Problems

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Solution: Writing \(c = \cos \theta, s = \sin \theta\) then de Moivre states that: \begin{align*} && \cos 9 \theta + i \sin 9 \theta &= (c +i s)^9 \\ &&&= c^9 + 9ic^8s - 36c^7s^2-84ic^6s^3+126c^5s^4 + 126ic^4s^5 -84c^3s^6 -36ic^2s^7+9cs^8+is^9 \\ &&&= (c^9-36c^7s^2+126c^5s^3-84c^3s^6+8cs^8)+i(9c^8s-75c^6s^3+126c^4s^5-36c^2s^7+s^9) \\ \Rightarrow && \tan 9\theta &= \frac{(9c^8s-75c^6s^3+126c^4s^5-36s^2c^7+s^9)}{(c^9-36c^7s^2+126c^5s^4-84c^3s^6+8cs^8)} \\ &&&= \frac{9t-75t^3+126s^5-36t^7+t^9}{1-36t^2+126t^4-84t^6+8t^8} \\ &&&= \frac{t(t^{2}-3)(t^{6}-33t^{4}+27t^{2}-3)}{(3t^{2}-1)(3t^{6}-27t^{4}+33t^{2}-1)} \end{align*} If we consider \(\tan 9\theta = 0\) it will have the roots \(\theta = \frac{n \pi}{9}, n \in \mathbb{Z}\), in particular, the numerator of our fraction for \(\tan 9 \theta\) will be zero for \(t = 0, \tan \frac{\pi}{9}, \tan \frac{2\pi}{9}, \tan \frac{3\pi}{9}, \tan \frac{4 \pi}{9}, \tan \frac{5\pi}{9}, \tan \frac{6 \pi}{9}, \tan \frac{7 \pi}{9}, \tan \frac{8\pi}{9}\). It's worth noting all other values of \(\theta\) will repeat these values. Also note that \(0,\tan \frac{\pi}{3}, \tan \frac{2\pi}{3}\) are the roots of \(t\) and \(t^2-3\) respectively. Therefore the other values are the roots of our sextic. However, also note that \(\tan \frac{8\pi}{9} = - \tan \frac{\pi}{9}\) and similar, therefore we can notice that all the roots in pairs can be mapped to \(\tan \frac{\pi}{9}, \tan \frac{2 \pi}{9}\) and \(\tan \frac{4 \pi}{9}\) and all those values are squared, so the roots of: \(x^3 - 33x^2+27x-3\) will be \(\tan^2 \frac{\pi}{9}, \tan^2 \frac{2 \pi}{9}\) and \(\tan^2 \frac{4 \pi}{9}\). The product of the roots will be \(3\), so \begin{align*} && \tan^2 \frac{\pi}{9} \tan^2 \frac{2 \pi}{9} \tan^2 \frac{4 \pi}{9} &= 3 \\ \Rightarrow && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \pm \sqrt{3} \\ \underbrace{\Rightarrow}_{\text{all positive}} && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \sqrt{3} \\ \end{align*} Notice that \(x^3 + y^3 +z^3 - 3xyz = (x+y+z)((x+y+z)^2-3(xy+yz+zx))\) Therefore \begin{align*} \tan^{6}\left(\frac{\pi}{9}\right)+\tan^{6}\left(\frac{2\pi}{9}\right)+\tan^{6}\left(\frac{4\pi}{9}\right) &= 33(33^2-3\cdot27) + 3 \cdot 3 \\ &= 33\,273 \end{align*}

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Solution:

1. The line \(OA\) is \(\lambda \begin{pmatrix} a^3 \\ a^2 \\ a \end{pmatrix}\). The line \(BC\) is \(\begin{pmatrix} b^3 \\ b^2 \\ b \end{pmatrix} + \mu \begin{pmatrix} c^3-b^3 \\ c^2-b^2 \\ c-b \end{pmatrix}\). If these lies are concurrent then there would be \(\mu\) and \(\lambda\) such that they are equal, and in particular, \begin{align*} && \frac{b^2 + \mu(c^2-b^2)}{b + \mu (c-b)} &= \frac{b^3 + \mu(c^3-b^3)}{b^2 + \mu (c^2-b^2)} \\ \Leftrightarrow && (b^2 + \mu(c^2-b^2))^2 &= (b+\mu(c-b))(b^3+\mu(c^3-b^3)) \\ && b^4 +2\mu b^2 (c^2-b^2) + \mu^2 (c^2-b^2) &= b^4 + \mu(c-b)b^3 + \mu b(c^3-b^3) + \mu^2 (c-b)(c^3-b^3) \\ \Leftrightarrow && 2\mu b^2 (c+b) + \mu^2(c-b)(c+b)^2 &= \mu (b^3 + b(c^2+bc+b^2)) + \mu^2 (c^3-b^3) \\ && \mu = 0 & \Rightarrow a = b \\ \Leftrightarrow && b^2c - bc^2 &= \mu (c^3-b^3-(c-b)(c+b)^2) \\ \Leftrightarrow && bc(b-c) &= \mu (c-b)(c^2+bc+b^2-c^2-2bc-b^2) \\ \Leftrightarrow && bc &= \mu (bc) \\ \Leftrightarrow && \mu &= 1 \\ && \mu = -1 & \Rightarrow a = c \end{align*} Therefore there are no solutions.
2. \begin{align*} \cos \angle AOB &= \frac{ab+a^2b^2+a^3b^3}{\sqrt{a^2+a^4+a^6}\sqrt{b^2+b^4+b^6}} \\ &= \frac{3}{\sqrt{1 + a^2 + a^4} \sqrt{1 + b^2 + b^4}} \\ &> 0 \end{align*} Therefore the angle satisfies \(\angle AOB < \tfrac12 \pi\). We cannot achieve \(0\), since that would require \(a = b = 1\), therefore it falls in the range \(0 < \angle AOB < \tfrac12 \pi\)

