Problems

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Solution: Case 1: \(x \leq -1\) \begin{align*} && -1-x+x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && -x-2 &= x + 2 \\ \Leftrightarrow && x = -2 \end{align*} Case \(-1 < x \leq 0\): \begin{align*} && x+1+x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && x&= x + 2 \\ \end{align*} No solutions Case \(0 < x \leq 1\): \begin{align*} && x+1-x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && -x&= x + 2 \\ \end{align*} No solutions Case \(1 < x \leq 2\): \begin{align*} && x+1-x+3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && 5x-6&= x + 2 \\ \Leftrightarrow && x = 2 \end{align*} Case \(2 < x\): \begin{align*} && x+1-x+3(x-1)-2(x-2) &= x + 2 \\ \Leftrightarrow && x+2&= x + 2 \\ \end{align*} Therefore the solutions are \(x \in \{-2\} \cup [2, \infty)\)

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Solution: Plugging in \(x = i\) we obtain \((1-i)^4 = (-2i)^2 = -4 \Rightarrow b = -4\). \begin{align*} x^4(1-x)^4 &= x^4(1-4x+6x^2-4x^3+x^4) \\ &= x^8-4x^7+6x^6-4x^5+x^4 \\ &= x^6(x^2+1) - x^6 -4x^7+6x^6-4x^5+x^4 \\ &= x^6(x^2+1) -4x^5(x^2+1)+4x^5 +5x^6-4x^5+x^4 \\ &= (x^6-4x^5)(x^2+1) +5x^4(x^2+1)-5x^4+x^4 \\ &= (x^6-4x^5+5x^4)(x^2+1) -4x^2(x^2+1)+4x^2 \\ &= (x^6-4x^5+5x^4-4x^2)(x^2+1) +4(x^2+1)-4 \\ &= (x^6-4x^5+5x^4-4x^2+4)(x^2+1) -4 \\ \end{align*} So \begin{align*} \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \d x &= \int_0^1 (x^6-4x^5+5x^4-4x^2+4) - \frac{4}{1+x^2} \d x \\ &= \frac17 - \frac46+1-\frac43+4 - \pi \\ &= \frac{22}7 - \pi \end{align*} \begin{align*} \int_0^1 x^4(1-x)^4 \d x &= B(5,5) \\ &= \frac{4!4!}{9!} \\ &= \frac1{630} \end{align*} Therefore since \(0 < \frac{x^4(1-x)^4}{1+x^2} < x^4(1-x)^4\) we must have that \begin{align*} && 0 &< \frac{22}7 - \pi \\ \Rightarrow && \pi & < \frac{22}{7} \\ && \frac{22}{7} - \pi &< \frac1{630} \\ \Rightarrow && \frac{22}{7} - \frac1{630} &< \pi \end{align*} which is what we wanted.

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Solution: Since this is true for all cubic polynomials, it must be true in particular for \(1, x, x^2, x^3\), therefore: \begin{align*} \int_{-1}^{1} 1 {\mathrm d}t &=a_{1}+a_{2} &=2\\ \int_{-1}^{1} x {\mathrm d}t &=a_{1}u_1+a_{2}u_2 &= 0 \\ \int_{-1}^{1} x^2 {\mathrm d}t &=a_{1}u_1^2+a_{2}u_2^2 &= \frac23\\ \int_{-1}^{1} x^3 {\mathrm d}t &=a_{1}u_1^3+a_{2}u_2^3 &= 0\\ \end{align*} \begin{align*} && \begin{cases} a_{1}+a_{2} &=2 \\ a_{1}u_1+a_{2}u_2 &= 0 \\ a_{1}u_1^2+a_{2}u_2^2 &= \frac23\\ a_{1}u_1^3+a_{2}u_2^3 &= 0\\ \end{cases} \\ \Rightarrow && \begin{cases} a_{1}(u_1^2 - \frac13) + a_{2}(u_2^2 - \frac13) &= 0 \\ a_{1}u_1(u_1^2 - \frac13) + a_{2}u_2(u_2^2 - \frac13) &= 0 \end{cases} \\ \Rightarrow && \begin{cases} u_i = \pm \frac1{\sqrt{3}} \\ a_i = 1\end{cases} \end{align*} Therefore we have: \[\int_{-1}^{1}{\mathrm P}(t)\,{\mathrm d}t ={\mathrm P} \l \frac1{\sqrt{3}} \r+{\mathrm P}\l -\frac1{\sqrt{3}} \r \] [Note: this question is actually asking about Gauss-Legendre polynomials, and could be done directly by appealing to standard results]

