Problems

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Solution:

1. If \(f(a) = a\) then the sequence is constant, ie \(a = p+a^2-pa \Rightarrow 0 = (a-p)(a-1)\). Therefore \(a = 1, p\) If there sequence has period \(2\) then there must be a solution to \(f(f(x)) = x\), ie \begin{align*} && x &= p+(f(x)-p)f(x) \\ &&&= p+(p+(x-p)x-p)(p+(x-p)x) \\ &&&= p + (x-p)x(p+(x-p)x) \\ &&&= p+(x^2-px)(x^2-px+p) \\ \Rightarrow && 0 &= x^4-2px^3+(p+p^2)x^2-(p^2+1)x+p \\ &&&= (x-1)(x-p)(x^2-(p-1)x+1) \end{align*} The first two roots (\(x = 1, p\)) are constant sequences, so we need the second quadratic to have a root, ie \((p-1)^2-4 \geq 0 \Rightarrow p \geq 3 , p \leq -1\). We also need this root not to be \(1\) or \(p\), ie \(1-(p-1)+1 = 3-p \neq 0\) and \(p^2-(p-1)p + 1 = 1+p \neq 0\) so \(p \neq -1, 3\). Therefore \(p > 3\) or \(p < -1\).
2. There exists a constant sequence iff there is a solution to \(f(x) = x\), ie \begin{align*} && x &= f(x) \\ &&&= q + (x-p)x \\ \Leftrightarrow && 0 &= x^2-(p+1)x + q \tag{has a solution} \\ \end{align*} But if it doesn't have a solution, clearly the RHS is always larger, and if it does have a solution then there is some point where the inequality doesn't hold. Suppose \(f(x) > x\) for all \(x\) then \(f(f(x)) > f(x) > x\) therefore there is no value where \(f(f(x)) = x\) which is required for any sequence of period 2. No, consider \(p = q = 0\) so \(f(x) = x^2\) then there cannot be a period \(2\) sequence by the first part, but also clearly \(u_n = 1\) is a valid constant sequence.

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Solution:

1. Consider \(f(M) = f(MI) = f(M)f(I)\). Since \(f(M) \neq 0\) we can divide by \(f(M)\) to obtain \(f(I) = 1\)
2. Let \(J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), then \(J^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I\). Therefore \(1 = f(I) = f(J^2) = f(J)f(J) \Rightarrow f(J) = \pm 1 \Rightarrow f(J) = -1\) since \(f(J) \neq 1\). \begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix}J &= \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ &= \begin{pmatrix} b & a \\ d & c \end{pmatrix} \\ J\begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ &= \begin{pmatrix} c & d \\ a & b \end{pmatrix} \\ J\begin{pmatrix} a & b \\ c & d \end{pmatrix}J &=\begin{pmatrix} d & c \\ b & a \end{pmatrix} \end{align*} Therefore \(f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f \left (J\begin{pmatrix} a & b \\ c & d \end{pmatrix} \right) = f(J) f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)\) and \(f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right) = f\left(J\begin{pmatrix} a & b \\ c & d \end{pmatrix}J \right) = f(J)f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)f(J) = f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)\) as required.
3. First consider \(O\) the matrix of \(0\), then \begin{align*} && JO &= O \\ \Rightarrow && f(JO) &= f(O) \\ \Rightarrow && f(J)f(O) &= f(O) \\ \Rightarrow && -f(O) &= f(O) \\ \Rightarrow && f(O) &= 0 \end{align*} Now consider \(K_{k} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\). Suppose \(A = \begin{pmatrix} a & b \\ ka & kb \end{pmatrix}\) then \begin{align*} K_{\frac1k}A &= \begin{pmatrix} 1 & 0 \\ 0 & \frac1k \end{pmatrix} \begin{pmatrix} a & b \\ ka & kb \end{pmatrix} \\ &= \begin{pmatrix} a & b \\ a & b \end{pmatrix} \end{align*} And so \(f(K_{\frac1k}A) = f\left ( \begin{pmatrix} a & b \\ a & b \end{pmatrix} \right) = - f \left ( \begin{pmatrix} a & b \\ a & b \end{pmatrix} \right) = 0\), therefore either \(f(K_{\frac1k}) = 0\) or \(f(A) = 0\), but we know that \(f(I) \neq 0\) therefore \(f(A) = 0\).
4. Let \(P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\), then \(P^2 = \begin{pmatrix} 1 &2 \\ 0 & 1 \end{pmatrix}\), \(P^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\), \(K_k^{-1}PK_k = K_k^{-1}\begin{pmatrix} 1 & k \\ 0 & k \end{pmatrix} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\). If \(A\) has an inverse then \(f(A) \neq 0\) since \(1 = f(I) = f(A)f(A^{-1})\), in particular, \(f(A)f(A^{-1}) = 1\). Using this for \(K_2\) we have: \(f(P)^2 = f(P^2) = f(K_2^{-1}PK_2) = f(P)\) therefore \(f(P) = 0, 1\), but since \(f(P)\) has an inverse, \(f(P) \neq 0\) so \(f(P) = 1\)

