Problems

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Solution: \begin{align*} \mathbb{P}(\text{fever tree grows}) &= \mathbb{P}(0 \leq L \leq 1) + \mathbb{P}(2 \leq L \leq 3) \\ &= \int_0^1 \frac12 -\frac18 x \d x + \int_2^3 \frac12 - \frac18 x \d x \\ &= \left [\frac12 x - \frac1{16}x^2 \right]_0^1+ \left [\frac12 x - \frac1{16}x^2 \right]_2^3 \\ &= \frac12 - \frac1{16}+\frac32-\frac9{16} - 1 + \frac{4}{16} \\ &= \frac58 \end{align*} The expected number of fever trees is just \(96 \cdot \frac58 = 60\).

1. \(f_Y(t)\) must match the distribution for \(L\), but limited to the points we care about, therefore it should be: $f_Y(t) = \begin{cases} ( \frac45 - \frac15t ) & \text{if } t \in [0,1]\cup[2,3] \\ 0 & \text{otherwise} \end{cases}$
   

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   \begin{align*} \mathbb{E}(Y) &= \frac12 \cdot \frac15 (4 - \frac12)+\frac52 \cdot (1 - \frac15 (4 - \frac12)) \\ &= \frac12 \cdot \frac7{10} + \frac52 \cdot \frac3{10} \\ &= \frac{22}{20} \\ &= \frac{11}{10} \end{align*}
2. Given the seeds are blown independently and the wind hasn't changed, it is reasonable to model the number of fever trees as \(B(96, \frac{5}{8})\), it is also acceptable to approximate this using a Normal distribution, ie \(N(60, 22.5)\), \(23\) is \(\frac{23-60}{\sqrt{22.5}}\) is a very negative number, so he should be extremely suspicious.

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Solution: \(A \sim 5\cdot 5 + U[0,10]\) \(B \sim 4 \cdot 5 + 3 \textrm{Po}(4)\) \(C \sim 2 \cdot 5 + B(5, \frac{4}{5}) \cdot 5\) \begin{align*} && \mathbb{P}(A \leq 27) &= \mathbb{P}(U \leq 2) = 0.2 \\ && \mathbb{P}(B \leq 27) &= \mathbb{P}(3 \textrm{Po}(4) \leq 7) \\\ &&&= \mathbb{P}(Po(4) \leq 2) \\ &&&= e^{-4}(1 + 4 + \frac{4^2}{2}) \\ &&&= 0.23810\ldots \\ && \mathbb{P}(C \leq 27) &= \mathbb{P}(5 \cdot B(5,\tfrac45) \leq 17) \\ &&&= \mathbb{P}(B(5,\tfrac45) \leq 3) \\ &&&= \binom{5}{0} (\tfrac15)^5 + \binom{5}{1} (\tfrac45)(\tfrac 15)^4+ \binom{5}{2} (\tfrac45)^2(\tfrac 15)^3 + \binom{5}3 (\tfrac45)^3(\tfrac 15)^2+\\ &&&= 0.26272 \end{align*} \begin{align*} \mathbb{P}(\text{came via B} | \text{at least 27 minutes}) &= \frac{\mathbb{P}(\text{came via B and at least 27 minutes})}{\mathbb{P}(\text{at least 27 minutes})} \\ &= \frac{\frac13 \cdot 0.23810\ldots }{\frac13 \cdot 0.2 + \frac13 \cdot 0.23810\ldots + \frac13 \cdot 0.26272} \\ &= 0.3397\ldots \\ &= 0.340 \, \, (3\text{ s.f.}) \end{align*}

1. \begin{align*} \mathbb{E}(A) &= 25 + 5 &= 30 \\ \mathbb{E}(B) &= 20 + 3\cdot 4 &= 32 \\ \mathbb{E}(C) &= 10 + 5 \cdot 4 &= 30 \end{align*} \(A\) and \(C\) are equally good.
2. \begin{align*} \mathbb{P}(A \leq 32) &= \mathbb{P}(U \leq 7) &= 0.7 \\ \mathbb{P}(B \leq 32) &= \mathbb{P}(Po(4) \leq 4) \\ &= e^{-4}(1 + 4 + 8 + \frac{4^3}{6}) &= 0.4334\ldots \\ \mathbb{P}(C \leq 32) &= \mathbb{P}(B(5,\tfrac45) \leq 4) \\ &= 1 - \mathbb{P}(B(5,\tfrac45) = 5) \\ &= 1 - \frac{4^5}{5^5} &=0.67232 \end{align*} So you should choose route \(A\).