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Solution: \begin{questionparts} \item \((ab)^2 = abab = e\) (since \(ab\) has order \(2\)), but \(a^2 = e, b^2 = e \Rightarrow a^{-1} = a, b^{-1} = b\) (since \(a\) and \(b\) have order 2) so \(ba = ab\) by multiplication on the left by \(a\) and right by \(b\). \item Suppose \((cd)^n = e \Leftrightarrow d(cd)^nc = dc \Leftrightarrow (dc)^n(dc) = e \Leftrightarrow (dc)^n = e\) Therefore any number for which \((cd)^n = e\) has the property that \((dc)^n = e\) and vice-versa, in particular the smallest number for either \(cd\) or \(dc\) will also be the smallest number for the other. \item Given \(c^{-1}bc=b^r\), then \(b^{rs} = (b^r)^s = (c^{-1}bc)^s =\underbrace{(c^{-1}bc)(c^{-1}bc) \cdots (c^{-1}bc)}_{s \text{ times}} = c^{-1}\underbrace{bb\cdots b}_{s \text{ times}}c = c^{-1}b^sc\) We proceed by induction on \(n\). When \(n = 0\), we have \(b^s = b^{sr^0}\) so the base case is true. Suppose it is true for some \(n = k\), ie \(c^{-k}b^sc^k = b^{sr^k}\). Now consider \(c^{-{k+1}}b^sc^{k+1} = c^{-1}c^{-k}b^sc^kc = c^{-1}b^{sr^k}c = (b^{sr^k \cdot r}) = b^{sr^{k+1}}\) (where the second to last equality was by the previous part). Therefore if our statement is true for \(n=k\) it is true for \(n = k+1\). Therefore, since it is also true for \(n=0\), by the principle of mathematical induction it is true for all non-negative integers \(n\).

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Solution: \begin{align*} \sum_{r = 0}^n \cos (\alpha + 2r \beta) &= \sum_{r = 0}^n \textrm{Re} \left ( \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (\sum_{r = 0}^n \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (e^{i \alpha} \sum_{r = 0}^n \ (e^{i 2 \beta})^r\right) \\ &= \textrm{Re} \left (e^{i \alpha} \frac{e^{2(n+1)\beta i}-1}{e^{2\beta i}-1} \right) \\ &= \textrm{Re} \left (e^{i \alpha} \frac{e^{(n+1)\beta i} (e^{(n+1)\beta i}-e^{-(n+1)\beta i})}{e^{\beta i}(e^{\beta i}-e^{-\beta i})} \right) \\ &= \textrm{Re} \left (\frac{e^{i \alpha} e^{(n+1)\beta i}}{e^{\beta i}} \frac{\sin (n+1) \beta}{\sin \beta} \right) \\ &= \textrm{Re} \left ( e^{i(\alpha + n \beta)}\frac{\sin (n+1) \beta}{\sin \beta} \right) \\ &= \frac{\cos (\alpha + n \beta)\sin (n+1) \beta}{\sin \beta} \end{align*} \begin{align*} \sum_{r = 0}^n \binom{n}{r} \cos (\alpha + 2r \beta) &= \sum_{r = 0}^n \textrm{Re} \left ( \binom{n}{r}\exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (\sum_{r = 0}^n \binom{n}{r} \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (e^{i \alpha}(e^{2\beta i}+1)^n \right) \\ &= \textrm{Re} \left (e^{i \alpha}e^{n\beta i}(e^{\beta i}+e^{-\beta i})^n \right) \\ &= \textrm{Re} \left (e^{i \alpha}e^{n\beta i}2^n \cos^n \beta \right) \\ &= 2^n \cos(\alpha + n \beta) \cos ^n \beta \end{align*} \begin{align*} \sum_{r = 0}^n \cos r \theta \sec^r \theta &= \sum_{r = 0}^n \textrm{Re} ( e^{i r \theta})\sec^r \theta \\ &= \textrm{Re} \left ( \sum_{r=0}^n e^{i r \theta} \sec^r \theta\right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n+1} \theta -1}{e^{i \theta}\sec \theta -1} \right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta}{e^{i \theta} -\cos \theta} \right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta}{i \sin \theta} \right) \\ &= \frac{1}{\sin \theta} \textrm{Im} \left ( e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta \right) \\ &= \frac{\sin(n+1) \theta \sec^{n} \theta}{\sin \theta} \end{align*}

