Problems

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Solution: \begin{align*} && \beta - \alpha &> q \\ \Rightarrow &&(\beta - \alpha)^2 &> q^2 \\ \Rightarrow && \beta^2 +\alpha^2 - 2\beta \alpha &> q^2 \\ \Rightarrow && \alpha^2+\beta^2-q^2 -2 \beta \alpha &> 0 \\ \Rightarrow && \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} - 2 &> 0 \end{align*} \begin{align*} && u_n(u_n+u_{n+2}) &= u_n \cdot \left (u_n + \frac {u_{n+1}^2 -q^2}{u_{n}}\right) \\ &&&= u_n^2 + u_{n+1}^2-q^2 \\ &&&= u_n^2 + u_{n+1}^2 - (u_n^2-u_{n-1}u_{n+1}) \\ &&&= u_{n+1}^2 + u_{n+1}u_{n-1} \\ &&&= u_{n+1}(u_{n-1}+u_{n+1}) \\ \\ && u_{n+1}-pu_n+u_{n-1} &= -pu_n+\frac{u_{n}(u_{n-2}+u_n)}{u_{n-1}} \\ &&&= \frac{u_n(u_{n}-pu_{n-1}+u_{n-2})}{u_{n-1}} \end{align*} Therefore if \(u_2 -pu_1 + u_0 = 0\) it is always zero, ie if \begin{align*} && u_2 &= p\beta - \alpha \\ && u_2 &= \frac{\beta^2-q^2}{\alpha} \\ \Rightarrow && \frac{\beta^2-q^2}{\alpha} &= p\beta - \alpha \\ \Rightarrow && p &= \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} \end{align*} If \(\beta > \alpha + q\) we must have that \(p > 2\), and so \(u_{n+1}-u_n = (p-1)u_n - u_{n-1} > u_n-u_{n-1} > 0\), therefore the sequence is strictly increasing. If \(\beta = \alpha + q\) the sequence follows \(u_{n+1} - 2u_n + u_{n-1} =0\) and so \(u_{n+1}-u_n = u_n - u_{n-1}\) for all \(n\) (which is still increasing - it's an arithmetic progression with common difference \(\beta - \alpha\)).

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Solution: \begin{align*} && s &= ut \\ \Rightarrow && a &= u \cos \alpha t\\ \Rightarrow && t &= \frac{a}{u \cos \alpha}\\ && s &= ut+ \frac12at^2 \\ \Rightarrow && -h &< -u\sin \alpha \frac{a}{u \cos \alpha}-\frac12 g \left (\frac{a}{u \cos \alpha} \right)^2 \\ &&&= -a \tan \alpha-\frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha \\ \Rightarrow && \frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha &< h -a\tan \alpha \\ \Rightarrow &&\frac{1}{u^2} &< \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \end{align*} \begin{align*} && s &= ut + \frac12a t^2 \\ \Rightarrow && 2h &= u\sin \alpha t + \frac12 gt^2 \\ \Rightarrow && t &= \frac{-u\sin \alpha \pm \sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && t &= \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && s &= ut \\ \Rightarrow && b &> u \cos \alpha t \\ \Rightarrow && \frac{b}{u \cos \alpha} &> \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g} \\ \Rightarrow && \sqrt{u^2 \sin^2 \alpha+4hg} &< \frac{bg}{u \cos \alpha} + u \sin \alpha \\ \end{align*} \begin{align*} \Rightarrow && u^2 \sin^2 \alpha+4hg &< \frac{b^2g^2}{u^2 \cos^2 \alpha} +u^2 \sin^2 \alpha + 2bg \tan \alpha \\ \Rightarrow && 4hg - 2bg \tan \alpha &< \frac{b^2g^2}{u^2 \cos^2 \alpha} \\ &&&< \frac{b^2g^2}{\cos^2 \alpha} \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \\ &&&= \frac{2b^2g(h-a\tan \alpha)}{a^2} \\ \Rightarrow && \tan \alpha \left (\frac{2b^2g}{a} - 2bg \right) &< \frac{2b^2gh}{a^2} - 4hg \\ \Leftrightarrow && \tan \alpha \left (\frac{2b^2g- 2abg}{a} \right) &< \frac{2b^2gh- 4hga^2}{a^2} \\ \Leftrightarrow && \tan \alpha \left (\frac{2bg(b- a)}{a} \right) &< \frac{2hg(b^2- 2a^2)}{a^2} \\ \Rightarrow && \tan \alpha &< \frac{h(b^2-2a^2)}{ab(b-a)} \end{align*}