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Solution:

1. Suppose A always plays \(1\), then B should always play \(2\) and every time they will win 1. Suppose A always plays \(2\) then B should always play \(3\) and every time they will win 1. If A always plays \(3\) then B should always play \(1\) and every time they will win 2.
2. \begin{array}{cccc} & b_1 & b_2 & b_3 \\ \frac13 & (0, \frac{b_1}{3}) & (1, \frac{b_2}{3}) & (-2, \frac{b_3}{3}) \\ \frac13 & (-1, \frac{b_1}{3}) & (0, \frac{b_2}{3}) & (1, \frac{b_3}{3}) \\ \frac13 & (2, \frac{b_1}{3}) & (-1, \frac{b_2}{3}) & (0, \frac{b_3}{3}) \\ \end{array} Therefore the expected value is: \(\frac{b_1}{3} - \frac{b_3}{3}\) and to maximise this he should always guess \(1\) (ie \(b_1 = 1, b_2 = 0, b_3 = 0\).)
3. \begin{array}{cccc} & b_1 & b_2 & b_3 \\ a_1 & (0, a_1b_1) & (1, a_1b_2) & (-2, a_1b_3) \\ a_2 & (-1, a_2b_1) & (0, a_2b_2) & (1, a_2b_3) \\ a_3 & (2, a_3b_1) & (-1, a_3b_2) & (0, a_3b_3) \\ \end{array} Therefore the expected value is: \((b_2-2b_3)a_1 + (b_3-b_1)a_2 + (2b_1-b_2)a_3\) We need \(b_2 \geq 2b_3, b_3 \geq b_1, 2b_1 \geq b_2\) so \(b_1 \leq b_3 \leq \frac12 b_2 \leq b_1\) so we could take \(b_1 = b_3 = \frac12 b_2\) or \(b_1 = b_3 = \frac14, b_2 = \frac12\) and all values would be \(0\). Therefore by choosing these values \(B\) can guarantee his expected value is \(0\) and therefore shouldn't expect to lose money in the long run.

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Solution: \begin{align*} \cos 3\theta &= \cos 2\theta\cos\theta - \sin 2\theta \sin \theta \\ &= (2\cos^2\theta-1)\cos \theta - 2\cos \theta \sin^2 \theta \\ &= 2\cos^3\theta-\cos \theta - 2\cos \theta(1- \cos^2 \theta) \\ &= 4\cos^3 \theta - 3\cos \theta \end{align*} \begin{align*} 0 &= 24x^{3}-72x^{2}+66x-19 \\ &= 24(y+1)^3-72(y+1)^2+66(y+1)-19 \\ &= 24(y^3+3y^2+3y+1)-72(y^2+2y+1)+66(y+1)-19\\ &= 24y^3+(72-144+66)y+(24-72+66-19) \\ &= 24y^3-6y-1 \\ &= 24b'^3z^3 - 6b'z - 1 \\ &= \frac{2}{\sqrt{3}}(4 z^3 -3z) - 1 \\ \end{align*} Therefore if \(b = \sqrt{3}, a = 1\), we have: \(4z^3 - 3z = \frac{\sqrt{3}}{2}\) So if \(z = \cos \theta \Rightarrow \cos 3\theta = \frac{\sqrt{3}}2 \Rightarrow 3 \theta = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6} \Rightarrow \theta = \frac{\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}\). Since \(\frac{\cos x}{\sqrt{3}} < 1\) we only need to approximate the first part to 2 significant figures. Therefore: \begin{align*} \sqrt{3} &\approx 1 + \frac{1}{1 + \frac{1}{2+\frac11}} = \frac{7}{4} = 1.75\\ \cos \tfrac{\pi}{18} &\approx \cos \frac{1}{6} \approx 1 - \frac{1}{2} \frac{1}{6^2} = \frac{71}{72} \approx 1 - \frac{1}{70} = 1 - 0.014 = 0.986 \\ \frac{\cos \tfrac{\pi}{18}}{\sqrt{3}} & \approx \frac{.986}{1.75} = 0.57 \\ \end{align*} Final answers: \(1.57, 0.803, 0.629\). We wouldn't be able to solve \(x^3 + 1= 0\) using this method, as we would have 2 non-real roots