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Solution: \(x = \log_bc\) means that \(b^x = c\) Therefore, we can write \(\frac{\log_ac}{\log_ab} = \frac{\log_ab^{x}}{\log_ab} = \frac{x \log_ab}{\log_ab} = x = \log_bc\), giving us the change of base rule. Rearranging the chance of base rule, we get \(\frac{1}{\log_bc} = \frac{\log_ab}{\log_ac}\)

1. Since \(\pi^2 < 10\), we have \begin{align*} \Leftrightarrow && \pi^2 &< 10 \\ \Leftrightarrow && 2\log_{10} \pi &< 1 \\ \Leftrightarrow && 2 &< \frac{1}{\log_{10} \pi} \\ \Leftrightarrow && 2 &< \frac{\log_{10} 2 + \log_{10} 5}{\log_{10} \pi} \\ \Leftrightarrow && 2 &< \frac{\log_{10} 2}{\log_{10} \pi} + \frac{\log_{10} 5}{\log_{10} \pi} \\ \Leftrightarrow && 2 &< \frac1{\log_2\pi}+ \frac1{\log_5\pi} \\ \end{align*}
2. Since \(e^2 < 8 \Rightarrow 2 < 3 \ln 2 \Rightarrow \ln 2 > \frac23\) \begin{align*} && \log_2 \frac{\pi}{e} &= \frac{\ln \frac{\pi}{e}}{\ln 2} \\ &&&= \frac{\ln \frac{\pi}{e}}{\ln 2} \\ &&&< \frac{3(\ln \pi - 1)}{2} \\ \Rightarrow && \frac{1}{5} &< \frac{3 \ln \pi - 1}{2} \\ \Rightarrow && \frac{2}{15} + 1 = \frac{17}{15}&< \ln \pi \end{align*}
3. From the inequalities given we can set up two linear inequalities in \(\ln 2\) and \(\ln 5\) \begin{align*} && 20 &< e^3 \\ \Rightarrow && 2\ln 2 + \ln 5 &< 3 \tag{*}\\ \\ && \frac{3}{10} &< \log_{10} 2 \\ \Rightarrow && \frac{3}{10} &< \frac{\ln 2}{\ln 10} \\ \Rightarrow && 3 \ln 2 + 3 \ln 5 &< 10 \ln 2 \\ \Rightarrow && -7 \ln 2 + 3 \ln 5 &< 0 \tag{**}\\ \\ \\ && \pi^2 & < 10 \\ \Rightarrow && 2 \ln \pi &< \ln 2 + \ln 5 \tag{***}\\ \end{align*} It would be nice to combine \((*)\) and \((**)\) to bound \(\ln 2+ \ln 5\), so we want to use a linear combination such that \begin{align*} && \begin{cases} 2x -7y &= 1 \\ x + 3y &= 1\end{cases} \\ \Rightarrow && \begin{cases} y &= \frac1{13} \\ x &= \frac{10}{13}\end{cases} \\ \\ \Rightarrow && \ln 2 + \ln 5 < \frac{30}{13} + 0 \\ \Rightarrow && \ln \pi < \frac{15}{13} \end{align*}

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Solution: \begin{questionparts} \item \begin{align*} && \dfrac{\d }{\d x} \left( \s(x)^3 + \c(x)^3 \right) &= 3\s(x)^2\s'(x) + 3\c(x)^2 \c'(x) \\ &&&= 3\s(x)^2\c(x)^2 - 3\c(x)^2\s(x)^2 \\ &&&= 0 \\ \\ \Rightarrow && \s(x)^3 + \c(x)^3 &= \text{constant} \\ &&&= \s(0)^3 + \c(0)^3 \\ &&&= 1 \end{align*} \item \begin{align*} \frac{\d }{\d x} \, \Big( \s(x) \c(x) \Big) &= \s'(x) \c(x) + \s(x)\c'(x) \\ &= \c(x)^3 - \s(x)^3 \\ &= \c(x)^3 - (1-\c(x)^3) \\ &= 2\c(x)^3 - 1 \\ \\ \dfrac{\d }{\d x} \left( \dfrac{\s(x)}{\c(x)} \right) &= \frac{\s'(x)\c(x) - \s(x)\c'(x)}{\c(x)^2} \\ &= \frac{\c(x)^3 + \s(x)^3}{\c(x)^2} \\ &= \frac{1}{\c(x)^2} \\ \end{align*} \item \begin{align*} \int \s(x)^2 \d x &= -\int -\s(x)^2 \d x \\ &= -\int \c'(x) \d x \\ &= - \s(x) +C \\ \\ \int \s(x)^5 \, \d x &= \int \s(x)^2 \s(x)^3 \d x \\ &= \int \s(x)^2 (1 - \c(x)^3) \d x \\ &= -\int \c'(x) (1 - \c(x)^3) \d x \\ &= - c(x) + \frac{\c(x)^4}{4} + C \end{align*} \item If \(u = \s(x), \frac{\d u}{\d x} = \c(x)^2\) \begin{align*} \int \frac{1}{(1-u^3)^{\frac{2}{3}}} \, \d u &= \int \frac{1}{(1-\s(x)^3)^{\frac{2}{3}}} \c(x)^2 \d x \\ &= \int 1 \d x \\ &= x + C \\ &= \s^{-1}(u) + C \\ \\ \int \frac{1}{{(1-u^3)^{\frac{4}{3}}}} \d u &= \int \frac1{(1-\s(x)^3)^{\frac43} }\c(x)^2 \d x \\ &= \int \frac1{(\c(x)^3)^{\frac43}} \c(x)^2 \d x \\ &= \int \frac1{\c(x)^2} \d x \\ &= \frac{\s(x)}{\c(x)} + C \\ &= \frac{u}{(1-u^3)^{\frac13}} + C \\ \end{align*} \begin{align*} && \int {(1-u^3)}^{\frac{1}{3}} \, \d u &= \int (1-s(x)^3)^{\frac13} c(x)^2 \d x \\ &&&= \int \c(x)^3 \d x = I\\ &&&= \int \c(x) s'(x) \d x \\ &&&= \left [\c(x) \s(x) \right] + \int \s(x)^2 s(x) \d x \\ &&&= \c(x) \s(x) + \int (1 - \c(x)^3) \d x + C \\ &&&= \c(x) \s(x) + x - I + C \\ \Rightarrow && I &= \frac{x + \c(x) \s(x)}{2} + k \\ \Rightarrow && &= \frac12 \l \s^{-1}(u) + u \sqrt[3](1-u^3)\r + k \end{align*}