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Solution:

Let \(M\) be the midpoint of \(AB\), then \begin{align*} \overset{\curvearrowright}{M}: && R_1 \sin \phi-\mu R_1 \cos \phi &= R_2 \sin \phi+\mu R_2 \cos \phi \\ \Rightarrow && (R_1-R_2) \tan \phi &= \mu(R_1+R_2) \end{align*} As required. \begin{align*} && \cos \phi = \frac{a}{r} &,\,\, \sin \phi = \frac{\sqrt{r^2-a^2}}{r} \\ \text{N2}(\rightarrow): && R_1\cos(\phi + \theta)+\mu R_1 \sin(\phi + \theta) &= R_2 \cos(\theta - \phi) + \mu R_2 \sin(\theta - \phi) \\ \Rightarrow && R_1(\cos \theta \cos \phi - \sin \theta \sin \phi)+ \mu R_1 (\sin \theta \cos \phi + \cos \theta \sin \phi) &= R_2 (\cos\theta \cos \phi + \sin \theta \sin \phi)+ \mu R_2 (\sin \theta \cos \phi - \cos \theta \sin \phi) \\ && R_1 (1 - \tan \theta \tan \phi)+\mu R_1 (\tan \theta + \tan \phi) &= R_2(1 + \tan \theta \tan \phi) +\mu R_2 (\tan \theta - \tan \phi) \\ && 0 &= (R_1-R_2)(1+\mu \tan \theta)+(R_1+R_2)(-\tan \theta \tan\phi+\mu \tan \phi) \\ \Rightarrow && \frac{R_1+R_2}{R_1-R_2} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\ \Rightarrow && \frac{\tan \phi}{\mu} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\ \Rightarrow && \tan^2 \phi &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\ \Rightarrow && \frac{r^2-a^2}{a^2} &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\ \Rightarrow && \tan \theta (r^2-a^2-a^2\mu^2) &= \mu a^2+\mu(r^2-a^2) \\ \Rightarrow && \tan \theta &= \frac{\mu r^2}{r^2-(1+\mu^2)a^2} \end{align*} Since \(\mu r^2 > 0\) we must also have \(r^2 > a^2(1+\mu^2)\) ie \(\\sec^2 \phi > 1 + \mu^2 = \sec^2 \lambda\) and the result follows.

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Solution: \begin{align*} \text{COM}: && \sum_{i=1}^n im \cdot \frac{u}{i} &= \left ( km + \sum_{i=1}^n im \right) v \\ \Rightarrow && nu &= \left ( k + \frac{n(n+1)}{2} \right) v \\ \Rightarrow && v &= \frac{2nu}{2k + n(n+1)} \end{align*} If \(2k = N(N+1)\), there will be no more collisions when \(v_n > \frac{u}{n+1}\), ie \begin{align*} && \frac{u}{n+1} &<\frac{2nu}{2k + n(n+1)} \\ \Leftrightarrow && N(N+1) + n(n+1) &< 2n(n+1) \\ \Leftrightarrow && N(N+1) &< n(n+1) \\ \end{align*} Therefore \(n = N+1\) and there will be \(N+1\) collisions. The loss of kinetic energy is: \begin{align*} && \text{initial k.e.} &= \sum_{k=1}^{N+1} \frac12 im \cdot \frac{u^2}{i^2} \\ &&&= \frac12 m u^2 \left ( \sum_{k=1}^{N+1} \frac{1}{i}\right) \\ && \text{final k.e.} &= \frac12 \left ( k + \frac{(N+1)(N+2)}{2}\right)m \left ( \frac{2(N+1)u}{N(N+1)+(N+1)(N+2)} \right)^2 \\ &&&= \frac12 m u^2 \frac{2(N+1)^2}{(N+1)(2N+2)} \\ &&&= \frac12 mu^2 \\ \Rightarrow && \Delta \text{ k.e.} &= \frac12 m u^2 \left ( \sum_{k=2}^{N+1} \frac{1}{i}\right) \end{align*}