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Solution: \begin{align*} && \cot x - \tan x &= 2 \cot 2x \\ \Rightarrow && x\cot x - x\tan x &= 2x\cot 2x \\ \Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n - \sum_{n=0}^{\infty}a_n x^{n+1} &= 1 + \sum_{n=1}^{\infty} b_n (2x)^n \\ \Rightarrow && \sum_{n=1}^{\infty}(1-2^n)b_nx^n &= \sum_{n=1}^{\infty} a_{n-1}x^n \\ \Rightarrow && a_{n-1} &= (1-2^n)b_n \quad \text{if }n \geq 1 \end{align*} \begin{align*} \cot x + \tan x &= 2 \cosec 2x \end{align*} So \begin{align*} && \cot x + \tan x &= 2 \cosec 2x \\ \Rightarrow && x \cot x + x\tan x &= 2x \cosec 2x \\ \Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n + \sum_{n=0}^{\infty} a_n x^{n+1} &= 1+\sum_{n=1}^\infty c_n (2x)^n \\ \Rightarrow && \sum_{n=1}^{\infty} \frac{1}{1-2^n}a_{n-1} +\sum_{n=1}^{\infty}a_{n-1}x^n &= \sum_{n=1}^{\infty} 2^nc_n x^n \\ \Rightarrow && c_n &= \frac{1}{2^n} \left ( 1 + \frac{1}{1-2^n} \right)a_{n-1} \\ &&&= \frac1{2^n} \frac{2^n-2}{2^n-1} a_{n-1}\\ &&&= \frac1{2^{n-1}}\frac{2^{n-1}-1}{2^n-1} a_{n-1} \end{align*}

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Solution: \begin{align*} && 2(u+iv) &= e^{a+iy} - e^{-a-iy} \\ && &=(e^a \cos y - e^{-a} \cos y) + (e^a \sin y + e^{-a} \sin y)i \\ &&&= 2 \sinh a \cos y + 2\cosh a \sin y i\\ \Rightarrow && \frac{u}{\sinh a} &= \cos y \\ && \frac{v}{\cosh a} &= \sin y \\ \Rightarrow && 1 &= \left(\frac{u}{\sinh a}\right)^{2}+\left(\frac{v}{\cosh a}\right)^{2} \end{align*} \begin{align*} && 2(u+iv) &= e^{x+ib} - e^{-x-ib} \\ &&&= 2\sinh x \cos b + 2\cosh x \sin b i \\ \Rightarrow && \frac{u}{\cos b} &= \sinh x \\ && \frac{v}{\sin b} &= \cosh x \\ \Rightarrow && 1 &= \left (\frac{v}{\sin b} \right)^2 - \left (\frac{u}{\cos b} \right)^2 \end{align*} Therefore all the points lie of a hyperbola, and since \(\frac{v}{\sin b} > 0 \Rightarrow v > 0\) it's one branch of the hyperbola. (And all points on it are reachable as \(x\) varies from \(-\infty < x < \infty\). \begin{align*} 2(u+iv) &= e^{a+ib} - e^{-a-ib} \\ &= 2 \sinh a \cos b + 2 \cosh a \sin b i \end{align*} so we can take \(u = \sinh a \cos b, v = \cosh a \sin b\). \begin{align*} \frac{\d }{\d u} && 0 &= \frac{2 u}{\sinh^2 a} + \frac{2v}{\cosh^2 a} \frac{\d v}{\d u} \\ \Rightarrow && \frac{\d v}{\d u} &= -\frac{u}{v} \coth^2 a \\ \\ && \frac{\d v}{\d u} \rvert_{(u,v)} &= -\frac{\sinh a \cos b}{\cosh a \sin b} \coth^2 a \\ &&&= -\cot b \coth a \\ \frac{\d }{\d u} && 0 &= \frac{2 v}{\sin^2 b} \frac{\d v}{\d u} - \frac{2u}{\cos^2 b} \\ \Rightarrow && \frac{\d v}{\d u} &= \frac{u}{v} \tan^2 b \\ && \frac{\d v}{\d u} \rvert_{(u,v)} &= \frac{\sinh a \cos b}{\cosh a \sin b} \tan^2 b \\ &&&= \tanh a \tan b \end{align*} Therefore they are negative reciprocals and hence perpendicular.