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Solution: \begin{align*} \binom{n}{r} + \binom{n}{r-1} &= \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} \\ &= \frac{n!}{(r-1)!(n-r)!} \left ( \frac{1}{r} + \frac{1}{n-r+1} \right) \\ &= \frac{n!}{(r-1)!(n-r)!} \frac{(n-r+1)+r}{r(n-r+1)} \\ &= \frac{n! (n+1)}{r! (n-r+1)!} \\ &= \frac{(n+1)!}{r!(n+1-r)!} \\ &= \binom{n+1}{r} \end{align*} Claim: \(\displaystyle (uv)^{(n)} = \sum_{r=0}^n \binom{n}{r} u^{(n-r)} v^{(r)}\) Proof: (By induction on \(n\)). Base case: \(n = 0\) is clear. Inductive step: Suppose it is true for \(n = k\), then consider \begin{align*} (uv)^{(k+1)} &= \left ( (uv)^{(k)} \right)' \\ &= \left ( \sum_{r=0}^k \binom{k}{r} u^{(k-r)} v^{(r)} \right)' \tag{by assumption} \\ &=\sum_{r=0}^k \binom{k}{r} \left ( u^{(k-r)} v^{(r)}\right)' \tag{linearity} \\ &=\sum_{r=0}^k \binom{k}{r} \left ( u^{(k-r+1)} v^{(r)} + u^{(k-r)}v^{(r+1)}\right) \\ &= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)} + \sum_{r=0}^{k} \binom{k}{r} u^{(k-r)}v^{(r+1)} \\ &= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)} + \sum_{r=1}^{k+1} \binom{k}{r-1} u^{(k-r+1)}v^{(r)} \\ &= u^{(k+1)}v + \sum_{r=1}^k \left (\binom{k}{r} + \binom{k}{r-1} \right)u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\ &= u^{(k+1)}v + \sum_{r=1}^k \binom{k+1}{r} u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\ &= \sum_{r=0}^{k+1} \binom{k+1}{r} u^{(k-r+1)}v^{(r)}\\ \end{align*} Therefore if our statement is true for \(n = k\) it is true for \(n = k+1\). Since it is true for \(n = 0\) by the principle of mathematical induction it is true for all integer \(n \geq 0\) Suppose \( y = \sin^{-1} x\), then \(y' = \frac{1}{\sqrt{1-x^2}}\), \(y'' = \frac{x}{(1-x^2)^{3/2}}\). Not that this means that \((1-x^2)y'' - xy' = 0\) (which is our formula when \(n = 0\)). Now apply Leibniz's formula to this. \begin{align*} 0 &= \left ( (1-x^2)y'' - xy' \right)^{(n)} \\ &= \left ( (1-x^2)y'' \right)^{(n)} -\left ( xy' \right)^{(n)} \\ &= \left ( (1-x^2)y^{(n+2)} - n\cdot 2x \cdot y^{(n+1)}-\binom{n}{2} \cdot 2 \cdot y^{(n)} \right )- \left (xy^{(n+1)}+ny^{(n)} \right) \\ &= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - \left ( n(n-1)+n \right)y^{(n)} \\ &= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - n^2y^{(n)} \\ \end{align*} as required