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Solution: None \begin{multicols}{2}

\columnbreak \begin{align*} N2(\uparrow): && T \cos \theta - T \cos \phi - mg &= 0 \\ N2(\rightarrow): && T \sin \theta + T \sin \phi &= m r \omega^2 \\ \\ && T \cos \theta - T \cos \phi &= mg \tag{\(*\)}\\ && T \sin \theta + T \sin \phi &= m r \omega^2 \tag{{\(**\)}} \end{align*} \end{multicols} Dividing \((*)\) by \((**)\) we obtain: \begin{align*} \frac{g}{r\omega^2} &= \frac{\cos \theta - \cos \phi}{\sin \theta + \sin \phi} \\ &= \frac{2 \sin \left ( \frac{\theta + \phi}2 \right )\sin \left (\frac{\phi - \theta}2 \right )}{2 \sin \left ( \frac{\theta + \phi}2 \right )\cos \left (\frac{\phi - \theta}2 \right )} \\ &= \tan \left ( \frac{\phi - \theta}2 \right ) \end{align*} as required. Squaring and adding \((*)\) and \((**)\) we obtain: \begin{align*} && m^2(g^2 + r^2 \omega^4) &= T^2(2 + \sin \theta \sin \phi - \cos \theta \cos \phi) \\ && &= T^2(2 - 2\cos (\theta + \phi)) \\ && &= T^2(2 - 2(1 - 2 \sin^2 \left ( \frac{\theta + \phi}2 \right ) )) \\ && &= T^2(4 \sin^2 \left ( \frac{\theta + \phi}2 \right )) \\ \Rightarrow && T &= \frac{m\sqrt{g^2 + r^2 \omega^4}}{2 \sin \left ( \frac{\theta + \phi}2 \right )} \\ \Rightarrow && T &= \frac{mg\sqrt{1 + \frac{r^2 \omega^4}{g^2}}}{2 \sin \left ( \frac{\theta + \phi}2 \right )} \end{align*} If \(r \omega^2 \ll g\) then \(\tan \l \frac{\phi - \theta}2 \r\) is very large, so \(\phi - \theta \approx \pi\) and so \(\phi + \theta \approx \pi\). We can then say that \[ T \approx \frac{mg}{2}\]

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Solution: \begin{align*} \text{N2}(\uparrow): && 1 \cdot v\frac{\d v}{\d s} &= -g - kv^2 \\ \Rightarrow && \int \frac{v}{g+kv^2} \d v &= \int -1 \d s \\ \Rightarrow && \frac{1}{2k}\ln(g+kv^2) &= -s + C \\ s =0, v = u: && \frac{1}{2k} \ln(g+ku^2) &= C \\ \Rightarrow && s &= \frac{1}{2k} \ln \frac{g+ku^2}{g+kv^2} \\ \Rightarrow && e^{-2ks} &= \frac{g+kv^2}{g+ku^2} \\ \Rightarrow && v^2 &= u^2e^{-2ks} + \frac{g}{k}(e^{-2ks}-1) \end{align*} The maximum height will be when \(v = 0\), ie \(\displaystyle s = \frac{1}{2k}\ln\left(1 + \frac{k}{g}u^2 \right)\). On the downward path the resistance will be going upwards, ie \begin{align*} \text{N2}(\uparrow): && 1 \cdot v\frac{\d v}{\d s} &= -g + kv^2 \end{align*} but our solution is solving a different differential equation, therefore unless \(k=0\) the equation will be different.

