Problems

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Solution:

1. \(\mathbb{P}(\text{head}) = \mathbb{P}(\text{head}|1)\mathbb{P}(\text{coin 1}) + \mathbb{P}(\text{head}|2)\mathbb{P}(\text{coin 2})+\mathbb{P}(\text{head}|3)\mathbb{P}(\text{coin 3}) = \frac13(p_1+p_2+p_3)\)
2. \(N_1 = X_1 + X_2\) where \(X_i \sim Bernoulli(p)\), therefore \(\mathbb{E}(N_1) = 2p\) and \(\textrm{Var}(N_1) = \textrm{Var}(X_1)+ \textrm{Var}(X_2) = p(1-p)+p(1-p) = 2p(1-p)\)
3. Let \(Y_i\) be the indicator for the \(i\)th coin is heads. Then \(\mathbb{E}(Y_i) = p\) and so \(\mathbb{E}(N_2) = 2p\). \begin{align*} && \textrm{Var}(N_2) &= \mathbb{E}(N_2^2) - [\mathbb{E}(N_2)]^2\\ &&&= 2^2 \cdot \left (\frac13 \left (p_1p_2+p_2p_3+p_3p_1 \right) \right) + 1 \cdot \left (\frac13 \left (p_1 (1-p_2) + (1-p_1)p_2 + p_2(1-p_3) +(1-p_2)p_3 + p_3(1-p_1) + (1-p_3)p_1 \right) \right) - [\mathbb{E}(N_2)]^2 \\ &&&= \frac43\left (p_1p_2+p_2p_3+p_3p_1 \right) + \frac13 \left ( 2(p_1+p_2+p_3) - 2(p_1p_2+p_2p_3+p_3p_1)\right)-[\mathbb{E}(N_2)]^2 \\ &&&= \frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) + \frac23 \left ( p_1+p_2+p_3 \right)-[\mathbb{E}(N_2)]^2\\ &&&= \frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) + \frac23 \left ( p_1+p_2+p_3 \right)-\left[\frac23(p_1+p_2+p_3)\right]^2\\ &&&= \frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) +2p(1-2p)\\ \end{align*}
4. \(\,\) \begin{align*} && \textrm{Var}(N_1) - \textrm{Var}(N_2) &= 2p(1-p) - \left (\frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) +2p(1-2p) \right) \\ &&&= 2p^2-\frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) \\ &&&= \frac23 \left ( \frac13(p_1+p_2+p_3)^2 -\left (p_1p_2+p_2p_3+p_3p_1 \right)\right)\\ &&&= \frac29 \left (p_1^2+p_2^2+p_3^2 -(p_1p_2+p_2p_3+p_3p_1) \right)\\ &&&= \frac19 \left ((p_1-p_2)^2+(p_2-p_3)^2+(p_3-p_1)^2 \right) &\geq 0 \end{align*} with equality iff \(p_1 = p_2 = p_3\)

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Solution:

1. Her probability of passing if she answers \(k \leq 2\) is \(0\), since she can attain at most \(4\) marks. If she attempts \(3\) questions, she needs to get all of them right, hence \(\mathbb{P}(\text{gets all }3\text{ correct}) = \frac{1}{n^3}\). If she attempts \(4\) questions, we can afford to get one wrong \begin{align*} && \mathbb{P}(\text{passes}|\text{attempts }4) &=\mathbb{P}(4/4) +\mathbb{P}(3/4) \\ &&&= \frac{1}{n^4} + 4\cdot\frac{1}{n^3} \cdot \frac{n-1}{n} \\ &&&= \frac{4n-3}{n^4} \end{align*} If she attempts \(5\) questions she can get \(5\) right (10), \(4\) right, \(1\) wrong (7), but \(3\) right will not work (\(6 - 2 = 4< 5\)), hence: \begin{align*} && \mathbb{P}(\text{passes}|\text{attempts }5) &=\mathbb{P}(5/5) +\mathbb{P}(4/5) \\ &&&= \frac{1}{n^5} + 5\cdot\frac{1}{n^4} \cdot \frac{n-1}{n} \\ &&&= \frac{5n-4}{n^5} \end{align*} If \(4n-3 > n \Leftrightarrow n \geq 2\) then \(4\) attempts is better than \(3\). If \(4n^2-3n > 5n-4 \Leftrightarrow 4n^2-8n+4 = 4(n-1)^2 > 0 \Leftrightarrow n \geq\) then \(4\) is better than \(5\), but \(n\) is \(\geq 2\) so, \(4\) is the best option.
2. \(\,\) \begin{align*} && \mathbb{P}(\text{passes}) &= \frac16 \cdot 0 + \frac16 \cdot 0 + \frac16 \cdot 0 + \frac16 \cdot \frac1{n^3} + \frac16 \frac{4n-3}{n^4} + \frac16 \frac{5n-4}{n^5} \\ &&&= \frac{n^2+4n^2-3n+5n-4}{6n^5} \\ &&&= \frac{5n^2+2n-4}{6n^5} \\ && \mathbb{P}(\text{answered }4|\text{passes}) &= \frac{\mathbb{P}(\text{answered }4\text{ and passes})}{ \mathbb{P}(\text{passes})} \\ &&&= \frac{\frac16 \cdot \frac{4n-3}{n^4}}{\frac{5n^2-2n-4}{6n^5} } \\ &&&= \frac{4n^2-3n}{5n^2+2n-4} \end{align*} Notice that the function takes all values for \(n\) between the roots of the denominator (which are either side of \(0\) and below \(3/4\). Therefore after \(3/4\) the function must be increase since otherwise we would have a quadratic equation with more than \(2\) roots.
3. \(\,\) \begin{align*} &&\mathbb{P}(C \text{ passes}) &= \binom{5}{3} \left ( \frac{n}{n+1} \right)^3 \left ( \frac{1}{n+1}\right)^2 \frac{1}{n^3} + \binom{5}{4} \left ( \frac{n}{n+1} \right)^4 \left ( \frac{1}{n+1}\right) \frac{4n-3}{n^4} +\\ &&&\quad \quad + \binom{5}{5} \left ( \frac{n}{n+1} \right)^5 \frac{5n-4}{n^5} \\ &&&= \frac{10}{(n+1)^5} + \frac{5(4n-3)}{(n+1)^5} + \frac{(5n-4)}{(n+1)^5} \\ &&&= \frac{10+20n-15+5n-4}{(n+1)^5}\\ &&&= \frac{25n-9}{(n+1)^5}\\ \end{align*}

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Solution: Let \(f(x) = x^4 + ax^3 + bx^2 + ax + 1\), and suppose \(f(k) = 0\). Since \(f(0) = 1\), \(k \neq 0\), therefore we can talk about \(k^{-1}\). \begin{align*} && f(k^{-1}) &= k^{-4} + ak^{-3} + bk^{-2} + ak^{-1} + 1 \\ &&&= k^{-4}(1 + ak + bk^2 + ak^3 + k^4) \\ &&&= k^{-4}(k^4+ak^3+bk^2+ak+1) \\ &&&= k^{-4}f(k) = 0 \end{align*} Therefore \(k^{-1}\) is also a root of \(f\)

1. If \(f\) has only on distinct root, we must have \(f(x) = (x+k)^4\) therefore \(k = k^{-1} \Rightarrow k^2 = 1 \Rightarrow k = \pm1\), or \(a = 4, b = 6\) or \(a = -4, b = 6\)
2. If \(f\) has exactly three distinct roots then one of the roots must be a repeated \(1\) or \(-1\), ie \(0 = f(1) = 1 + a + b + a + 1 = 2 + b +2a \Rightarrow b = -2a-2\) or \(0 = f(-1) = 1 -a + b -a + 1 \Rightarrow b = 2a - 2\)
3. If \(b = 2a-2\), we have \begin{align*} && f(x) &= 1 + ax + (2a-2)x^2 + ax^3 + x^4 \\ &&&= (x^2+2x+1)(1+(a-2)x+x^2) \\ \Rightarrow && x &= \frac{2-a \pm \sqrt{(a-2)^2 - 4}}{2} \\ &&&= \frac{2-a \pm \sqrt{a^2-4a}}{2} \end{align*}

\(f\) has exactly three distinct real roots iff \(b = \pm 2a - 2\) and \(b \neq 6\)

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Solution:

Consider the function \(g(t) = f(tx_1 + (1-t)x_2) - tf(x_1) - (1-t)f(x_2)\), notice that \(g(0) = g(1) = 0\). Since \(g''(x) < 0\) over the whole interval, we must have two things: 1. \(g'(x)\) is increasing. 2. It \(g'(x) = 0\) can have at most one solution. Therefore \(g'(x)\) is initially \(0\), we have exactly one turning point. Therefore the function is initially decreasing and then increasing, therefore it is always negative and our inequality holds.

