Problems

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Solution:

1. \(y = e^x(2x^2-5x+2) = e^x(2x-1)(x-2)\), we also have \(y' = e^x(2x^2-5x+2 + 4x-5) = e^x(2x^2-x-3) = e^x(2x-3)(x+1)\) \(y(-1) = \frac{9}{e}\), \(y(\frac32) = -e^{3/2}\)
   

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   If \(k < -e^{3/2}\) there are no solutions. If \(k = -e^{3/2}\) there is a unique solution. If \(-e^{3/2} < k \leq 0\) there are two solutions. If \(0 < k < \frac{9}{e}\) there are three solutions. Otherwise there is a unique solution.
2. 

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Solution:

1. \begin{align*} \cos 15^{\circ} &= \cos (45^{\circ} - 30^{\circ}) \\ &= \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} \\ &= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2} \\ &= \frac{\sqrt{3}+1}{2\sqrt{2}} \\ \\ \sin15^{\circ} &= \sin(45^{\circ} - 30^{\circ}) \\ &= \sin45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \\ &= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\frac{1}{2} \\ &= \frac{\sqrt{3}-1}{2\sqrt{2}} \\ \end{align*}
2. \begin{align*} \cos 3 \alpha &= \cos 2\alpha \cos \alpha - \sin2\alpha \sin \alpha \\ &= (2\cos^2 \alpha -1)\cos \alpha - 2 \cos \alpha \sin^2 \alpha \\ &= 2\cos^3 \alpha - \cos \alpha - 2\cos \alpha (1-\cos^2 \alpha) \\ &= 4\cos^2 \alpha - 3\cos \alpha \end{align*} Therefore if \(x = \cos \alpha\) then \(4x^3 - 3x-\cos3\alpha = 0\). \begin{align*} 0 &= 4x^3 - 3x-\cos3\alpha \\ &= 4x^3 - 3x - 4\cos^3\alpha+ 3\cos \alpha \\ &= 4(x-\cos\alpha)(x^2+x\cos\alpha+\cos^2\alpha)-3(x-\cos\alpha)\\ &= (x - \cos \alpha)(4x^2+4x\cos\alpha+4\cos^2\alpha-3) \end{align*} Therefore the other roots will be solutions to the second quadratic which are: \begin{align*} \frac{-4\cos \alpha \pm \sqrt{16\cos^2\alpha - 16(4\cos^2\alpha-3)}}{8} &= \frac{-\cos \alpha \pm \sqrt{3(1-\cos^2\alpha)}}{2} \\ &= \frac{-\cos \alpha \pm \sqrt{3} \sin \alpha}{2} \end{align*}
3. Suppose \(y^3-3y-\sqrt{2} = 0\) then \(4\l \frac{y}{2} \r ^3-3(\frac{y}{2}) -\frac{\sqrt{2}}{2} = 0\) or alternatively, if \(x = \frac{y}{2}\), \(4x^3-3x-\cos 45^{\circ} = 0\). Therefore \(x = \cos 15^{\circ}, \frac{-\cos 15^{\circ} \pm \sqrt{3} \sin 15^{\circ}}{2}\) Therefore \(y =2\cos 15^{\circ}, -\cos 15^{\circ} \pm \sqrt{3} \sin 15^{\circ}\) or \(y = \frac{\sqrt{6}+\sqrt{2}}{2}\), \begin{align*} y &= -\frac{\sqrt{3}+1}{2\sqrt{2}} \pm \frac{3-\sqrt{3}}{2\sqrt{2}} \\ &= \frac{-4}{2\sqrt{2}}, \frac{-2\sqrt{3}}{2\sqrt{2}} \\ &= -\sqrt{2}, -\frac{\sqrt{6}-\sqrt{2}}{2} \end{align*}

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Solution:

The orange guard will always see \(2b+b-a = 3b-a\) The blue guard will see \(b + \frac{b(b-a)}{a} = \frac{b^2}{a}\) if \(b < 3a\) and \(3b + \frac{b(b-3a)}{(b-a)} = \frac{2b(2b-3a)}{b-a}\). Therefore the blue guard always sees more if \(b > 3a\). He sees more in the other case if \begin{align*} && \frac{b^2}{a} &> 3b - a \\ \Leftrightarrow && \frac{b^2}{a^2} &> 3\frac{b}{a} - 1 \\ \Leftrightarrow && x^2 - 3x + 1 &> 0\\ \Leftrightarrow && x > \frac{3 + \sqrt{5}}{2} \text{ or } x < \frac{3-\sqrt{5}}{2} \end{align*} Since \(b > a\) we must have \(b > \frac{3+\sqrt{5}}2 a\) There is an alternative interpretation which is that the orange guard is in the top left corner, ie In this case the green guard will always see \(2b + \frac{2b(b-a)}{b+a} = \frac{4b^2}{b+a}\) Comparing \(\frac{4b^2}{b+a}\) with \(\frac{b^2}{a}\) we can see the former is larger if \(3a > b\). Comparing \(\frac{4b^2}{b+a}\) with $$