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Solution: \begin{align*} T\begin{pmatrix}1\\ 2 \end{pmatrix} &= \frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}1\\ 2 \end{pmatrix} \\ &= \frac25\begin{pmatrix}9 - 4\\ -2+12 \end{pmatrix} \\ &= \begin{pmatrix}2\\ 4 \end{pmatrix} \\ &= 2 \begin{pmatrix}1\\ 2 \end{pmatrix} \end{align*} \begin{align*} T\begin{pmatrix}1\\ 2 \end{pmatrix} &= \frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}2\\ -1 \end{pmatrix} \\ &= \frac25\begin{pmatrix}18+2\\ -4-6 \end{pmatrix} \\ &= \begin{pmatrix}8\\ -4 \end{pmatrix} \\ &= 4 \begin{pmatrix}2\\ -1 \end{pmatrix} \end{align*} Consider $T^{-1} = \frac{5}{2} \frac{1}{50}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix}\(, so \)T^{-1} \begin{pmatrix}X\\ Y \end{pmatrix} = \begin{pmatrix}x\\ y \end{pmatrix}$ and so: \begin{align*} x^2 + y^2 & = \begin{pmatrix}x& y \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix} \\ &= \begin{pmatrix}X& Y \end{pmatrix} (T^{-1})^T T^{-1} \begin{pmatrix}X\\ Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix} \begin{pmatrix}X\\ Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix} \begin{pmatrix}6X+2Y\\ 2X+9Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}6(6X+2Y)+2(2X+9Y)\\ 2(6X+2Y)+9(2X+9Y) \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}40X+30Y\\ 30X +85Y \end{pmatrix} \\ &= \frac{1}{80}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}8X+6Y\\ 6X +17Y \end{pmatrix} \\ &= \frac{1}{80} \l 8X^2 + 12XY + 17Y^2\r \end{align*} Therefore \(8X^2 + 12XY + 17Y^2 = 80\). The area will be \(\det T \cdot \pi = \frac{4}{25} \cdot 50 \cdot \pi = 8 \pi\). Differentiating we obtain \(2 \cdot 8 \cdot X \cdot \frac{dX}{dY} + 2 \cdot 6 \cdot X + 2 \cdot 6 \cdot Y \cdot \frac{dX}{dY} + 2 \cdot 17 Y \Rightarrow \frac{dX}{dY} = -\frac{6X + 17Y}{8X+6Y}\), at a maximum (or minimum, \(6X = -17Y\)). Therefore \begin{align*} \Rightarrow && 8X^2 + 12 \cdot \frac{6}{17}X^2 + 17 ( -\frac{6}{17} X)^2 &= 80 \\ \Rightarrow && \frac{100}{17}X^2 &= 80 \\ \Rightarrow &&X^2 &= \frac{17 \cdot 4}{5} \\ \Rightarrow && |X| = 2 \sqrt {\frac{17}{5}} \end{align*} The point \(\frac15 (3,4)\) maps to \begin{align*} \frac{2}{5}\frac{1}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}3\\ 4 \end{pmatrix} &= \frac{2}{25} \begin{pmatrix}19\\ 18 \end{pmatrix} \end{align*} So the point is \((\frac{38}{25}, \frac{36}{25})\), with gradient \(\frac{dY}{dX} = -\frac{8X+6Y}{6X + 17Y}\) which is \(-\frac{8 \cdot 19+6 \cdot 18}{6\cdot 19 + 17 \cdot 18} = -\frac{13}{21}\) therefore the equation is \(21Y+13X = 50\)

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Solution: \begin{align*} && 0 &= xy^{2}+x^{3}+a^{2}y-a^{3} \\ \frac{\d }{\d x} : && 0 &= y^2+2xyy' + 3x^2+a^2y' \\ \Rightarrow && y' &= -\frac{y^2+3x^2}{a^2+2xy} \\ \\ \frac{\d^2 }{\d x^2}: && 0 &= 2yy'+2yy'+2x(y')^2+2xyy''+6x+a^2y'' \\ \Rightarrow && y'' &= -\frac{4yy'+2x(y')^2+6x}{a^2+2xy} \\ \\ P: && y &= a \\ && y' &= -\frac{a^2}{a^2} = -1 \\ && y'' &= -\frac{-4a}{a^2} = \frac{4}{a} \\ \Rightarrow && y &\approx a - h+\frac{2}{a}h^2 \\ \\ Q: && y &= 0 \\ && y' &= -\frac{3a^2}{a^2} = -3 \\ && y'' &= -\frac{18a+6a}{a^2} = -\frac{24}{a} \\ \Rightarrow && y &\approx 0-3h-\frac{12}{a}h \\ \\ R: && y &= -a \\ && y' &= -\frac{a^2+3a^2}{a^2-2a^2} = 4 \\ && y'' &= -\frac{-16a+32a+6a}{a^2-2a^2} = \frac{22}{a} \\ \Rightarrow && y &\approx -a+4h + \frac{11}{a}h^2 \end{align*} Alternatively: \begin{align*} && 0 &= xy^{2}+x^{3}+a^{2}y-a^{3} \\ P(0,a): && y &\approx a + c_1h + c_2h^2 \\ && 0 &= h(a+c_1h)^2 + a^2(a + c_1h + c_2h^2)-a^3 \\ &&&= a^3-a^3 + (a^2+a^2c_1)h+(2ac_1+a^2c_2)h^2 \\ \Rightarrow && c_1 &=-1, c_2 =\frac{2}{a} \\ \Rightarrow && y &\approx a - h + \frac{2}{a}h^2 \\ \\ Q(a,0): && y &\approx c_1h + c_2h^2 \\ && 0 &= (a+h)(c_1h)^2+(a+h)^3+a^2(c_1h + c_2h^2 )-a^3 \\ &&&= a^3-a^3+(3a^2+a^2c_1)h + (ac_1^2+3a+a^2c_2)h^2 + \cdots \\ \Rightarrow && c_1 &=-3, c_2 = -\frac{12}{a} \\ \Rightarrow && y &\approx -3h -\frac{12}{a}h \\ \\ R(a,-a): && y &\approx -a + c_1h + c_2h^2 \\ && 0 &= (a+h)(-a + c_1h+c_2h^2)^2+(a+h)^3+a^2(-a + c_1h + c_2h^2)-a^3 \\ &&&= (a^2-2a^2c_1+3a^2+a^2c_1)h+(-2ac_1+c_1^2+\cdots)h^2 \\ \Rightarrow && c_1 &=4, c_2 = \frac{11}{a} \\ \Rightarrow && y &\approx -a + 4h + \frac{11}{a} \end{align*}

