Problems

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Solution: Writing out the possibilities in order of the largest coin used (and then second largest and so-on): \begin{align*} && 10 &= 10 \\ &&&= 5 + 5 \\ &&&= 5 + 2 + 2 + 1 \\ &&&= 5 + 2 + 1 + 1 + 1 \\ &&&= 5 + 1 + 1 + 1 + 1 + 1\\ &&&= 2 + 2 + 2 + 2 + 2 = 5 \cdot 2\\ &&&= 4 \cdot 2 + 2 \cdot 1 \\ &&&= 3 \cdot 2 + 4 \cdot 1\\ &&&= 2 \cdot 2 + 6\cdot 1\\ &&&= 1 \cdot 2 + 8\cdot 1 \\ &&&= 10 \cdot 1 \end{align*} For 20p, we have \begin{align*} && 20 &= 20 \\ &&&= 10 + \text{all 11 ways} \\ &&&= 4\cdot 5 \\ &&&= 3\cdot 5 +\text{3 ways} \\ &&&= 2\cdot5 + \text{6 ways} \\ &&&= 1\cdot 5 + \text{8 ways} \\ &&&= k\cdot 2 + (20-2k)\cdot 1 \quad \text{11 ways} \end{align*} ie 41 ways

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Solution:

1. \begin{align*} && f(x)&=\tan^{-1}x+\tan^{-1}\left(\frac{1-x}{1+x}\right) \\ \Rightarrow && f'(x) &= \frac{1}{1+x^2} + \frac{1}{1+\l \frac{1-x}{1+x} \r^2} \cdot \l \frac{-2}{(1+x)^2}\r \\ &&&= \frac1{1+x^2}- \frac{2}{(1+x)^2+(1-x)^2} \\ &&&= \frac1{1+x^2} - \frac{2}{2+2x^2} \\ &&&= 0 \end{align*} Therefore $f(x) = \begin{cases} c_1 & \text{if } x < -1 \\ c_2 & \text{if } x > -1 \end{cases}$ \(f(0) = \tan^{-1} 0 + \tan^{-1} 1 = \frac{\pi}{4}\) \(\lim_{x \to \infty} f(x) = -\frac{\pi}{2} + \tan^{-1} -1 = -\frac{3\pi}{4}\) therefore $f(x) = \begin{cases} -\frac{3\pi}{4}& \text{if } x < -1 \\ \frac{\pi}{4} & \text{if } x > -1 \end{cases}$
2. \begin{align*} && x &= y \sin y^2 \\ \Rightarrow && \frac{\d x}{\d y} &= \sin y^2 + 2y^2 \cos y^2 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{1}{\sin y^2+2y^2 \cos y^2} \\ &&&=\frac{1}{\frac{x}{y}+2y^2 \sqrt{1-\sin^2y^2}} \\ &&&= \frac{y}{x + 2y^3 \sqrt{1-\frac{x^2}{y^2}}} \\ &&&= \frac{y}{x+2y^2 \sqrt{y^2-x^2}} \end{align*}

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Solution:

1. Notice that \(a_n = 3^{3^{n-1}}\) in particular, \(a_7 = 3^{3^6}\). Using Fermat's little theorem, we can see that \(3^4 \equiv 1 \pmod{5}\) and so we need to figure out \(3^6 \pmod{4}\), which is clearly \(1\). Therefore \(3^{3^6} \equiv 3^{4k+1} \equiv 3 \pmod{5}\). Therefore the units digit is \(3\).
2. Notice that \(3^5 > 100\) and \(3^3 > 10\). Therefore \begin{align*} a_7 &= 3^{3^6} \\ &= (3^3)^{3^5} \\ &> 10^{3^5} \\ &> 10^{100} \end{align*}
3. \begin{align*} \frac{a_7+1}{2a_7} &= \frac12 + \frac1{2a_7} \\ &= 0.5 + 0.\underbrace{000\cdots}_{\text{at least }99\text{ zeros}} \\ &= 0.50 \end{align*} Since \(a_7 > 10^{100}, \, \frac{1}{2a_7} < 10^{-100}\)

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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 \[ T \approx \frac{mg}{2}\]