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Solution: At the point \((2t, t^2)\) the gradient is \(t\). Suppose \(\tan \theta = t\), then the point \(b\) away in each direction is \(\binom{2t}{t^2} \pm b \binom{\cos \theta}{\sin \theta}\), ie one end can be written in the form \((x,y) = (2\tan \theta - b \cos \theta, \tan^2 \theta - b \sin \theta)\). Notice we must have \(2\tan \alpha- b \cos \alpha= 0 \Rightarrow b = 2 \frac{\sin \alpha}{\cos ^2 \alpha}\), therefore the coordinates are \((2 \tan \alpha - 2 \tan \alpha, \tan^2 \alpha - 2\tan^2 \alpha) = (0, -\tan^2 \alpha)\) and \((4 \tan \alpha, 3\tan^2 \alpha)\)

The area we can find by calculating the integrate of \(\tan^2 \alpha + \frac14x^2\) between \(0\) and \(2 \tan \alpha\) and then subtracting the triangle, ie \begin{align*} &&A &= 2\tan^3 \alpha + \frac1{12} (2 \tan \alpha)^3 - \frac12 \cdot 2 \tan \alpha \cdot (2 \tan^2 \alpha) \\ &&&= \left (2 + \frac23 -2\right) \tan^3 \alpha \\ &&&= \frac23 \tan^3 \alpha \end{align*}

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Solution:

1. \(\,\) \begin{align*} && f(x) &= \int_1^3 (t-1)^{x-1} \d t \\ &&&= \left [ \frac1x(t-1)^{x} \right]_1^3 \\ &&&= \frac{2^x}{x} \end{align*}
   

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2. \(\,\) \begin{align*} && g(x) &= \int_{-1}^1 \frac{1}{\sqrt{1-2xt+x^2}} \d t \\ &&&= \left [ -\frac{1}{x}(1 +x^2 - 2xt)^{\frac12} \right]_{-1}^1 \\ &&&= \frac1x \left ( \sqrt{1+x^2+2x}-\sqrt{1+x^2-2x}\right) \\ &&&= \frac1x \left ( |1+x|-|1-x| \right) \end{align*}
   

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Solution: Let \(O\) be the origin, and let \(\mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}'\) be the four points. The conditions give us \begin{align*} && \mathbf{a} \cdot \mathbf{b} &= 0 \\ && |\mathbf{a}| &= |\mathbf{b}| \\ && \mathbf{a}' \cdot \mathbf{b}' &= 0 \\ && |\mathbf{a}'| &= |\mathbf{b}'| \\ \end{align*} So \begin{align*} \text{midpoint }AB \text{ to midpoint } BA' &= (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}'))\cdot (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}')) \\ &= \frac12(\mathbf{a}-\mathbf{a}')\cdot \frac12(\mathbf{a} - \mathbf{a}') \\ \text{midpoint }BA' \text{ to midpoint } A'B' &= (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}')) \cdot (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}'))\\ &= \frac12(\mathbf{b}-\mathbf{b}')\cdot \frac12(\mathbf{b} - \mathbf{b}') \\ &= \frac14 (|\mathbf{b}|^2 + |\mathbf{b}'|^2 - 2\mathbf{b}\cdot\mathbf{b}')\\ &= \frac14(|\mathbf{a}|^2 + |\mathbf{a}'|^2 - 2\mathbf{b}\cdot\mathbf{b}') \\ \text{midpoint }A'B' \text{ to midpoint } B'A &= (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a})) \cdot (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a}))\\ &= \frac12(\mathbf{a}'-\mathbf{a})\cdot \frac12(\mathbf{a}' - \mathbf{a}) \\ \text{midpoint }B'A \text{ to midpoint } AB &= (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b})) \cdot (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b}))\\ &= \frac12(\mathbf{b}'-\mathbf{b})\cdot \frac12(\mathbf{b}' - \mathbf{b}) \\ \end{align*} So it's sufficient to prove \(\mathbf{a}\cdot \mathbf{a}' = \mathbf{b}\cdot \mathbf{b}'\) but this is clear from looking at a diagram for 1 second. Given the length of the square is what it is, we want to minimise \(\mathbf{b}\cdot \mathbf{b}'\) which is when they are vertically opposite each other, ie \(\angle BOA' = 90^\circ\)