If \((x,y)\) lies on the curve, then viewing it as a quadratic in \(y\) we must have \(\Delta = (a^2)^2-4\cdot x \cdot (x^3-a^3) \geq 0 \Rightarrow a^4-4x^4+4xa^3 \geq 0 \Rightarrow 4x^4-4a^3x-a^4 \leq 0\)

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Solution: Suppose \(Py'' + Qy' + Ry = \frac{\d}{\d x}(p y' + qy)\), then \begin{align*} Py'' + Qy' + Ry &= \frac{\d}{\d x}(p y' + qy) \\ &= py'' + p'y' + qy' + q' y \\ &= py'' + (p'+q)y' + q' y \end{align*} Therefore \(P = p, Q = p'+q, R = q'\), Therefore \(q = Q-P'\) and \(R = Q'-P''\) or \(P'' -Q'+R = 0\). \((\Rightarrow)\) Suppose it can be written in that form, then the logic we have applied shows that equation is true. \((\Leftarrow)\) Suppose \(P''-Q'+R = 0\), then letting \(p = P, q = Q-P'\) we find functions of the form which will be expressed correctly. Notice that if \(P = x-x^4, Q = (1-7x^3), R = -9x^2\) then: \begin{align*} P'' - Q' + R &= -12x^2+21x^2-9x^2 \\ &= 0 \end{align*} Therefore we can write our second order ODE as: \begin{align*} && (x^{3}+3x)\mathrm{e}^{x} &= \frac{\d}{\d x} \left ( (x-x^4) y' +(1-7x^3-(1-4x^3))y \right) \\ &&&= \frac{\d}{\d x} \left ( (x-x^4) y' -3x^3y \right) \end{align*} Suppose \(z = (x-x^4)y' - 3x^2y\), then \(z = (2-2^4) \cdot 0 - 3 \cdot 2^2 \cdot 2e^2 = -24e^2\) when \(x = 2\). and we have: \begin{align*} && \frac{\d z}{\d x} &= (x^3+3x^2)e^x \\ \Rightarrow && z &= \int (x^3+3x^2)e^x \d x \\ &&&= x^3 e^x+c \\ \Rightarrow && -48e^2 &= e^2(8) + c \\ \Rightarrow && c &= -56e^2 \\ \Rightarrow && z &= e^x(x^3)-56e^2 \\ \end{align*} So our differential equation is: \begin{align*} && (x-x^4)y'-3x^3 y &= x^3e^x -5 6e^2 \\ \Rightarrow && (1-x^3)y' -3x^2y &= x^2 e^x - \frac{6e^2}{x} \\ \Rightarrow && \frac{\d }{\d x} \left ( (1-x^3)y \right) &= x^2e^x - \frac{56e^2}{x} \\ \Rightarrow && (1-x^3)y &= (x^2-2x+2)e^x - 56e^2 \ln x + k \\ \underbrace{\Rightarrow}_{x=2} && (1-2^3)2e^2 &= (2^2-2\cdot2 + 2)e^2 -56e^2 \ln 2 + k \\ \Rightarrow && k &= -16e^2+56 \ln 2 \cdot e^2 \\ \Rightarrow && y &= \frac{(x^2-2x+2)e^x - 56e^2 \ln x -16e^2+56 \ln 2 \cdot e^2}{(1-x^3)} \end{align*}

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Solution:

1. \begin{align*} v &= \mathrm{sech} u \\ &= \frac{1}{\mathrm{cosh } u} \\ &= \frac{1}{\sqrt{1+\mathrm{sinh}^2 u}} \tag{\(u > 0\)} \\ &= \frac{1}{\sqrt{1+\tan^2 \theta}} \\ &= \frac{1}{\sqrt{\mathrm{sec}^2 \theta}} \\ &= \cos \theta \tag{\(0 < \theta < \tfrac{\pi}{2}\)} \end{align*}
2. \begin{align*} && \tan \theta &= \textrm{sinh} u \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && \sec^2 \theta \cdot \frac{\d \theta}{\d u} &= \cosh u \\ \Rightarrow && \frac{\d \theta}{\d u} &=\cosh u \cdot \cos^2 \theta \\ &&&= \frac{1}{v} \cdot v^2 \\ &&&=v \end{align*}
3. \begin{align*} \sin 2 \theta &= 2 \sin \theta \cos \theta \\ &= 2 \sin \theta \cdot \frac{\d \theta}{\d u} \\ &= -2 \frac{\d v}{\d \theta} \cdot \frac{\d \theta}{\d u} \tag{\(\cos \theta = v\)} \\ &= -2 \frac{\d v}{\d u} \end{align*} \begin{align*} && \sin 2 \theta &= -2 \frac{\d v}{\d u} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && 2 \cos 2 \theta \cdot \frac{\d \theta}{\d u} &= -2 \frac{\d^2 v}{\d u^2} \\ \Rightarrow && \cos 2 \theta &= - \frac{\d^2 v}{\d u^2} \frac{1}{v} \\ &&&= -\frac{\d ^2v}{\d u^2} \cosh u \end{align*}
4. \begin{align*} && \frac{\d u}{\d \theta} &= \frac{1}{v} \\ \Rightarrow && \frac{\d^2 u}{\d \theta^2} &= -\frac{1}{v^2} \frac{\d v}{\d \theta} \\ &&&= \frac{1}{v^2} \sin \theta \\ && \frac{\d v}{\d \theta} &= -\sin \theta \\ \Rightarrow && \frac{\d^2 v}{\d \theta^2} &= -\cos \theta \\ &&&= - v \\ \end{align*} Therefore \begin{align*} \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2} &= \frac{1}{v} \cdot \left (-v\right) + \left ( - \sin \theta \right ) \cdot \left (\frac{1}{v^2} \sin \theta \right) + \frac{1}{v^2} \\ &= -1 + \frac{1-\sin^2 \theta}{v^2} \\ &= -1 + \frac{\cos^2 \theta}{v^2} \\ &= -1 + 1 \\ &= 0 \end{align*}

