = 2, \theta = \tfrac13 \pi\), then, \(2e^{i \tfrac13 \pi} = 1 + \sqrt{3}i\) \begin{align*} \textrm{Im}(z_{n+1})) &= \textrm{Im} \left ( \frac{(2e^{i \tfrac13 \pi})^{n}-1}{2e^{i \tfrac13 \pi}-1}\right) \\ &= \textrm{Im} \left ( \frac{(2e^{i \tfrac13 \pi})^{n}-1}{\sqrt{3}i}\right) \\ &= -\frac{1}{\sqrt{3}}\textrm{Re} \left (2^n e^{i \frac{n}{3} \pi} \right) + \frac{1}{\sqrt{3}} \end{align*} Which clearly changes sign infinitely many times, ie crosses the origin infinitely many times.

Solution:

+ - ↺

1. Since the area not visible is increasing as \(p\) increases, we would like \(p\) to be as large as possible, ie \(p = 4\).
2. Similarly, he can see the most when \(p =0\)

Solution: \begin{align*} && \mathbf{0} &= \sum_i \mathbf{a}_i \tag{ii} \\ \Rightarrow && 0 &= \mathbf{a}_i \cdot \mathbf{0} \\ &&&= \sum_j \mathbf{a}_i \cdot \mathbf{a}_j \\ &&&= (n-1)\beta + \alpha^2 \tag{i} \\ \Rightarrow && (n-1)\beta &= -\alpha^2 \end{align*} Suppose we have \(\mathbf{a}_j = \binom{x}{y}\), \(j \neq 1\) then \(x = \beta\). We also must have \(\beta^2 + y^2 = 1\), so there are at most two values for \(y\), ie two extra vectors. ie \(n = 2, 3\). If \(n = 2 \Rightarrow \mathbf{a}_2 = - \mathbf{a}_1\). If \(n = 3\) \begin{align*} && \mathbf{0} &= \binom{1}{0} + \binom{\beta}{y} + \binom{\beta}{-y} \\ \Rightarrow && \beta = -1/2 \\ \Rightarrow && y &= \pm \frac{\sqrt{3}}{2} \end{align*} Suppose $\mathbf{a}_{1}=\begin{pmatrix}1\\ 0\\ 0 \end{pmatrix}\(, \)\mathbf{a}_{2}=\begin{pmatrix}\cos \theta \\ \sin \theta \\ 0 \end{pmatrix}$ (since we need \(\mathbf{a}_2 \cdot \mathbf{a}_2 = 1\)). \(\beta = \cos \theta\)). We can have \(\cos \theta = - 1\). Suppose we have \(\mathbf{a}_j =\begin{pmatrix}x\\ y \\ z \end{pmatrix}\), so \(x = \cos \theta\), and \(y^2 + z^2 = \sin^2 \theta\), so we can write it as: \(\mathbf{a}_j =\begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix}\). We must also have \(\beta = \begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix} \cdot \begin{pmatrix} \cos \theta \\ \sin \theta\\ 0 \end{pmatrix} = \cos^2 \theta + \sin^2 \theta \cos \phi = \cos \theta\), so \(\cos \phi = \frac{\cos \theta - \cos^2\theta}{1-\cos^2 \theta} = \frac{\cos \theta}{1+\cos \theta}\). Therefore there is one value for \(\cos \phi\), so at most two values for \(\sin \phi\), Therefore we can have either \(2, 3,4\) or \(5\) different values in the set. \(n = 2\), we've already handled. If \(n = 3\), then \(\beta = -\frac12\), \(\cos \phi = -1\), so we can only have two different values for \(\sin \theta\), ie: \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ \frac{\sqrt{3}}{2} \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ -\frac{\sqrt{3}}{2} \\ 0 \end{pmatrix} \right \}\) Finally, if \(n = 4\), we have \(\beta = -\frac13\), \(\cos \phi = \frac{-1/3}{2/3} = -\frac12\). \(\sin \theta = \pm \frac{\sqrt{3}}{2}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix} \right \}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix} \right \}\) If \(n = 5\), then \(\beta = -\frac14\), \(\cos \phi = \frac{-1/4}{3/4} = -\frac13\). \(\sin \theta = \frac{\sqrt{15}}{4}\), \(\sin \phi = \frac{2\sqrt{2}}{3}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{12} \\ \frac{\sqrt{30}}{6} \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{12} \\ -\frac{\sqrt{30}}{6} \end{pmatrix}, \right \}\)

Solution:

1. \begin{align*} && g(x) &= \frac{2x^3+3}{x^4+4} \\ \Rightarrow && g'(x) &= \frac{6x^2(x^4+4) - (2x^3+3)(4x^3)}{(x^4+4)^2} \\ &&&= \frac{-2x^6-12x^3+24x^2}{(x^4+4)} \\ &&&= \frac{-2x^2(x^4+6x-12)}{(x^4+4)} \end{align*} So \(g'(x)\) is not \(0\) for \(x = \pm 1\). We can also note that \(g(-1) = \frac1{5} \neq 1\) Even if the other turning point was \(1\), we would also need to check the behaviour as \(x \to \pm \infty\). We can also note that \(g(-1) = \frac{1}{5} < \frac34\) so the conclusion is also not true.
2. There are several errors. \[ \int (1-x)^{-3} \d x = \underbrace{\frac{1}{4}}_{\text{correct constant}}(1-x)^{-4} + \underbrace{C}_{\text{constant of integration}} \] We cannot integrate through the asymptote at \(1\). There is a sense in which we could argue \(\displaystyle \int_{-1}^3 (1-x)^{-3} \d x = 0\), specifically using Cauchy principal value \begin{align*} \mathrm {p.v.} \int_{-1}^3 (1-x)^{-3} &=\lim_{\epsilon \to 0} \left [ \int_{-1}^{1-\epsilon} (1-x)^{-3} \d x+ \int_{1+\epsilon}^{3} (1-x)^{-3} \d x\right] \\ &=\lim_{\epsilon \to 0} \left [ \left[ \frac14 (1-x)^{-4}\right]_{-1}^{1-\epsilon}+ \left[ \frac14 (1-x)^{-4}\right]_{1+\epsilon}^3\right] \\ &=\lim_{\epsilon \to 0} \left [ \frac14 \epsilon^{-4}-\frac14 \frac1{2^4} + \frac14 \frac1{2^4} - \frac14 \epsilon^{-4} \right] \\ &= \lim_{\epsilon \to 0} 0 \\ &= 0 \end{align*} However, in many normal ways of treating this integral it would be undefined.