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(\text{hall switch on}) &= \underbrace{p \cdot \frac34 }_{\text{already on and not flipped}}+ \underbrace{(1-p) \cdot \frac14}_{\text{not on and flipped}} \\ &&&= \frac14 +\frac12 p\\ && \mathbb{P}(\text{kitchen on}) &= \frac12 \\ \Rightarrow && \mathbb{P}(\text{pool is on}) &= \frac18 + \frac14p \end{align*} \begin{align*} && \mathbb{P}(\text{flipped hall switch} | \text{pool on}) &= \frac{\mathbb{P}(\text{flipped hall and pool on})}{\mathbb{P}(\text{pool on})} \\ &&&= \frac{(1-p)\frac14 \cdot \frac 12}{\frac18 + \frac14 p} \\ &&&= \frac{1-p}{1+2p} \end{align*}
2. The number of days the swimming pool light was pressed is \(X = B\left (7, \frac{1-p}{1+2p} \right)\), and we have that \(\mathbb{P}(X = 2) < \mathbb{P}(X = 3) > \mathbb{P}(X=4)\) (since the binomial is unimodal). Let \(q = \frac{1-p}{1+2p} \) \begin{align*} && \mathbb{P}(X = 2) &< \mathbb{P}(X = 3) \\ \Rightarrow && \binom{7}{2} q^2(1-q)^5 &< \binom{7}{3}q^3(1-q)^4 \\ \Rightarrow && 21(1-q) &< 35q \\ \Rightarrow && 21 &< 56q \\ \Rightarrow && \frac{3}{8} &< \frac{1-p}{1+2p} \\ \Rightarrow && 3+6p &< 8-8p \\ \Rightarrow && 14p &< 5\\ \Rightarrow && p &< \frac5{14} \\ \\ && \mathbb{P}(X = 3) &> \mathbb{P}(X = 4) \\ \Rightarrow && \binom{7}{3} q^3(1-q)^4 &> \binom{7}{4}q^4(1-q)^3 \\ \Rightarrow &&(1-q)&> q \\ \Rightarrow && \frac12 &> q \\ \Rightarrow && \frac12 &> \frac{1-p}{1+2p} \\ \Rightarrow && 1+2p &> 2-2p \\ \Rightarrow && 4p &> 1\\ \Rightarrow && p &> \frac1{4} \end{align*} Therefore \(\frac14 < p < \frac{5}{14}\) as required.

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Solution: A circle radius \(y\) has a number of supermarkets \(X\) where \(X \sim Po(k \pi y^2)\). \[ \mathbb{P}(X = 0) = e^{-k\pi y^2} \frac{1}{0!} = e^{-k\pi y^2} \] The probability \(\mathbb{P}(Y < y) = 1-\mathbb{P}(Y \geq y) = 1-e^{-k\pi y^2}\), and in particular \(f_Y(y) = 2k\pi y e^{-k\pi y^2}\) (by differentiating). \begin{align*} && \mathbb{E}(Y) &= \int_0^\infty yf_Y(y) \d y \\ &&&= \int_0^\infty 2\pi y^2 k e^{-\pi k y^2} \d y \\ \sigma^2 = \frac{1}{2k\pi}:&&&= \pi k \sqrt{2 \pi}\sigma \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \sigma }y^2 e^{-\frac12 \cdot 2\pi k y^2} \d y \\ &&&=\pi k \sqrt{2 \pi}\sigma \mathbb{E}\left (N(0, \sigma^2)^2 \right) \\ &&&= \pi k \sqrt{2 \pi}\sigma\sigma^2 \\ &&&= \pi k \sqrt{2 \pi} \frac{1}{(2k\pi)^{3/2}} \\ &&&= \frac{1}{2\sqrt{k}} \end{align*} \begin{align*} && \mathbb{E}(Y^2) &= \int_0^\infty y^2f_Y(y) \d y \\ &&&= \int_0^\infty 2\pi y^3 k e^{-\pi k y^2} \d y \\ &&&= \int_0^{\infty}y^2 2y \pi k e^{-\pi k y^2} \d y \\ \\ &&&= \left [-y^2 e^{-\pi k y^2}\right]_0^{\infty}+\int_0^\infty 2ye^{-\pi k y^2} \d y \\ &&&= \left [-\frac{1}{\pi k}e^{-\pi k y^2} \right]_0^{\infty} \\ &&&= \frac{1}{\pi k} \\ \Rightarrow && \textrm{Var}(Y) &= \mathbb{E}(Y^2) - \left [ \mathbb{E}(Y)\right]^2 \\ &&&= \frac{1}{\pi k} - \frac{1}{4k} \\ &&&= \frac{4 - \pi}{4\pi k} \end{align*}