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Solution:

Clearly we can achieve \(0\), \(1\), and \(2\) intersections by never entering the range, entering too flat, or entering before hitting the second branch. To achieve \(3\) we can go at a flat enough slope that we hit somewhere near the top of the second branch, and since the gradient there will be \(\approx 0\), and our gradient is positive, we must intersect before that point as well, ie \(3\) intersections. Clearly we cannot intersect the second branch \(3\) times or the first branch twice, therefore there are at most \(3\) intersections. To intersect the graph only once, we need to:

• be below \(((4n+1)\tfrac{\pi}{2}, 1)\) and
• not touch the second gradient

The first condition means that \(k (4n+1)\tfrac{\pi}{2} < 1 \Rightarrow k < \frac{2}{(4n+1)\pi}\). For the second condition, consider a point on the curve \(\sin x\) whose tangent line goes through the origin, ie \(\frac{y - \sin t}{x - t} = \cos t \Rightarrow y = (\cos t)x - t \cos t+\sin t\) ie \(\sin t = t \cos t\). For this point \(t\) to be in the required interval, we need \((4n+5) \tfrac{\pi}{2} -t \in (0, \frac{\pi}{2})\), so let's call this value \(\delta\). Then our result is: The gradient needs to be steeper than \(\cos t = \cos \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) = \sin \delta\) and \(\cos \delta =\left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \sin \delta \). If \(n\) is large, then, \begin{align*} && 1 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1 &\approx (4n+5) \tfrac{\pi}{2} \delta \\ \Rightarrow && \delta &\approx \frac{2}{(4n+5)\pi} \end{align*}. To higher order: \begin{align*} && 1-\frac12 \delta^2 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1-\frac12 \delta^2 &\approx (4n+5) \tfrac{\pi}{2} \delta - \delta^2 \\ \Rightarrow && 0 &\approx 1 - (4n + 5)\tfrac{\pi}{2} \delta + \frac12 \delta^2 \\ \Rightarrow && \delta &\approx (4n+5) \tfrac{\pi}{2} - \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2} \\ &&&= \frac{2}{(4n+5) \tfrac{\pi}{2} + \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2}} \end{align*}.

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Solution:

Consider the moment of inertia of the lamina. The MoI about the centre of mass is \(\frac1{12}M((2a)^2 + (2a)^2) = \frac23Ma^2\). //el axis theorem, tells us the moment of inertia about \(O\) is \(I_O = I_G + Md^2_{OG} = \frac23Ma^2 + M2a^2 = \frac83Ma^2\) Moment of inertia of particle is \(6Ma^2\) Total moment of inertial is: \(\frac{26}{3}Ma^2\). Conservation of angular momentum states that \(6M \frac{\sqrt{2}}2Va = \frac{26}{3}Ma^2 \omega \Rightarrow \omega = \frac{9\sqrt{2}V}{26a}\) Consider the centre of mass (in the frame drawn) \begin{array}{c|c|c} \text{Shape} & \text{Mass} & \text{COM} \\ \hline \text{Square} & M & (0,-\sqrt{2}a) \\ \text{Particle} & 6M & (-\frac{\sqrt{2}}2a, -\frac{\sqrt{2}}{2}a) \\ \text{combined} & 7M & \left ( \frac{-3\sqrt{2}}{7} a, -\frac{4\sqrt{2}}{7}a \right) \end{array} The lamina/particle system will complete full circles if it still has positive angular velocity at the peak, ie: \begin{align*} && \underbrace{\frac12 I \omega^2}_{\text{initial rotational energy}} + mgh_{start} &\geq mgh_{top} \\ && \frac 12 \frac{26}{3} Ma^2 \frac{9^2 \cdot 2 V^2}{26^2 a^2} - (7M)g\frac{4\sqrt{2}}{7}a &\geq (7M)g\frac{5\sqrt{2}}{7}a \\ \Rightarrow && \frac{V^2 \cdot 27}{26} &\geq 9\sqrt{2}ga \\ \Rightarrow && V^2 & \geq \frac{26\sqrt{2}}{3}ga \end{align*}