Solution:

1. \begin{align*} y = \pi - x, \d y = -\d x: && \int_0^{\pi} x f(\sin x) &= \int_{y = \pi}^{y = 0}(\pi - y) f(\sin(\pi-y))- \d y \\ &&&= \int_0^{\pi} (\pi -y) f(\sin y) \d y \\ \Rightarrow && 2 \int_0^{\pi} x f(\sin x)\d x &= \pi \int_0^{\pi} f(\sin x) \d x \\ &&&= \pi \int_0^{\pi/2} f(\sin x ) \d x + \pi \int_{\pi/2}^{\pi} f(\sin x ) \d x \\ &&&= \pi \int_0^{\pi/2} f(\sin x ) \d x +\pi \int_{y=\pi/2}^{y=0} f(\sin (\pi-y) ) (-\d y) \\ &&&= 2 \pi \int_0^{\pi/2} f(\sin x) \d x \\ \Rightarrow && \int_0^{\pi} x f(\sin x)\d x &= \pi \int_0^{\pi/2} f(\sin x) \d x \end{align*} Therefore if \(f(x) = \frac1{2+\sin x}\), letting \(t = \tan \frac{x}{2}\) we have \(\sin x = \frac{2 t}{1+t^2}, \frac{dt}{\d x} = \frac12 (1+t^2)\) \begin{align*} && \int_0^{\pi} \frac{x}{2 + \sin x } \d x &= \pi \int_0^{\pi/2} \frac{1}{2 + \sin x} \d x \\ &&&= \pi \int_{t = 0}^{t = 1} \frac{1}{2+\frac{2t}{1+t^2}} \frac{2}{1+t^2} \d t \\ &&&=\pi \int_0^1 \frac{2}{2t^2+2t+2} \d t\\ &&&=\pi \int_0^1 \frac{1}{(t+\tfrac12)^2 + \tfrac34} \d t\\ &&&= \pi \left [\frac{1}{\sqrt{3/4}} \tan^{-1} \frac{u}{\sqrt{3/4}} \right ]_{u=1/2}^{3/2} \\ &&&= \frac{2 \pi}{\sqrt{3}} \left ( \tan^{-1} \sqrt{3} - \tan^{-1} \frac1{\sqrt{3}} \right) \\ &&&= \frac{2 \pi}{\sqrt{3}} \left ( \frac{\pi}{3} - \frac{\pi}{6} \right) \\ &&&= \frac{\pi^2}{3\sqrt{3}} \end{align*}
2. Let \(u = (\sin^{-1} t)^2, \frac{\d u}{\d t} = 2(\sin^{-1} t) \frac{1}{\sqrt{1-t^2}}\) \begin{align*} \int_{0}^{1}\frac{(\sin^{-1}t)\cos\left[(\sin^{-1}t)^{2}\right]}{\sqrt{1-t^{2}}}\,\mathrm{d}t &= \int_{u=0}^{\pi^2/4} \frac12 \cos u \d u \\ &= \frac12 \sin \frac{\pi^2}{4} \end{align*}

Solution:

1. Suppose \(n > 4\) then the following inequalities are both true \begin{align*} 3 < x^3 < 4 & \Rightarrow 3^5 < x^{15} < 4^{5}\\ 5 < x^5 < 6 & \Rightarrow 5^{3} < x^{15} < 6^3 \end{align*} But \(3^5 = 243\) and \(6^3 = 216\) so \(243 < x^{15} < 216\) whichis a contradiction.
2. This question is wrong. Consider \(n = 2\), then \(\frac{1}{2+1} + \frac{1}{2+2} = \frac13+\frac14 = \frac{7}{12} < 1\). The question should be about \(n \geq 4\). \begin{align*} \frac{1}{n+1}+\frac1{n+2}+\cdots + \frac{1}{2n} > \frac{n}{2n} &= \frac12 \\ \frac{1}{2n+1}+\frac1{2n+2}+\cdots + \frac{1}{3n} > \frac{n}{3n} &= \frac13 \\ \frac{1}{4n+1}+\frac1{4n+2}+\cdots + \frac{1}{4n} > \frac{n}{4n} &= \frac14 \\ \sum_{k=1}^{n^2-n} \frac{1}{n+k} > \frac{13}{12} &> 1 \end{align*} We have a stronger result, \(\frac1{n+1} + \cdots + \frac1{4n} > 1\) for \(n > 4\) so we can take \(N = 4^{10}\) since, since there will be \(9\) sequences from \(\frac{1}{4^{i}+1} \to \frac{1}{4^{i+1}}\) and we will have \(\frac1{1}\) at the start to give use the extra \(1\).