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Solution: Notice that \(\ln f(x) = \ln (x - a) + \ln (x-b) - \ln (x-c) - \ln (x-d)\) therefore: \begin{align*} \frac{\d}{\d x}: && \frac{f'(x)}{f(x)} &= (x-a)^{-1}+(x-b)^{-1}-(x-c)^{-1} - (x-d)^{-1} \\ &&&= \frac{1}{x} \left ( (1-\frac{a}{x})^{-1}+(1-\frac{b}{x})^{-1}-(1-\frac{c}{x})^{-1} - (1-\frac{d}{x})^{-1}\right) \end{align*} Multiplying by \(x\) gives the desired result. When \(|x|\) is very large then: \begin{align*} \frac{x f'(x)}{f(x)} &\approx 1 + \frac{a}{x} + o(\frac{1}{x^2})+ 1 + \frac{b}{x} + o(\frac{1}{x^2})-(1 + \frac{c}{x} + o(\frac{1}{x^2}))-(1 + \frac{d}{x} + o(\frac{1}{x^2})) \\ &= \frac{a+b-c-d}{x} + o(x^{-2}) \end{align*} Dividing by \(x\) we obtain \(\frac{f'(x)}{f(x)} \approx \frac{a+b-c-d}{x^2} + o(x^{-3})\) if \(|x|\) is sufficiently large this will be dominated by the \(\frac{a+b-c-d}{x^2}\) term which will have the same sign as \((a+b-c-d)\). When \(|x|\) is very large all of the brackets will have the same sign, and therefore \(f(x)\) will be positive, and so \(f'(x)\) must have the same sign as \(a+b-c-d\). To prove this result, we could set \(f(x) = k\) and rearrange to form a quadratic in \(x\). We could then check the discriminant is non-zero. Case 1: \(a < c < d < b\) and \(a+b > c+d \Rightarrow\) not all values reached and approx asymtope from below on the right and above on the left.

Case 2: \(a < c < d < b\) and \(a + b < c + d \Rightarrow\) not all values hit, but approach from above on the right and below on the left Case 2: \(a < c < b < d \Rightarrow\) all values hit Case 3: \(a < b < c < d \Rightarrow \) not all values hit

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Solution:

1. Suppose our curve is \(r(\theta)\), then \(y = r \sin \theta, x = r \cos \theta\) and \begin{align*} && \frac{\d y}{\d \theta} &= \frac{\d r}{\d \theta} \sin \theta + r \cos \theta \\ && \frac{\d x}{\d \theta} &= \frac{\d r}{\d \theta} \cos \theta - r \sin \theta \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{\d y}{\d \theta} \Bigg / \frac{\d x}{\d \theta} \\ &&&= \frac{\frac{\d r}{\d \theta} \sin \theta + r \cos \theta}{\frac{\d r}{\d \theta} \cos \theta - r \sin \theta} \\ &&&= \frac{\frac{\d r}{\d \theta} \tan\theta + r }{\frac{\d r}{\d \theta} - r \tan\theta} \end{align*} as required.
2. Set up a system of polar coordinates such that the origin is at \(O\) and all points in the plane containing \(L\) are represented by \((r, \theta)\). The constraint we have is that the angle of the normal, is \(\frac12 \theta\). Let \(\tan \tfrac12 \theta = t\), then \(\tan \theta = \frac{2t}{1-t^2}\) \begin{align*} && \tan \frac12 \theta &= -\frac{\frac{\d r}{\d \theta} - r \tan\theta}{\frac{\d r}{\d \theta} \tan\theta + r } \\ \Rightarrow && t &= -\frac{r'-r\frac{2t}{1-t^2}}{r' \frac{2t}{1-t^2}+r} \\ &&&= \frac{2tr-(1-t^2)r'}{2tr'+(1-t^2)r} \\ \Rightarrow && (2t^2+1-t^2)r' &= (2t-t+t^3)r \\ && (1+t^2)r' &= t(t^2+1) r \\ \Rightarrow && r' &= t r \\ \Rightarrow && \frac{\d r}{\d \theta} &= \tan \tfrac12 \theta r \\ \Rightarrow && \int \frac1r \d r &= \int \tan \frac12 \theta \d \theta \\ && \ln r &= -2\ln \cos \tfrac12 \theta+C \\ \Rightarrow && r\cos^2 \frac12 \theta &= C \\ \Rightarrow && r + r\cos \theta &= D \\ \Rightarrow && r &= D-x \\ \Rightarrow && x^2 + y^2 &= D^2 - 2Dx + x^2 \\ \Rightarrow && y^2 &= D^2-2Dx \end{align*} Therefore it is a parabola