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Solution:

1. Let \(n! + 5 = p\). If \(n \geq 5\) then \(5\) divides the LHS and so must also divide the RHS. Since \(n!+5 > 5\) this means the RHS cannot be prime. Therefore consider \(n = 1, 2, 3, 4\). \begin{align*} n = 1: && 1! + 5 = 6 &&\text{ nope} \\ n=2: && 2! + 5 = 7 && \checkmark \\ n=3: && 3! + 5 = 11 && \checkmark \\ n=4: && 4! + 5 = 29 && \checkmark \end{align*} Therefore the solutions are \((2,7), (3,11), (4,29)\).
2. Suppose \(1! \times 3! \times \cdots \times (2n-1)! = m!\). If \(n \geq 7\) then \(m! > (4n)!\) (by the first theorem) in particular \(m > 4n\). Therefore (by the second theorem) the RHS is divisible by some prime which cannot divide the LHS. Therefore consider \(n = 1,2,3,4,5,6\) \begin{align*} n = 1: && 1! = 1 = 1! && \checkmark \\ n = 2: && 1! \times 3! = 6 = 3! && \checkmark \\ n = 3: && 1! \times 3! \times 5! = 6! && \checkmark \\ n = 4: && 1! \times 3! \times 5! \times 7! = 6! \times 7! = 10! && \checkmark \\ n = 5: && 1! \times 3! \times 5! \times 7! \times 9! = 10! 9! > 11! && \text{would need a factor of } 11\text{ so no} \\ n = 6: && 1! \times 3! \times 5! \times 7! \times 9! \times 11! = 10! 11! 9! > 13! && \text{would need a factor of } 13\text{ so no} \\ \end{align*} Therefore all solutions are \((1,1), (2,3), (3,6), (4,10)\)

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Solution:

1. \begin{align*} \frac{\left(\cot \theta + i\right)^{2n+1} -\left(\cot \theta - i\right)^{2n+1}}{2i} &= \frac{1}{\sin^{2n+1} \theta}\frac{\left(\cos \theta + i \sin \theta \right)^{2n+1} -\left(\cos\theta - i\sin \theta\right)^{2n+1}}{2i} \\ &= \frac{1}{\sin^{2n+1} \theta} \frac{e^{i(2n+1) \theta} - e^{-i(2n+1) \theta} }{2i} \\ &=\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} \end{align*} Notice that: \begin{align*} (\cot \theta + i)^{2n+1} - (\cot \theta - i)^{2n+1} &= \sum_{k=0}^{2n+1} \binom{2n+1}{k}(i)^k \cdot \cot^{2n+1-k} \theta - \sum_{k=0}^{2n+1} \binom{2n+1}{k}(-i)^k \cdot \cot^{2n+1-k} \theta \\ &= \sum_{k=0}^{2n+1} \binom{2n+1}{k} \l i^k - (-i)^k \r \cdot \cot^{2n+1-k} \theta \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} \l i^{2l+1} - (-i)^{2l+1} \r \cdot \cot^{2n+1-(2l+1)} \theta \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} 2i \cdot \cot^{2(n-l)} \theta \\ &= 2i\sum_{l=0}^{n} \binom{2n+1}{2l+1} \cot^{2(n-l)} \theta \\ \end{align*} Therefore if \(\theta\) satisfies \(\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} = 0\) then \(z = \cot^2 \theta\) satisfies the equation. But \(\theta = \frac{m \pi}{2n+1}, m = 1, 2, \ldots, n\) are \(n\) distinct all the roots must be \(\cot^2 \frac{m \pi}{2n+1}\).
2. Notice that the sum of the roots will be \(\displaystyle \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} = \frac{(2n+1)\cdot 2n \cdot (2n-1)}{3! \cdot (2n+1)} = \frac{n \cdot (2n-1)}{3}\) and so \[ \sum_{m=1}^n \cot^{2}\left(\frac{m\pi}{2n+1}\right) =\frac{n\left(2n-1\right)}{3}. \]
3. For \(0 < \theta < \frac{1}{2}\pi\), \begin{align*} && 0 < \sin \theta < \theta < \tan \theta \\ \Leftrightarrow && 0 < \cot \theta < \frac{1}{\theta} < \frac{1}{\sin \theta} \\ \Leftrightarrow && 0 < \cot^2 \theta < \frac{1}{\theta^2} < \cosec^2 \theta = 1 + \cot^2 \theta\\ \end{align*} Therefore \begin{align*} && \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} < \sum_{n=1}^N \frac{(2N+1)^2}{n^2 \pi^2} < N + \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} \\ \Rightarrow && \frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \sum_{n=1}^N \frac{1}{n^2 \pi^2} < \frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\ \Rightarrow && \lim_{N \to \infty}\frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} < \lim_{N \to \infty}\frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\ \Rightarrow && \frac{1}{6} \leq \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} \leq \frac16 \\ \Rightarrow && \sum_{n=1}^N \frac{1}{n^2} = \frac{\pi^2}{6} \end{align*}

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Solution:

1. Consider the equilateral triangle \(CYA\), notice that \(YM\) is a vertical line of symmetry, and \(\angle ACM = 30^\circ\) therefore \(\frac{AC/2}{CM} = \cos 30^\circ \Rightarrow |CM| = \frac{b}{2} \cdot \frac{2}{\sqrt{3}} = \frac{b}{\sqrt{3}}\). Similarly \(|CL| = \frac{a}{\sqrt{3}}\)
2. \(\,\) \begin{align*} && |LM|^2 &= |CM|^2 + |CL|^2 - 2 \cdot |CM| \cdot |CL| \cdot \cos \angle MCL \\ &&&= \frac{b^2}{3} + \frac{a^2}{3} - 2 \frac{ab}{3} \cos \left (\angle CMA + \angle CAB + \angle BCL \right) \\ &&&= \frac13 \left (b^2 + a^2 - 2ab \cos \left ( \frac{\pi}{3} + \angle CAB \right) \right) \\ &&&= \frac13 \left (b^2 + a^2 - ab \cos \left ( \angle CAB \right) + \sqrt{3}ab \sin \angle CAB \right) \\ &&&= \frac13 \left (b^2 + a^2 - ab \cos \angle CAB + 2\sqrt{3} \Delta\right) \\ &&&= \frac13 \left (b^2 + a^2 - ab \left (\frac{a^2+b^2-c^2}{2ab} \right) + 2\sqrt{3} \Delta\right) \\ &&&= \frac13 \left ( \frac12(a^2+b^2+c^2) + 2\sqrt{3}\Delta \right) \\ \Rightarrow && 6|LM|^2 &= a^2 + b^2 + c^2 + 4\sqrt{3} \Delta \end{align*} However, nothing in our reasoning here was special about \(LM\), therefore \(LN\) and \(MN\) also equal this value, and we find that the triangle is equilateral. The area of equilateral triangle [LMN] is \(\frac{\sqrt{3}}4 |LM|^2\), ie \begin{align*} &&& \text{areas are equal} \\ \Leftrightarrow && \Delta &= \frac{\sqrt{3}}4 |LM|^2 \\ &&&= \frac{\sqrt{3}}4 \frac{a^2+b^2+c^2+4\sqrt{3}\Delta}{6} \\ &&&= \frac{\sqrt{3}}{24} (a^2+b^2+c^2) + \frac12 \Delta \\ \Leftrightarrow && \Delta &= \frac{\sqrt{3}}{12}(a^2+b^2+c^2)\\ \Leftrightarrow && 4\sqrt{3}\Delta &=a^2+b^2+c^2\\ \end{align*}
3. \(\,\) \begin{align*} && (a-b)^2 &= -2ab(1 - \cos(C - 60^{\circ})) \\ \Leftrightarrow && a^2+b^2 - 2ab &=-2ab + 2ab \cos(C - 60^{\circ}) \\ \Leftrightarrow && a^2+b^2 &= ab \cos C+\sqrt{3}ab\sin C \\ \Leftrightarrow && a^2+b^2 &= ab \frac{a^2+b^2-c^2}{2ab} + 2\sqrt{3} \Delta \\ \Leftrightarrow && a^2+b^2+c^2 &= 4\sqrt{3}\Delta \end{align*} Since the LHS is non-positive, and the RHS is positive, the only way they can be equal is if they are both \(0\), ie \(a=b\) and \(C = 60^{\circ}\) ie \(ABC\) is equilateral.