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Solution: \(f'(x) = 6ax-18x^2\), therefore \(f\) has turning points at \(0\) and \(\frac{a}3\) (ie decreasing for \(x \leq 0\) and \(x \geq \frac{a}{3}\) and increasing otherwise). Therefore possible maxima are \(f(-\tfrac13), f(\frac{a}{3}), f(1)\) where we consider \(\frac{a}{3}\) if \(a \leq 3\) and \(1\) otherwise. \(f(-\frac13) = \frac{a}{3} + \frac{2}{9} = \frac{3a+2}{9}\) \(f(\frac{a}{3}) = \frac{a^3}{3} - \frac{2a^3}{9} = \frac{a^3}{9}\) \(f(1) = 3(a-2)\) Comparing \(\frac{a^3}{9}\) to \(\frac{3a+2}{9}\) we have a double root at \(a = -1\) and a single root at \(a = 2\), therefore \(\frac{a^3}9\) is larger if \(a \geq 2\) Comparing \(3(a-2)\) to \(\frac{3a+2}9\) we have a cross-over at \(a = \frac{7}3\). Therefore we have: \begin{align*} M(a) &= \begin{cases} \frac{3a+2}{9} & 0 \leq a \leq 2 \\ \frac{a^3}{9} & 2 \leq a \leq 3 \\ 3(a-2) & 3 \leq a \end{cases} \end{align*}

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Solution:

1. \(\,\) \begin{align*} && S & = 1 +\quad 2\quad \;\;+ \quad 3 \quad+ \cdots + \quad n \\ && S &= n + (n-1) + (n-2) + \cdots + 1 \\ && 2S &= (n+1) + (n+1) + \cdots + (n+1) \\ \Rightarrow && S &= \frac12n(n+1) \end{align*}
2. \(\,\) \begin{align*} && (N-m)^{k} + m^k&= \sum_{i=0}^k \binom{k}{i} N^{k-i} (-m)^{i} + m^k \\ &&&= N\sum_{i=0}^{k-1} \binom{k}{i}N^{k-i-1}(-m)^i -m^k+m^k \\ &&&= N\sum_{i=0}^{k-1} \binom{k}{i}N^{k-i-1}(-m)^i \end{align*} which is clearly divisible by \(N\).

\begin{align*} 2S &= 2\sum_{i=1}^n i^k \\ &= \sum_{i=0}^n (\underbrace{(n-i)^k + i^k}_{\text{divisible by }n}) \\ \end{align*} Therefore \(2S\) is divisible by \(n\) and so if \(n\) is odd, \(n\) divides \(S\) and if \(n\) is even, \(\frac{n}{2}\) divides \(S\). Also notice \begin{align*} 2S &= 2\sum_{i=1}^n i^k \\ &= \sum_{i=1}^{n} (\underbrace{(n+1-i)^k + i^k}_{\text{divisible by }n+1}) \\ \end{align*} Therefore if \(n+1\) is odd, \(n+1 \mid S\) otherwise \(\frac{n+1}{2} \mid S\), and in either case \(\frac{n(n+1)}{2} \mid S\) (since they are both coprime) but this is the same as \(1 + 2 + \cdots + n \mid S\)

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Solution: The bullet fired at time \(t\) will hit the ground at time \(t+\frac{2u \sin (\frac13\pi - \lambda t)}{g}\). To find the last time a bullet hits the ground, we can differentiate, noting that \begin{align*} && T(t) &= t + \frac{2u \sin \alpha}{g} \\ \Rightarrow && T'(t) &= 1 - \frac{2u\lambda}{g} \cos \alpha \\ && T''(t) &= \frac{2u \lambda^2}{g} \sin \alpha > 0 \end{align*} If \(k = \frac{g}{2\lambda u} \in [\frac12, \frac12\sqrt{3}]\) then notice that this turning point is always achieved, and will be a maximum. It will be when \(\cos \alpha = k, \sin \alpha = \sqrt{1-k^2}\). The distance will be \(u \cos \alpha \cdot \frac{2 u \sin \alpha}{g} = \frac{2ku^2\sqrt{1-k^2}}{g}\). If \(k < \frac12\) then the last bullet to hit the ground will be the last bullet fired, ie \(\frac{2u^2 \sin \frac16\pi \cos \frac16\pi}{g} = \frac{u^2 \sin \frac13 \pi}{g} = \frac{\sqrt{3}u^2}{2g}\)

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Solution:

1. \(\mathbb{P}(r \text{ need surgery}|n \text{ casualties}) = \binom{n}{r} \left ( \frac14\right)^r \left ( \frac34\right)^{n-r}\)
2. \(\,\) \begin{align*} && \mathbb{P}(r \text{ need surgery}) &= \sum_{n=r}^{\infty} \mathbb{P}(r \text{ need surgery} |n \text{ casualties}) \mathbb{P}(n \text{ casualties}) \\ &&&= \sum_{n=r}^{\infty} \binom{n}{r}\left ( \frac14\right)^r \left ( \frac34\right)^{n-r} \frac{e^{-8} 8^n}{n!} \\ &&&= \sum_{n=r}^{\infty} \frac{n!}{(n-r)!r!}\left ( \frac14\right)^r \left ( \frac34\right)^{n-r} \frac{e^{-8} 8^n}{n!} \\ &&&= \frac{e^{-8}8^r}{r!}\left ( \frac14\right)^r \sum_{n=r}^{\infty} \frac{8^{n-r}}{(n-r)} \left ( \frac34\right)^{n-r} \\ &&&= \frac{e^{-8}8^r}{r!}\left ( \frac14\right)^r \sum_{n=r}^{\infty} \frac{6^{n-r}}{(n-r)} \\ &&&= \frac{e^{-8}2^r}{r!} e^6 \\ &&&= \frac{e^{-2}2^r}{r!} \end{align*} Therefore the number requiring surgery is \(Po(2)\) with mean \(2\).
3. \(\,\) \begin{align*} && \mathbb{P}(X_1 = 8| X_1 + X_2 =12) &= \frac{\mathbb{P}(X_1 = 8,X_2 =4)} {\mathbb{P}(X_1+X_2 = 12)}\\ &&&= \frac{\frac{e^{-2}2^8}{8!} \cdot \frac{e^{-2}2^4}{4!}}{\frac{e^{-4}4^{12}}{12!}} \\ &&&= \frac{12!}{8!4!} \frac{1}{2^{12}} \\ &&&= \binom{12}4 \left ( \frac12 \right)^4\left ( \frac12 \right)^8 \\ &&&= \frac{495}{4096} \end{align*}

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Solution:

1. \(\,\) \begin{align*} \mathbb{P}(A) &= \mathbb{P}(\text{the first 6 arises on the \(n\)th throw.}) \\ &= \mathbb{P}(\text{\(n-1\) not 6s, followed by a 6.})\\ &= \left ( \frac56\right)^{n-1} \cdot \frac16 = \frac{5^{n-1}}{6^n} \end{align*}
2. There is nothing special about \(5\) or \(6\), so which comes first is \(50:50\), therefore this probability is \(\frac12\)
3. There is nothing special about \(4\), \(5\) or \(6\) so this is the probability that \(6\) appears last out of these three numbers, hence \(\frac13\)
4. \(\,\) \begin{align*} \mathbb{P}(D) &= \mathbb{P}(\text{exactly one 5 arises before the first 6.}) \\ &=\sum_{n=2}^{\infty} \mathbb{P}(\text{exactly one 5 arises before the first 6 which appears on the \(n\)th roll.}) \\ &= \sum_{n=2}^{\infty} \binom{n-1}{1} \left ( \frac46 \right)^{n-2} \frac16 \cdot \frac16 \\ &= \frac1{36} \sum_{n=2}^{\infty} (n-1) \left ( \frac23 \right)^{n-2} \\ &= \frac1{36} \sum_{n=1}^{\infty} n \left ( \frac23 \right)^{n-1} \\ &= \frac1{36} \frac{1}{\left ( 1- \frac23 \right)^2} = \frac14 \end{align*}
5. \(\,\) \begin{align*} \mathbb{P}(D \cup E) &= \mathbb{P}(D) + \mathbb{P}(E) - \mathbb{P}(D \cap E) \\ &= \frac12 - \mathbb{P}(D \cap E) \\ &=\frac12 - \sum_{n=3}^{\infty} \mathbb{P}(\text{exactly one 5 and one 4 arises before the first 6 which appears on the \(n\)th roll.}) \\ &=\frac12 - \sum_{n=3}^{\infty} 2\binom{n-1}{2} \left ( \frac36 \right)^{n-3}\cdot \frac16 \cdot \frac16 \cdot \frac16 \\ &=\frac12 - \frac2{6^3}\sum_{n=3}^{\infty} \frac{(n-1)(n-2)}{2} \left ( \frac12 \right)^{n-3} \\ &=\frac12 - \frac2{6^3}\frac{1}{(1-\tfrac12)^3}\\ &= \frac12 - \frac{2}{27} \\ &= \frac{23}{54} \end{align*}