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Solution: Conservation of energy, tells us that \(2m \cdot g \cdot 4a = 8amg\) is equal to \(\frac12 m v_{wedge}^2 + \frac12(2m)v_{particle}^2\). Conservation of momentum (horizontally) tells us that \(m v_{wedge}+2mv_{particle, \rightarrow} = 0 \Rightarrow v_{particle, \rightarrow} = -\frac12 v_{wedge}\).

We know that the particle must remain on the slope, and so \(v_{particle,\downarrow} = \frac{4}{3} \frac{3}{2} v_{wedge} = 2v_{wedge}\). In conclusion, we have: \begin{align*} && 8amg &= \frac12 m v_{wedge}^2 + \frac12 (2m)\left ((-\tfrac12 v_{wedge})^2 + (2v_{wedge})^2 \right ) \\ &&&= \frac{19}{4}mv_{wedge}^2 \\ \Rightarrow && v_{wedge}^2 &= \frac{32}{19}ag \end{align*}. To calculate the time the ball bounces for, note that: \(s = ut + \frac12 at^2 \Rightarrow 0 = 2v_{wedge} - \frac12 gt \Rightarrow t = \frac{4v_{wedge}}{g}\). During this time, the wedge (and ball) who horizontally are moving apart with speed \(\frac32 v_{wedge}\) we have they move apart by: \begin{align*} && DE &= \underbrace{\frac32 v_{wedge}}_{\text{speed they move apart}} \cdot \underbrace{\frac{4v_{wedge}}{g}}_{\text{time they are moving apart for}} \\ &&&= \frac{6}{g} v_{wedge}^2 \\ &&&= \frac{6}{g}\frac{32}{19}ag \\ &&&= \frac{192}{19}a \end{align*}

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Solution:

\begin{align*} %\text{COE}: && \frac12 m u^2 - mga &= \frac12mv^2 + mga\cos \theta \\ \text{N2}(\swarrow): && R+mg\cos\theta &= \frac{m v^2}{a} \\ R = 0: && v^2 &= ag\cos \theta \\ \end{align*} So the particle will become a projectile moving tangent to the circle with \(v^2 = ag \cos \theta\). Therefore the velocity will be \(\displaystyle \sqrt{ag \cos \theta}\binom{-\cos \theta}{\sin \theta}\). We have: \begin{align*} && \mathbf{s} &= a\binom{\sin \theta}{\cos \theta}+\sqrt{ag \cos \theta}\binom{-\cos \theta}{\sin \theta} t + \frac12 \binom{0}{-g} t^2 \\ \Rightarrow && a^2 &= \mathbf{s} \cdot \mathbf{s} \\ &&&= a^2 + ag\cos \theta t^2 + \frac1{4} g^2t^4 -ag \cos \theta t^2 - \sqrt{ag \cos \theta} \sin \theta g t^3 \\ \Rightarrow && 0 &= \frac14 g t - \sqrt{ag \cos \theta} \sin \theta \\ \Rightarrow && t &= \frac{4\sqrt{a g \cos \theta} \sin \theta}{g} \end{align*} At this time, the vertical position will be: \begin{align*} && s_y &= a \cos \theta + \sqrt{ag \cos \theta} \sin \theta \frac{4\sqrt{a g \cos \theta} \sin \theta}{g} - \frac12 g \left ( \frac{4\sqrt{a g \cos \theta} \sin \theta}{g} \right)^2 \\ &&&= a \cos \theta + 4a\cos \theta \sin^2 \theta - 8a\cos \theta \sin^2 \theta \\ &&&= a \cos \theta - 4 a \cos \theta \sin^2 \theta \\ &&&= a \cos \theta (1-4 \sin^2 \theta) \\ \underbrace{\Rightarrow}_{s_y < 0} && 0 &> 1 - 4 \sin^2 \theta \\ \Rightarrow && \sin\theta &> \frac12 \\ \Rightarrow && \theta & > \frac{\pi}{6} \end{align*}

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Solution: \begin{align*} &&\mathbf{M} &= \begin{pmatrix}\lambda \\ 0\\ 1 \end{pmatrix} \times \begin{pmatrix}a\\ a \\ a \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} \times \begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix} + \begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix} \times \begin{pmatrix} a \\ 0 \\ a \end{pmatrix} \\ &&&= \begin{pmatrix} -a \\ -a(\lambda -1) \\ \lambda a \end{pmatrix} + \begin{pmatrix} -2a \\ 0 \\ -a \end{pmatrix} + \begin{pmatrix} \mu a \\ -a(1-\nu) \\ -a \mu \end{pmatrix} \\ &&&=a \begin{pmatrix} \mu - 3 \\ \nu - \lambda \\ \lambda-1-\mu \end{pmatrix} \\ \Rightarrow && \mu &= 3, \lambda = 4, \nu = 4 \end{align*}