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Solution: \begin{align*} x_{n+2} &= \frac{ax_{n+1}-1}{x_{n+1}+b} \\ &= \frac{a \frac{ax_n - 1}{x_n+b}-1}{\frac{ax_n - 1}{x_n+b}+b} \\ &= \frac{a(ax_n-1)-(x_n+b)}{ax_n-1+b(x_n+b)} \\ &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \end{align*}

1. If \(x_{n+2} = x_n\) then \begin{align*} && x_n &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \\ \Rightarrow && 0 &=(a+b)x_n^2+(b^2-a^2)x_n+(a+b) \\ &&&= (a+b)(x_n^2+(a-b)x_n + 1) \end{align*} If \(x_{n+1} = x_n\) then \(x_n^2+(a-b)x_n + 1\) and since our sequence has period \(2\) rather than \(1\) it must be the case this is non-zero. Therefore \(a+b =0\).
2. \begin{align*} x_{n+4} &= \frac{(a^2-1)x_{n+2}-(a+b)}{(a+b)x_{n+2}+b^2-1} \\ &= \frac{(a^2-1)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} -(a+b)}{(a+b)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} +b^2-1} \\ &= \frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \end{align*} If \(x_{n+4} = x_n\) then \begin{align*} x_n &=\frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \\ 0 &= (a^2+b^2-2)(a+b)x_n^2 + \l (b^2-1)^2-(a^2-1)^2 \r x_n+(a^2+b^2-2)(a+b) \\ &= (a^2+b^2-2)(a+b)x_n^2+(b^2-a^2)(a^2+b^2-2)x_n + (a^2+b^2-2)(a+b) \\ &= (a^2+b^2-2)(a+b)(x_n^2+(b-a)x_n + 1) \end{align*} Since we do not want \(x_n\) to be periodic with period \(1\) we must have the quadratic in \(x_n\) \(\neq 0\). If \(a+b = 0\) then \(x_n\) is periodic with period \(2\) since \(x_{n+2} = \frac{(a^2-1)x_n}{((-a)^2-1)} = x_n\). Therefore it is necessary that \(a^2+b^2-2 = 0\). If \(a^2+b^2-2= 0\) then \begin{align*} x_{n+4} &= \frac{((a^2-1)^2-(a+b)^2)x_n}{(b^2-1)^2-(a+b)^2} \\ &=\frac{((a^2-1)^2-(a+b)^2)x_n}{((2-a^2)-1)^2-(a+b)^2} \\ &=\frac{((a^2-1)^2-(a+b)^2)x_n}{((1-a^2)^2-(a+b)^2} \\ &= x_n \end{align*} Therefore it is sufficient too. So our conditions are \(a+b \neq 0, \, \, x_n^2+(a-b)x_n + 1 \neq 0\) and \(a^2+b^2-2 = 0\)

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Solution:

1. Let \(f(x) = 1, g(x) = e^x, a = 0, b = t\), so \begin{align*} && \left ( \int_0^t e^x \d x \right)^2 &\leq \left (\int_0^t 1^2 \d x \right) \cdot \left (\int_0^t (e^x)^2 \d x \right) \\ \Rightarrow && (e^t-1)^2 &\leq t \cdot (\frac12e^{2t} - \frac12) \\ \Rightarrow && \frac{e^t-1}{e^t+1} & \leq \frac{t}{2} \end{align*}
2. Let \(f(x) = x, g(x) = e^{-\frac14 x^2}, a = 0, b = 1\) \begin{align*} && \left ( \int_0^1 xe^{-\frac14 x^2} \d x \right)^2 &\leq \left (\int_0^1 x^2 \d x \right) \cdot \left (\int_0^1 (e^{-\frac14x^2})^2 \d x \right) \\ \Rightarrow && \left ( \left [-2e^{-\frac14x^2} \right]_0^1 \right)^2 & \leq \frac{1}{3} \int_0^1 e^{-\frac12 x^2} \d x \\ \Rightarrow && \int_0^1 e^{-\frac12 x^2} \d x & \geq 12(1-e^{-\frac14})^2 \end{align*}
3. Let \(f(x) = 1, g(x) = \sqrt{\sin x}, a = 0, b = \tfrac12 \pi\), then \begin{align*} && \left ( \int_0^{\frac12 \pi} \sqrt{\sin x} \d x \right)^2 &\leq \left (\int_0^{\frac12 \pi} 1^2 \d x \right) \cdot \left (\int_0^{\frac12 \pi}|\sin x| \d x \right) \\ &&&= \frac{\pi}{2} \cdot 1 \\ \Rightarrow && \int_0^{\frac12 \pi} \sqrt{\sin x} \d x & \leq \sqrt{\frac{\pi}{2}} \end{align*} Let \(f(x) =(\sin x)^{\frac14}, g(x) = \cos x, a = 0, b = \tfrac12 \pi\), so \begin{align*} && \left ( \int_0^{\frac12 \pi} (\sin x)^{\frac14} \cos x \d x \right)^2 &\leq \left (\int_0^{\frac12 \pi} \cos^2 x \d x \right) \cdot \left (\int_0^{\frac12 \pi}\sqrt{\sin x} \d x \right) \\ \Rightarrow &&\left ( \left [\frac45 (\sin x)^{\frac54} \right]_0^{\frac12 \pi} \right)^2 & \leq \frac{\pi}{4} \int_0^{\frac12 \pi}\sqrt{\sin x} \d x \\ \Rightarrow && \frac{64}{25\pi} &\leq \int_0^{\frac12 \pi}\sqrt{\sin x} \d x \end{align*}