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Solution: \begin{align*} &&z &= y^n \left( \frac{\d y}{\d x}\right)^{2} \\ \Rightarrow && \frac{\d z}{\d x} &= ny^{n-1}\left( \frac{\d y}{\d x}\right)^{3} + y^{n} \cdot 2 \left( \frac{\d y}{\d x}\right) \left( \frac{\d^2 y}{\d x^2}\right) \\ &&&= y^{n-1} \left( \frac{\d y}{\d x}\right) \left (n \left( \frac{\d y}{\d x}\right)^2 + 2y \frac{\d^2 y}{\d x^2} \right) \end{align*}

1. Let \(z = y (y')^2\), then \begin{align*} && \frac{\d z}{\d x} &= y' \sqrt{y} \\ &&&= \sqrt{z} \\ \Rightarrow && \int z^{-1/2} \d z &= x+C \\ \Rightarrow && 2\sqrt{z} &= x + C \\ x = 0, z=0: && C &= 0 \\ \Rightarrow && y(y')^2 &= \frac14 x^2 \\ \Rightarrow &&\sqrt{y} \frac{\d y}{\d x} &= \frac{1}{2}x\\ \Rightarrow && \int \sqrt{y} \d y &= \int \frac{1}{2}x\d x \\ \Rightarrow && \frac23y^{3/2} &=\frac14x^2 + K \\ x = 0, y = 1: && K &= \frac23 \\ \Rightarrow && y &= \left (\frac38 x^2 + 1 \right)^{2/3} \end{align*}
2. Let \(z = y^{-2} (y')^2\) \begin{align*} && \frac{\d z}{\d x} &= y^{-3} \frac{\d y}{\d x} \left (-2 \left( \frac{\d y}{\d x}\right) + 2y \frac{\d^2 y}{\d x^2} \right) \\ &&&= y^{-3} \frac{\d y}{\d x} 2y^2 \\ &&&= 2y^{-1}(y') = 2 \sqrt{z} \\ \Rightarrow && 2\sqrt{z} &= 2x + C \\ x = 0, z = 0: && C&= 0 \\ \Rightarrow && z &= x^2 \\ \Rightarrow && \frac{\d y}{\d x} &= xy \\ \Rightarrow && \ln |y| &= \frac12 x^2 + K \\ x =0 , y =1; && K &= 0 \\ \Rightarrow && y &= e^{\frac12 x^2} \end{align*}

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Solution:

1. \begin{align*} (1-x)&(1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) \\ &= (1-x^2)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) \\ &= (1-x^4)(1+x^4) \cdots (1+x^{2^n}) \\ &= 1-x^{2^{n+1}} \\ \end{align*} Therefore, \begin{align*} && \frac{1}{1-x} - \frac{x^{2^{n+1}}}{1-x} &= (1+x)(1+x^2)\cdots(1+x^{2^n}) \\ \Rightarrow && \frac{1}{1-x} &=(1+x)(1+x^2)\cdots(1+x^{2^n})+ \frac{x^{2^{n+1}}}{1-x} \\ \Rightarrow && -\ln (1-x) &= \sum_{r=0}^{\infty} \ln (1+x^{2^r}) \\ \Rightarrow && \ln(1-x) &= - \sum_{r=0}^{\infty} \ln (1+x^{2^r}) \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{1}{1-x} &= \sum_{r=0}^{\infty} \frac{2^r x^{2^r-1}}{1+x^{2^r}} \\ &&&= \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \cdots \end{align*}
2. Consider \begin{align*}(1+x+x^2)&(1-x+x^2)(1-x^2+x^4)\cdots(1-x^{2^n}+x^{2^{n+1}}) \\ &= (1+x^2 + x^4)(1-x^2+x^4) \cdots (1-x^{2^n}+x^{2^{n+1}}) \\ &= (1-x^{2^{n+1}}+x^{2^{n+2}}) \\ \end{align*} Therefore, \begin{align*} && \frac{1}{1+x+x^2} &= (1-x+x^2)(1-x^2+x^4)\cdots(1-x^{2^n}+x^{2^{n+1}}) + \frac{x^{2^{n+1}}}{1+x+x^2} -\frac{x^{2^{n+2}}}{1+x+x^2} \\ \Rightarrow && -\ln(1+x+x^2) &= \sum_{r=0}^\infty \ln (1 - x^{2^r}+x^{2^{r+1}}) \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && -\frac{1+2x}{1+x+x^2} &= \sum_{r=0}^{\infty} \frac{ -2^r x^{2^r-1}+2^{r+1}x^{2^{r+1}-1}}{1 - x^{2^r}+x^{2^{r+1}}} \\ &&&= \frac{-1+2x}{1-x+x^2}+\frac{-2x+4x^3}{1-x^2+x^4} + \frac{-4x^3+8x^7}{1-x^4+x^8} + \cdots \end{align*} Which is the desired result when we multiply both sides by \(-1\)