1. To find the force we add all vectors: \begin{align*} \mathbf{F} &= \begin{pmatrix}\lambda \\ 0\\ 1 \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} + \begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix} \\ &= \begin{pmatrix}4\\ 0\\ 1 \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} + \begin{pmatrix}1\\ 3 \\ 4 \end{pmatrix} \\ &= \begin{pmatrix} 4 \\ 3 \\ 7 \end{pmatrix} \end{align*} Since the moment about \(O\) is \(0\), we have the moment about \(B\) is: \begin{align*} \mathbf{M} &= \begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix} \times \begin{pmatrix} 4 \\ 3 \\ 7\end{pmatrix} \\ &= \begin{pmatrix} 7a \\ 0 \\ -4a\end{pmatrix} \end{align*}
2. \begin{align*} \mathbf{0} &= \mathbf{r} \times \begin{pmatrix} 4 + 2 \\ 3+3 \\ 7+2 \end{pmatrix} \\ &= \mathbf{r} \times \begin{pmatrix} 6 \\ 6 \\ 9 \end{pmatrix} \\ \end{align*} Therefore \(\mathbf{r} = t\begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}\) (ie a line) \begin{align*} \text{Work done} &= \text{Force} \cdot \text{distance} \end{align*} Since they are parallel, it's just the magnitude of the force, which is \(3\sqrt{2^2+2^2+3^2} = 3\sqrt{17}\)

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Solution: \begin{align*} \mathbb{P}(M = m, N = n) &= \left ( \frac{3}{12} \right)^m \left ( \frac{9}{12} \right)^n \frac{3}{12} \\ &= \frac{3^n}{4^{m+n+1}} \end{align*}

1. \begin{align*} \mathbb{P}(M = m) &= \sum_{n = 1}^{\infty} \mathbb{P}(M=m,N=n) \\ &= \sum_{n = 1}^{\infty} \frac{3^n}{4^{m+n+1}} \\ &= \frac{1}{4^{m+1}} \sum_{n = 1}^{\infty} \left ( \frac34\right)^n \\ &= \frac{1}{4^{m+1}} \frac{3/4}{1/4} \\ &= \frac{3}{4^{m+1}} \\ \\ \mathbb{P}(N = n) &= \sum_{m = 0}^{\infty} \mathbb{P}(M=m,N=n) \\ &= \sum_{m = 0}^{\infty} \frac{3^n}{4^{m+n+1}} \\ &= \frac{3^n}{4^{n+1}} \sum_{m = 0}^{\infty} \left ( \frac14\right)^n \\ &= \frac{3^n}{4^{n+1}} \frac{1}{3/4} \\ &= \frac{3^{n-1}}{4^{n}} \\ \end{align*}
2. \(M+1 \sim Geo(\frac34) \Rightarrow \mathbb{E}(M) = \frac43 -1 = \frac13\) \(N \sim Geo(\frac14) \Rightarrow \mathbb{E}(N) = 4\)

\(M,N\) are independent since \(\mathbb{P}(M = m, N =n ) = \mathbb{P}(M=m)\mathbb{P}(N=n)\) \begin{align*} \mathbb{P}(N > k) &= \sum_{n=k+1}^{\infty} \mathbb{P}(N = n) \\ &= \sum_{n=k+1}^{\infty} \frac{3^{n-1}}{4^{n}} \\ &= \frac{3^k}{4^{k+1}} \sum_{n = 0}^{\infty} \left ( \frac34\right)^n \\ &= \frac{3^k}{4^{k+1}} \frac{1}{1/4} \\ &= \frac{3^k}{4^k} \end{align*} \begin{align*} \mathbb{P}(N > M) &= \sum_{m=0}^{\infty} \mathbb{P}(N > m) \mathbb{P}(M = m) \\ &= \sum_{m=0}^{\infty} \left (\frac34 \right)^m \frac{3}{4^{m+1}}\\ &=\sum_{m=0}^{\infty} \frac{3^{m+1}}{4^{2m+1}}\\ &= \frac{3}{4} \frac{1}{13/16} \\ &= \frac{12}{13} \\ \\ \mathbb{P}(N=M) &= \sum_{m=1}^{\infty} \mathbb{P}(N=m, M=m) \\ &= \sum_{m=1}^{\infty} \frac{3^m}{4^{2m+1}} \\ &= \frac{3}{64} \sum_{m=0}^{\infty} \left ( \frac{3}{16} \right)^m \\ &= \frac{3}{64} \frac{1}{13/16} \\ &= \frac{3}{52}\\ \\ \mathbb{P}(N < M) &= 1 - \frac34 - \frac3{52} \\ &= 1 - \frac{48}{52} - \frac{3}{52} \\ &= 1 - \frac{51}{52} \\ &= \frac{1}{52} \end{align*}