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Solution:

1. \(\,\) \begin{align*} \bf{COE}: && \frac12 m \cdot 5ag &= mg\cdot 2a + \frac12 m v^2 \\ \Rightarrow && v^2 &= ag \\ && v &= \sqrt{ag} \end{align*} If it just clears the second wall, we must have: \begin{align*} && 0 &= \sqrt{ag} \sin \theta t - \frac12 gt^2 \\ \Rightarrow && t &= \frac{2\sqrt{ag}\sin \theta}{g} \\ && a &= \sqrt{ag} \cos \theta t \\ &&&=\sqrt{ag} \cos \theta \frac{2\sqrt{ag}\sin \theta}{g} \\ &&&= a \sin 2 \theta \\ \Rightarrow && \theta &= 45^{\circ} \end{align*} Imagine firing the particle backwards from the top of the wall at \(45^\circ\) then \begin{align*} && -2a &= \sqrt{ag}\cdot \left ( -\frac1{\sqrt{2}} \right) t - \frac12 g t^2 \\ \Rightarrow && 0 &= gt^2+\sqrt{2ag} t -4a \\ &&&= (\sqrt{g}t -\sqrt{2} \sqrt{a})(\sqrt{g}t +2\sqrt{2} \sqrt{a}) \\ \Rightarrow && t &= \sqrt{\frac{2a}{g}} \\ \Rightarrow && s &= \left ( -\frac1{\sqrt{2}} \right) \sqrt{ag} \sqrt{\frac{2a}{g}} \\ &&&= -a \end{align*} Therefore the \(A\) is \(a\) from the wall.
2. When it passes over the first wall, \begin{align*} \bf{COE}: && \frac52amg &= (2a+h)mg + \frac12 m v^2 \\ \Rightarrow && v^2 &= (a-2h)g \end{align*} Now imagine firing a particle with this speed in any direction. The question is asking whether we can ever travel \(2a\) without descending more than \(h\). \begin{align*} && a &= \sqrt{(a-2h)g} \cos \beta t \\ \Rightarrow && t &= \frac{a}{\sqrt{(a-2h)g} \cos \beta}\\ && -h &= \sqrt{(a-2h)g} \sin \beta t - \frac12 g t^2 \\ &&&= a \tan \beta - \frac12 \frac{a^2}{(a-2h)} \sec^2 \beta \\ &&&= a \tan \beta - \frac{a^2}{2(a-2h)}(1+ \tan^2 \beta )\\ \Rightarrow && 0 &= \frac{a^2}{2(a-2h)} \tan^2 \beta-a \tan \beta + \frac{a^2-2ah+4h^2}{2(a-2h)} \\ && \Delta &= a^2 - \frac{a^2}{a-2h} \frac{a^2-2ah+4h^2}{a-2h} \\ &&&= \frac{a^2}{(a-2h)^2}\left ( a^2-4ah+4h^2-a^2+2ah-4h^2\right) \\ &&&= \frac{a^2}{(a-2h)^2}\left ( -2ah\right) < 0 \\ \end{align*} So there are no solutions if \(h > 0\)