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Solution:

1. 1. \(\,\)
      

      + - ↺
   2. \(\,\)
      

      + - ↺
   3. \(\,\)
      

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   4. \(\,\)
      

      + - ↺
2. Suppose \(m = 12\) \begin{align*} && y &= 4-x^2 \\ && 12x &= k-y^2 \\ \Rightarrow && 12 x&=k-(4-x^2)^2 \\ &&&= k-16+8x^2-x^4 \\ \Rightarrow && 0 &= x^4- 8x^2+12x+16-k \end{align*} When the curves touch, we will have repeated root, ie \(a\) is a root of \(4x^3-16x+12 \Rightarrow a^3-4a+3 =0\). \begin{align*} &&0 &= a^3-4a+3 \\ &&&= (a-1)(a^2+a-3) \\ \Rightarrow &&a &= 1, \frac{-1 \pm \sqrt{13}}{2} \end{align*} \begin{align*} && 0 &= a^4-8a^2+12a+16-k \\ \Rightarrow && k &= a(a^3-8a+12)+16 \\ &&&= a(4a-3-8a+12)+16 \\ &&&= -4a^2+9a+16 \\ \\ \Rightarrow && a = 1& \quad k = 21 \\ && k &= -4(3-a)+9a+16 = 13a+4\\ && a = \frac{-1-\sqrt{13}}2& \quad k = \frac{-5 - 13\sqrt{13}}{2} < 0 \\ && a = \frac{-1+\sqrt{13}}2& \quad k = \frac{-5 + 13\sqrt{13}}{2} \\ \end{align*} So we have type (a), and (d).

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Solution:

1. \begin{align*} \sum_{n=1}^{\infty} \frac{n+1}{n!} &= \sum_{n=1}^\infty \left ( \frac{1}{(n-1)!} + \frac{1}{n!} \right) \\ &= \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=1}^\infty \frac{1}{n!} \\ &= \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=0}^\infty \frac{1}{n!} - 1 \\ &= e + e - 1 \\ &= 2e-1 \end{align*} \begin{align*} \sum_{n=1}^{\infty} \frac{(n+1)^2}{n!} &= \sum_{n=1}^{\infty} \frac{n(n-1) + 3n + 1}{n!} \\ &= \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + 3 \sum_{n=1}^\infty \frac1{(n-1)!} + \sum_{n=1}^\infty \frac{1}{n!} \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} + 3 \sum_{n=0}^\infty \frac1{n!} + \sum_{n=0}^\infty \frac{1}{n!} -1 \\ &= 5e-1 \end{align*} \begin{align*} \sum_{n=1}^\infty \frac{(2n-1)^3}{n!} &= \sum_{n=1}^\infty \frac{8n^3-12n^2+6n-1}{n!} \\ &= \sum_{n=1}^\infty \frac{8n(n-1)(n-2)+12n^2-10n-1}{n!} \\ &= \sum_{n=1}^\infty \frac{8n(n-1)(n-2)+12n(n-1)+2n-1}{n!} \\ &= 8 e+12e+2e-(e-1) \\ &=21e+1 \end{align*}
2. \begin{align*} \frac{n^2+1}{(n+1)(n+2)} &= \frac{n^2+3n+2-3n-1}{(n+1)(n+2)}\\ &= 1 - \frac{3n+1}{(n+1)(n+2)} \\ &= 1 + \frac{2}{n+1} - \frac{5}{n+2} \\ -\log(1-x) &= \sum_{n=1}^\infty \frac1{n}x^{n} \\ \log(2) &= \sum_{n=1}^\infty \frac{2^{-n}}{n} \\ \sum_{n=0}^\infty \frac{(n^2+1)2^{-n}}{(n+1)(n+2)} &= \sum_{n=0}^{\infty} 2^{-n} + 2 \sum_{n=0}^\infty \frac{2^{-n}}{n+1}-5 \sum_{n=0}^\infty \frac{2^{-n}}{n+2} \\ &= 2 + 2\log2-5 \sum_{n=2}^\infty \frac{2^{-n+2}}{n} \\ &= 2 + 2 \log 2 - 5 \left (2\log 2 - 2 \right) \\ &= 12-8\log2 \end{align*}