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Solution: \((x+1)^2+T-1 = 0\) so the larger root is \(-1 + \sqrt{1-T}\) \begin{align*} && \mathbb{P}(X > -1/3) &= \mathbb{P}(-1 + \sqrt{1-T} > -1/3) \\ &&&= \mathbb{P}(\sqrt{1-T} > 2/3)\\ &&&= \mathbb{P}(1-T > 4/9)\\ &&&= \mathbb{P}\left (T < \frac59 \right) = \frac59 \end{align*} Similarly, for \(t \in [-1,0]\) \begin{align*} && \mathbb{P}(X \leq t) &= \mathbb{P}(-1 + \sqrt{1-T} \leq t) \\ &&&= \mathbb{P}(\sqrt{1-T} \leq t+1)\\ &&&= \mathbb{P}(1-T \leq (t+1)^2)\\ &&&= \mathbb{P}\left (T \geq 1-(t+1)^2\right) = (t+1)^2 \\ \Rightarrow && f_X(t) &= 2(t+1) \\ \Rightarrow && \E[X] &= \int_{-1}^0 x \cdot f_X(x) \d x \\ &&&= \int_{-1}^0 x2(x+1) \d x \\ &&&= \left [\frac23x^3+x^2 \right]_{-1}^0 \\ &&&= -\frac13 \\ && \E[X^2] &= \int_{-1}^0 x^2 \cdot f_X(x) \d x \\ &&&= \int_{-1}^0 2x^2(x+1) \d x \\ &&&= \left [ \frac12 x^4 + \frac23x^3\right]_{-1}^0 \\ &&&= \frac16 \\ \Rightarrow && \var[X] &= \E[X^2] - \left (\E[X] \right)^2 \\ &&&= \frac16 - \frac19 = \frac1{18} \end{align*} Notice that by the central limit theorem \(\frac{Y}{800} \approx N( -\tfrac13, \frac{1}{18 \cdot 800})\). Also notice that \(\Phi^{-1}(0.08) \approx -1.4 \approx -\sqrt{2}\) Therefore we are looking for roughly \(800 \cdot (-\frac13 -\frac{1}{\sqrt{18 \cdot 800}} \sqrt{2})) = -267-9 = -276\)

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Solution: \begin{array}{c|c|c|c} & A & B & C \\ \hline 1 & \frac{1}{10} \cdot \frac{2}{10} & \frac{1}{10} \cdot \frac{1}{10} & \frac{1}{10} \cdot \frac{7}{10} \\ 2 & \frac{2}{10} \cdot \frac{3}{10} &\frac{2}{10} \cdot \frac{2}{10} &\frac{2}{10} \cdot \frac{5}{10} \\ 3 & \frac{3}{10} \cdot \frac{4}{10} &\frac{3}{10} \cdot \frac{3}{10} &\frac{3}{10} \cdot \frac{3}{10} \\ 4 & \frac{4}{10} \cdot \frac{5}{10} &\frac{4}{10} \cdot \frac{4}{10} &\frac{4}{10} \cdot \frac{1}{10} \\ \hline & \frac{2+6+12+20}{100} & \frac{1 + 4 + 9 + 16}{100} & \frac{7 + 10 + 9 + 4}{100} \end{array} Therefore \(\mathbb{P}(A) = \frac{4}{10}, \mathbb{P}(B) = \frac{3}{10}, \mathbb{P}(C) = \frac{3}{10}\), in particular, \begin{align*} \mathbb{P}(A | \text{not }C) &= \frac{4}{7} \\ \mathbb{P}(B | \text{not }C) &= \frac{3}{7} \\ \end{align*} If he switches the light on, his probability of survival is \(\frac47 \cdot \frac23 + \frac37 \cdot \frac12 = \frac{25}{42}\), if he doesn't his probability is \(\frac12 \cdot \frac47 +\frac37= \frac{5}{7} = \frac{30}{42}\) therefore he shouldn't switch the light on.