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Solution: \begin{align*} \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) &= \frac{r+1}{r} \l \frac{r!(n-1)!}{(n+r-1)!} - \frac{r!n!}{(n+r)!} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \l 1 - \frac{n}{n+r} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \frac{r}{n+r} \\ &= \frac{(r+1)!n!}{(n+r)!} \\ &= \frac{1}{^{n+r}\C_{r+1}} \end{align*} \begin{align*} \sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} &= \sum_{n=1}^{\infty} \l \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) \r \\ &= \frac{r+1}{r} \sum_{n=1}^{\infty} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \sum_{n=1}^{N} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \l \frac{1}{^{1+r-1}\C_{r}} - \frac{1}{^{N+r}\C_{r}}\r \\ &= \frac{r+1}{r} \frac{1}{^{1+r-1}\C_{r}} \tag{since \(\frac{1}{^{N+r}\C_{r}} \to 0\)} \\ &= \frac{r+1}{r} \end{align*} When \(r = 2\), we have: \begin{align*} && \frac{3}{2} &= \sum_{n=1}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &=\frac{1}{^{1+2}\C_{3}} + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &= 1 + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ \Rightarrow && \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} &= \frac12 \end{align*} \begin{align*} \frac{1}{^{n+1}\C_{3}} &= \frac{3!}{(n+1)n(n-1)} \\ &= \frac{3!}{n^3-n} \\ &> \frac{3!}{n^3} \end{align*} \begin{align*} \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &= \frac{5!}{(n+1)n(n-1)} - \frac{5!}{(n+2)(n+1)n(n-1)(n-2)} \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l 1- \frac{1}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l \frac{n^2-5}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2(n^2-5)}{(n^2-1)(n^2-4)} \\ &< \frac{5!}{n^3} \end{align*} Since \(k(k-5) < (k-1)(k-4) \Leftrightarrow 0 < 4\), this only makes sense if \(n \geq 3\) \begin{align*} &&\frac{3!}{n^3} &< \frac{1}{^{n+1}\C_{3}} \tag{if \(n \geq 3\)} \\ \Rightarrow &&\sum_{n=3}^\infty \frac{3!}{n^3} &< \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{3!}{n^3} &< \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \sum_{n=2}^\infty \frac{1}{^{n+2}\C_{2+1}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \frac{1}{2} = \frac{29}{4} \\ \Rightarrow && \sum_{n=1}^\infty \frac{1}{n^3} &< \frac{29}{24} = \frac{116}{96} \\ \end{align*} \begin{align*} && \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &< \frac{5!}{n^3} \\ \Rightarrow && \sum_{n=3}^\infty \l \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} \r &< \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{20}{^{n+1}\C_3} - \sum_{n=3}^\infty \frac{1}{^{n+2}\C_{5}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=2}^\infty \frac{20}{^{n+2}\C_{2+1}} - \sum_{n=1}^\infty \frac{1}{^{n+4}\C_{4+1}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \frac{20}{2} - \frac{4+1}{4} &< \sum_{n=1}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{115}{96} &< \sum_{n=1}^\infty \frac{1}{n^3} \\ \end{align*}

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Solution:

1. \begin{align*} && a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} &= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} \\ &&&= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \frac{\ln f(x)}{\ln a} \, dx} \\ &&&= e^{ \frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \\ &&&= F(y) \end{align*}
2. \(\,\) \begin{align*} && H(y) &= e^{\frac1y \int_0^y \ln h(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln (f(x)g(x))\d x} \\ &&&= e^{\frac1y \int_0^y \left ( \ln f(x)+\ln g(x) \right)\d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x +\frac1y \int_0^y \ln g(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x }e^{\frac1y \int_0^y \ln g(x) \d x} \\ &&&= F(y)G(y) \end{align*}
3. Suppose \(f(x) = b^x\), then \begin{align*} && F(y) &= b^{\frac1y \int_0^y \log_b f(x) \d x} \\ &&&= b^{\frac1y \int_0^y x \d x}\\ &&&= b^{\frac1y \frac{y^2}{2}} \\ &&&= b^{\frac{y}2} = \sqrt{b^y} \end{align*}
4. Suppose the geometric mean of \(f(x)\) is \(\sqrt{f(y)}\) then the geometric mean of \(f(x)^2\) is \(f(y)\) by the the second part.

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Solution:

1. \(S \cup T\) is the set of odd positive integers. \(S \cap T\) is the empty set.
2. Suppose we have two integers in \(S\), say \(4n+1\) and \(4m+1\), so \((4n+1)(4m+1) = 16nm + 4(n+m) + 1 = 4(4nm + n+m) + 1\) which clearly is in \(S\). The product of any two integers in \(T\) is in \(S\), since \((4n+3)(4m+3) = 16nm + 12(n+m) + 9 = 4(4nm+3(n+m) + 2) + 1\)
3. Suppose \(p \in T\) is not prime, then it must have prime factors. They must all be odd, so they are in \(T\) or \(S\). If they are all in \(S\) then \(p\) must also be in \(S\) (as the product of numbers in \(S\)) but this is a contraction. Therefore \(p\) must have a prime factor in \(T\).
4. • \(3, 7, 11, 19\) are all primes in \(T\). Notice therefore that any product of two of them must be \(S\)-prime. But then \((3 \times 7) \times (11 \times 19)\) and \((3 \times 11) \times (7 \times 19)\) are distinct factorisations into \(S\)-prime factors.