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Solution: Let \(u = f(x)\) then \(\frac{\d u}{\d x} = f'(x)\) and \begin{align*} \int_a^b f(x) \d x &\underbrace{=}_{\text{IBP}} \left [ xf(x) \right]_a^b - \int_a^b x f'(x) \d x \\ &\underbrace{=}_{u = f(x)} b \beta - a \alpha - \int_{u = f(a) = \alpha}^{u = f(b) = \beta} g(u) \d u \\ &= b \beta - a \alpha - \int_{\alpha}^{\beta} g(u) \d u \end{align*}

\[ \underbrace{\int_{a}^{b}\mathrm{f}(x)\,\mathrm{d}x}_{\text{red area}}=\underbrace{b\beta}_{\text{whole area}}-\underbrace{a\alpha}_{\text{area in green}}-\underbrace{\int_{\alpha}^{\beta}\mathrm{g}(y)\,\mathrm{d}y}_{\text{area in blue}}, \] \begin{align*} 2\pi \int_a^b x f(x) \d x &\underbrace{=}_{\text{IBP}}\pi \left [ x^2 f(x) \right]_a^b - \pi \int_a^b x^2 f'(x) \d x \\ &\underbrace{=}_{x = g(u)} \pi (b^2 \beta - a^2 \alpha) - \pi \int_{u = f(a) = \alpha}^{u = f(b) = \beta} [g(u)]^2 \d u \\ &= \pi(b^2 \beta - a^2 \alpha) - \pi \int_\alpha^\beta [g(u)]^2 \d u \end{align*} This is the volume outside the function in the volume of revolution about the \(y\) axis between \( \alpha\) and \(\beta\).

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Solution: Let \(z = x^\alpha, \frac{\d z}{\d x}=\alpha x^{\alpha-1} \), then \begin{align*} \frac{\d y}{\d x} &= \frac{\d y}{\d z} \frac{\d z}{\d x} \\ &= \alpha x^{\alpha-1}\frac{\d y}{\d z} \\ \\ \frac{\d^2 y}{\d x^2} &= \frac{\d }{\d x} \left ( \alpha x^{\alpha-1}\frac{\d y}{\d z} \right) \\ &= \alpha (\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha x^{\alpha-1} \frac{\d ^2 y}{\d z^2} \frac{\d z}{\d x} \\ &= \alpha(\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha^2 x^{2\alpha-2} \frac{\d ^2y}{\d z^2} \end{align*} \begin{align*} && 0 &=x\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}-b\frac{\mathrm{d}y}{\mathrm{d}x}+x^{2b+1}y \\ &&&= x \left ( \alpha(\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha^2 x^{2\alpha-2} \frac{\d ^2y}{\d z^2}\right) - b \left ( \alpha x^{\alpha-1}\frac{\d y}{\d z} \right) + x^{2b+1}y \\ &&&= \alpha^2 x^{2\alpha-1} \frac{\d^2 y}{\d z^2} +\left (\alpha(\alpha-1)x^{\alpha-1}-b\alpha x^{\alpha-1} \right) \frac{\d y}{\d z} + x^{2b+1} y \\ \end{align*} If we set \(\alpha = b +1\) the middle term disappears, so we get \begin{align*} && 0 &= (b+1)^2 x^{2b+1} \frac{\d^2 y}{\d z^2} + x^{2b+1} y \\ \Rightarrow && 0 &= (b+1)^2 \frac{\d^2 y}{\d z^2} + y \\ \Rightarrow && y &= A \sin \left (\frac{z}{b+1} \right) + B \cos \left (\frac{z}{b+1} \right) \\ &&&= \boxed{A \sin \left (\frac{x^{b+1}}{b+1} \right) + B \cos \left (\frac{x^{b+1}}{b+1} \right)} \\ \\ \lim_{x \to 0}: && y &\to B \\ && \frac{\d y}{\d x} &= A x^b \cos\left (\frac{x^{b+1}}{b+1} \right) - B x^b \sin\left (\frac{x^{b+1}}{b+1} \right) \\ b>0: && \frac{\d y}{\d x} &\to 0 \\ \end{align*} So there are infinitely many different solutions with \(B = 1\) and \(A\) is anything it wants to be. If \(b = 0\) \(y' \to A\) so \(A =0 \) and unique. If \(b < 0\) \(x^b \to \infty\) so we need \(A = 0\), unique. However, we also need \(y' \to 0\), so we need to check \(y' = -x^b \sin \left ( \frac{x^{b+1}}{b+1}\right) \to 0\), \begin{align*} y' &= -x^b \sin \left ( \frac{x^{b+1}}{b+1}\right) \\ &\approx -x^b \left ( \frac{x^{b+1}}{b+1}\right) \\ &= - \frac{x^{2b+1}}{b+1} \end{align*} so we need \(2b+1>0 \Rightarrow b > -\frac12\). Therefore the solution is unique on \((-\frac12,0]\)