1. \(\,\) \begin{align*} && f \left ( \frac{u+v+w}{3} \right) &= f \left ( \frac{2}{3}\cdot \frac{u+v}2+\frac{1}{3}w \right) \\ &&&\geq \frac23 f \left ( \frac{u+v}{2} \right) + \frac13 f(w) \\ &&&\geq \frac23 \left (\frac12 f(u) + \frac12 f(v) \right) + \frac13 f(w) \\ &&&= \frac{f(u)+f(v)+f(w)}{3} \end{align*}
2. Notice that if \(A, B, C\) are angles in a triangle then they add to \(\pi\) \(0 < A,B,C < \pi\). We also have \(f(x) = \sin x \Rightarrow f''(x) = - \sin x < 0\) on this interval. Therefore \(\sin A + \sin B + \sin C \leq 3 \sin \frac{A+B+C}{3} = 3 \sin \frac{\pi}{3} = \frac{3\sqrt{3}}2\)
3. Also notice that \begin{align*} && f(x) &= \ln ( \sin x) \\ \Rightarrow && f'(x) &= \frac{\cos x}{ \sin x} \\ && f''(x) &= -\textrm{cosec}^2 x < 0 \\ \\ \Rightarrow && \ln( \sin A) + \ln (\sin B) + \ln (\sin C) &\leq 3 \ln \left (\sin \left ( \frac{A + B+ C}{3} \right) \right) \\ &&&= 3 \ln \left ( \frac{\sqrt{3}}{2} \right) = \ln \frac{3\sqrt{3}}{8} \\ \Rightarrow && \sin A \sin B \sin C &\leq \frac{3\sqrt{3}}8 \end{align*}

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Solution:

1. \(\,\) \begin{align*} && f(x) &= \frac1{1+\tan x} \\ && f'(x) &=-(1+\tan x)^{-2} \cdot \sec^2 x \\ &&&= - (\cos x+ \sin x)^{-2} \\ &&&= - (1 + 2 \sin x \cos x)^{-1} \\ &&&= - \frac{1}{1+\sin 2x} \end{align*} \(\sin 2x \in [0, 1]\) so \(1+\sin 2x \in [1,2]\) and \(f'(x) \in [-1, -\tfrac12]\)
   

   + - ↺
2. \(\displaystyle \int_{-1}^1 g(x) \d x = 2g(0) \)
3. Let \(g(x) = \frac{1}{1 + \tan^k x}\) then \(g(x)\) has rotational symmetry of order \(2\) about the point \((\frac{\pi}{4}, \frac12)\) which we can see since \begin{align*} g(x) + g(\tfrac12\pi - x) &= \frac{1}{1 + \tan^k x} + \frac{1}{1 + \tan^k(\tfrac12\pi - x)} \\ &= \frac{1}{1+\tan^k x} + \frac{1}{1+\cot^k x} \\ &= \frac{1}{1+\tan^k x} + \frac{\tan^k x}{\tan^k x + 1} \\ &= 1 = 2 \cdot \tfrac12 \end{align*} Therefore \[ \int_{\frac16\pi}^{\frac13\pi} \frac 1 {1+\tan^kx} \, \d x = \frac{\pi}{6} \cdot \frac12 = \frac{\pi}{12}\]

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Solution:

1. \(\,\) \begin{align*} && 0 &= \cos x + 3 \cos 2x + 3 \cos 3x + \cos 4 x \\ &&&= \cos x + \cos 4x + 3 \left (\cos 2x + \cos 3 x \right) \\ &&&= 2 \cos \frac{5x}{2} \cos \frac{3x}{2} + 6 \cos \frac{x}{2}\cos\frac{5x}{2} \\ &&&= 2 \cos \frac{5x}{2} \left (\cos \frac{3x}{2} + 3 \cos \frac{x}{2} \right)\\ &&&= 2 \cos \frac{5x}{2} \left ( \cos \frac{2x}{2}\cos \frac{x}{2} - \sin \frac{2x}{2} \sin \frac{x}{2}+3 \cos \frac{x}{2} \right) \\ &&&= 2 \cos \frac{5x}{2} \left ( \left (2\cos^2 \frac{x}{2} - 1 \right)\cos \frac{x}{2} - 2\sin \frac{x}{2} \cos \frac{x}{2} \sin \frac{x}{2}+3 \cos \frac{x}{2} \right) \\ &&&= 2 \cos \frac{5x}{2} \left ( 4\cos^3 \frac{x}{2} \right) \\ &&&= 8 \cos \frac{5x}{2} \cos^3 \frac{x}{2} \\ \Rightarrow && \frac{x}{2} &= \frac{\pi}{2}, \frac{3\pi}{2}, \cdots \\ && \frac{5x}{2} &= \frac{\pi}{2}, \frac{3\pi}{2}, \cdots \\ \Rightarrow && x &= \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5} \end{align*}
2. \(\,\) \begin{align*} && 1 &= \cos (x + y) + \cos(x-y) - \cos 2x \\ &&&= 2 \cos x \cos y - 2\cos^2 x + 1 \\ \Rightarrow && 0 &= \cos x (\cos y - \cos x) \\ \Rightarrow && 0 &=\cos x \left ( \cos y + \cos (\pi - x) \right) \\ &&&= 2\cos x \cos \frac{y+x-\pi}{2} \cos \frac{y-x+\pi}{2} \\ \Rightarrow && x &= \frac{\pi}{2} \\ && y+x - \pi&= \pi ,3\pi, \cdots \\ && y-x + \pi&=\pi, 3 \pi, \cdots \\ \Rightarrow && x &= \frac{\pi}{2} \\ && y+x &= 2\pi \Rightarrow x = y = \pi \\ && y&= x \end{align*} So the only solutions are \(x =y\) and \(x = \frac{\pi}{2}\)
3. \(\,\) \begin{align*} && \frac32 &= \cos x + \cos y - \cos (x+y) \\ &&&= 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} - 2 \cos^2 \frac{x+y}{2} + 1 \\ \Rightarrow && \frac14 &= \cos \frac{x+y}{2} \left ( \cos \frac{x-y}{2} - \cos \frac{x+y}{2} \right) \\ \Rightarrow && 0 &= \cos^2 \frac{x+y}{2} - \cos \frac{x-y}{2}\cos \frac{x+y}{2} + \frac14 \\ \Rightarrow && \cos \frac{x+y}{2} &= \frac{\cos \frac{x-y}{2} \pm \sqrt{\cos^2 \frac{x-y}{2}-1}}{2} \\ \Rightarrow && \cos \frac{x-y}{2} &= \pm 1\\ && \cos \frac{x+y}{2} &= \pm \frac12 \\ \Rightarrow && x-y &= -4\pi, 0, 4\pi, \cdots \\ \Rightarrow && x &= y \\ \Rightarrow && \cos x &= \frac12 \\ \Rightarrow && x &= \frac{\pi}{3} \\ \Rightarrow && (x, y) &= \left ( \frac{\pi}{3}, \frac{\pi}{3}\right) \end{align*}

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Solution:

1. \begin{align*} && \frac1{1+x} &= 1 - x+ x^2 - x^3+ \cdots \\ \Rightarrow && \int_0^x \frac{1}{1+t} \d t &= \int_0^x \sum_{n=0}^{\infty} (-t)^n \d t \\ &&&= \left [\sum_{n=0}^{\infty} -\frac{(-t)^{n+1}}{n+1} \right]_0^x \\ \Rightarrow &&\ln(1+x)&=- \sum_{n=1}^\infty \frac{(-x)^n}{n} \end{align*}
2. \begin{align*} && e^{-ax} &= \sum_{n=0}^\infty \frac{(-a)^n}{n!} x^n \\ \Rightarrow && \int_0^{\infty} \frac{1}{x} \left (1-e^{-ax} \right)e^{-x} \d x &= \int_0^{\infty} \frac{1}{x} \left (-\sum_{n=1}^\infty \frac{(-a)^n}{n!}x^n \right)e^{-x} \d x \\ &&&= -\int_0^{\infty} \sum_{n=1}^\infty \frac{(-a)^n}{n!} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} \int_0^{\infty} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} (n-1)! \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n} \\ &&&= \ln (1+a) \end{align*}
3. \begin{align*} && \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x &= \int_0^1 \frac{x^p(1 - x^{q-p})}{\ln x} \, \d x \\ e^{-u} = x, \d x = -e^{-u} \d u: &&&=\int_{u=\infty}^{0} \frac{e^{-pu}-e^{-qu}}{-u} (-e^{-u})\d u \\ &&&= \int_0^\infty \frac{e^{-u}(e^{-qu}-e^{-pu})}{u} \d u \\ &&&= \int_0^\infty \frac{e^{-(1+q)u}(1-e^{-(p-q)u})}{u} \d u \\ v = (1+q)u, \d v = (1+q) \d u: &&&=\int_0^{\infty} \frac{e^{-v}(1-e^{-\left(\frac{p-q}{1+q}\right)v}}{v}\d v \\ &&&= \ln \left(1 + \frac{p-q}{1+q} \right) \\ &&&= \ln \left ( \frac{1+p}{1+q} \right) \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 \((1,1), (2,3), (3,6), (4,10)\)