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Solution:

1. \(\frac{(-N)(-N-1)\cdots(-N-n+1)}{n!} = \binom{N+n-1}{n}\), so \[ (1-x)^{-N} = \sum_{n=0}^{\infty} \binom{N+n-1}{n} x^n\] \begin{align*} && (1-x)^{-N-1} &= (1-x)^{-1}(1-x)^{-N} \\ &&&= (1 + x + x^2 + \cdots)\left ( \sum_{n=0}^{\infty} \binom{N+n-1}{n} x^n\right)\\ [x^{n}]: && \binom{N+1+n-1}{n} &= \sum_{j=0}^n \underbrace{1}_{x^{n-j} \text{ from 1st bracket}}\cdot\underbrace{\binom{N+j-1}{j}}_{x^j\text{ from second bracket}} \\ \Rightarrow && \binom{N+n}{n} &= \sum_{j=0}^n \binom{N+j-1}{j} \end{align*}
2. Consider \((1+x)^{m+n} = (1+x)^m(1+x)^n\) and consider the coefficient of \(x^r\) from each side. On the left hand side this is clearly \(\binom{m+n}{r}\) on the right hand side we can take \(x^j\) from \((1+x)^m\) and \(x^{n-j}\) from \((1+x)^n\) and \(j\) can take any value from \(0\) to \(r\), ie \[ \binom{m+n} r = \sum _{j=0}^r \binom m j \binom n {r-j} \]
3. Consider \((1-x)^{-(N+m+1)} = (\)

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Solution: \begin{align*} && \mathbb{P}(A \cup B \cup C) &= \mathbb{P}(A \cup B) + \mathbb{P}(C) - \mathbb{P}((A \cup B) \cap C) \tag{applying with \(A\cup B\) and \(C\)} \\ &&&= \mathbb{P}(A \cup B) + \mathbb{P}(C) - \mathbb{P}((A \cap C) \cup (B \cap C)) \\ &&&= \mathbb{P}(A)+\mathbb{P}(B) - \mathbb{P}(A\cap B) + \mathbb{P}(C) - \mathbb{P}((A \cap C) \cup (B \cap C)) \tag{applying with \(A\) and \(B\)}\\ &&&= \mathbb{P}(A)+\mathbb{P}(B) - \mathbb{P}(A\cap B) + \mathbb{P}(C) - \left ( \mathbb{P}(A \cap C) +\mathbb{P}(B \cap C) - \mathbb{P}( (A \cap C) \cap (B \cap C) )\right) \\ &&&= \mathbb{P}(A)+\mathbb{P}(B) +\mathbb{P}(C)- \mathbb{P}(A\cap B)- \mathbb{P}(A \cap C) -\mathbb{P}(B \cap C)+\mathbb{P}( A \cap B \cap C) \end{align*} \[ \mathbb{P}(A_1 \cup A_2 \cup A_3 \cup A_4) = \sum_i \mathbb{P}(A_i) - \sum_{i \neq j} \mathbb{P}(A_i \cap A_j) + \sum_{i \neq j \neq j} \mathbb{P}(A_i \cap A_j \cap A_k) - \mathbb{P}(A_1 \cap A_2 \cap A_3 \cap A_4) \]

1. \(\mathbb{P}(E_i) = \frac{1}{n}\)
2. \(\mathbb{P}(E_i \cap E_j) = \frac{1}{n} \cdot \frac{1}{n-1} = \frac{1}{n(n-1)}\)
3. \(\mathbb{P})(E_i \cap E_j \cap E_k) = \frac{1}{n(n-1)(n-2)}\)

First notice that the probability that \(k\) (or more) cards are in the correct place is \(\frac{(n-k)!}{n!}\) (place the other \(n-k\) cards in any order. We are interested in: \begin{align*} \mathbb{P} \left ( \bigcup_{i=1}^n E_i \right) &= \sum_{i} \mathbb{P}(E_i) - \sum_{i \neq j} \mathbb{P}(E_i \cap E_j) + \sum_{i \neq j \neq k} \mathbb{P}(E_i \cap E_j \cap E_k) - \cdots \\ &= \sum_i \frac1n - \sum_{i\neq j} \frac{1}{n(n-1)} + \sum_{i \neq j \neq k} \frac{1}{n(n-1)(n-2)} - \cdots + (-1)^{k+1} \sum_{i_1 \neq i_2 \neq \cdots \neq i_k} \frac{(n-k)!}{n!} + \cdots\\ &= 1 - \binom{n}{2} \frac{1}{n(n-1)} + \binom{n}{3} \frac{1}{n(n-1)(n-2)} - \cdots + (-1)^{k+1} \binom{n}{k} \frac{(n-k)}{n!} + \cdots \\ &= 1 - \frac12 + \frac1{3!} - \cdots + (-1)^{k+1} \frac{n!}{k!(n-k)!} \frac{(n-k)!}{n!} + \cdots \\ &= 1 - \frac1{2!} + \frac1{3!} - \cdots + (-1)^{k+1} \frac{1}{k!} + \cdots + (-1)^{n+1} \frac{1}{n!} \end{align*} The probability exactly one card is in the right place is the probability none of the other \(n-1\) are in the right place, which is: \(\frac1n \left (1 - \left (1 - \frac1{2!} + \frac1{3!} - \cdots + (-1)^{k+1} \frac{1}{k!} + \cdots + (-1)^{n} \frac{1}{(n-1)!} \right) \right)\) but there are also \(n\) cards we can choose to be the card in the right place, hence \(\frac{1}{2!} - \frac{1}{3!} + \cdots +(-1)^n \frac{1}{(n-1)!}\)