Problems

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Solution:

First note our triangles are 7-24-25 and 3-4-5 triangles, so we can easily calculate \(\sin\) and \(\cos\) of our angles. \begin{questionparts} \item \begin{align*} \text{N2}(\uparrow, P): && 2T - Mg &= 0 \\ \\ \text{N2}(\perp AQ): && R_A - 5mg \cdot \frac{24}{25} &= 0 \\ \Rightarrow && R_A &= \frac{24}{5}mg \\ \text{N2}(\parallel AQ): && T - \mu R_A - 5mg \cdot \frac{7}{25} &= 0 \\ \Rightarrow && T &= \frac15 mg \l 7+24 \mu \r \\ \text{N2}(\perp BR): && R_B - 3mg \cdot \frac{3}{5}&= 0 \\ \Rightarrow && R_B &= \frac{9}{5}mg \\ \text{N2}(\parallel AQ): && T - \mu R_B - 3mg \cdot \frac{4}{5} &= 0 \\ \Rightarrow && T &= \frac15 mg \l 12+9 \mu \r \\ \\ \Rightarrow && 12 + 9 \mu &= 7 + 24 \mu \\ \Rightarrow && \mu &= \frac{5}{15} = \frac13 \\ \\ \Rightarrow && Mg &= 2 \cdot \frac15 \cdot mg \cdot (7 + 24 \cdot \frac13) \\ &&&= 6mg \\ \Rightarrow && M &= 6m \end{align*} \item Assuming \(\mu = \frac13\) \begin{align*} &&9m \ddot{p} &= 9mg - 2T \\ &&5m \ddot{a} &= T - 3mg \\ &&3m \ddot{b} &= T - 3mg \\ &&2\ddot{p} &= \ddot{a}+\ddot{b} \\ \Rightarrow &&30m\ddot{p} &= 8T - 24mg \\ &&9m\ddot{p} &= 9mg - 2T \\ \Rightarrow && 66m \ddot{p} &=12mg \\ \Rightarrow && \ddot{p} &= \frac{2}{11}g \\ && T &= 9mg - 9m \frac{2}{11} g = \frac{9^2}{11}mg\\ && \ddot{a} &= \frac{3}{22} g \\ && \ddot{b} &= \frac{5}{22}g \end{align*}

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Solution: The probability it becomes faulty in each year is: \begin{align*} \mathbb{P}(\text{faulty in Y}1) &= \int_0^1 \frac{2t}{(1+t^2)^2} \, dt \\ &= \left [ -\frac{1}{(1+t^2)} \right]_0^1 \\ &= 1 - \frac{1}{2} = \frac{1}{2} \\ \mathbb{P}(\text{faulty in Y}2) &= \frac{1}{2} - \frac{1}{5} = \frac{3}{10} \\ \mathbb{P}(\text{faulty in Y}3) &= \frac{1}{5} - \frac{1}{10} = \frac{1}{10} \end{align*} The probability of failing for the first time after exactly \(3\) years is: \begin{align*} \mathbb{P}(\text{faulty in Y1, }PPF) &+ \mathbb{P}(\text{faulty in Y2, }PF) + + \mathbb{P}(\text{faulty in Y3, }F) \\ &= \frac12 (1-p)^2p + \frac3{10}(1-p)p + \frac1{10}p \\ &= \frac{p}{10} \l 5(1-p)^2 + 3(1-p) + 1 \r \\ &= \frac{p}{10} \l 5 - 10p + 5p^2 + 3 -3p +1 \r \\ &= \frac{p}{10} \l 9 - 13p + 5p^2 \r \end{align*} as required. The probability that a randomly chosen fire extinguisher that was destroyed exactly three years after its manufacture was faulty 18 months after its manufacture is: \begin{align*} \mathbb{P}(\text{faulty 18 months after} | \text{fails after 3 tries}) &= \frac{\mathbb{P}(\text{faulty 18 months after and fails after 3 tries})}{\mathbb{P}(\text{fails after exactly 3 tries})} \end{align*} We can compute \(\mathbb{P}(\text{faulty 18 months after and fails after 3 tries})\) by looking at \(2\) cases, fails between \(12\) months and \(18\) years, and between \(0\) years and \(1\) year. \begin{align*} \mathbb{P}(\text{faulty between 1y and 18m}) &= \int_{1}^{\frac32} \frac{2t}{(1+t^2)^2} \, dt \\ &= \left [ -\frac{1}{(1+t^2)} \right]_{1}^{\frac32} \\ &= \frac12 - \frac{4}{13} = \frac{5}{26} \\ \end{align*} So the probability is: \begin{align*} \mathbb{P} &= \frac{\frac{5}{26}(1-p)p + \frac12(1-p)^2p}{\frac{p}{10} \l 9 - 13p + 5p^2 \r} \\ &= \frac{\frac{25}{13}(1-p) + 5(1-p)^2}{9 - 13p + 5p^2} \\ &= \frac{5}{13} \frac{(1-p)\l 5 + 13(1-p) \r}{9 - 13p + 5p^2} \\ &= \frac{5}{13} \frac{(1-p)\l 18 - 13p \r}{9 - 13p + 5p^2} \\ \end{align*}

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Solution: We are choosing any \(5\) digit number from \(\{1,2,3,4,5\}\). There are \(5^5\) such numbers. \begin{align*} && \mathbb{E}(\text{different digits}) &= \frac1{5^5} \left (1 \cdot 5 + 2 \cdot \binom{5}{2}(2^5-2)+3 \cdot \binom{5}{3}(3^5-3 \cdot 2^5+3)+4 \cdot \binom{5}{4}(4^5 - 4 \cdot 3^5+6 \cdot 2^5-4) + 5 \cdot 5! \right) \\ &&&= \frac{2101}{625} = 3.3616 \end{align*}

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Solution: \(\displaystyle (1-x^6)^{-2} = \sum_{n=0}^{\infty} (n+1)x^{6n}\)

1. \(\,\) \begin{align*} && f(x) &= (1-x^6)^{-2}(1-x^3)^{-1} \\ &&&= \left ( \sum_{n=0}^{\infty} (n+1)x^{6n} \right) \left ( \sum_{n=0}^{\infty} x^{3n} \right) \\ [x^{24}]: && c_{24} &= 1 + 2+ 3+4+5 = 15 \end{align*} Clearly \(n\) must be a multiple of \(3\). If \(n = 6k\) then we have \(1 + 2 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}\) If \(n = 6k+3\) then we have \(1 + 2 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}\) the same way, we just must always get one extra \(x^3\) term from the second expansion.
2. We can obtain \(x^{24}\) from the product of \((1-x^6)^{-2}(1-x^3)^{-1}\) and \((1-x)^{-1}\) in the following ways: \begin{array}{c|c|c} (1-x^6)^{-2}(1-x^3)^{-1} & (1-x)^{-1} & \text{product} \\ \hline 15x^{24} & x^0 & 15x^{24} \\ 10x^{21} & x^3 & 10x^{24} \\ 10x^{18} & x^6 & 10x^{24} \\ 6x^{15} & x^9 & 6x^{24} \\ 6x^{12} & x^{12} & 6x^{24} \\ 3x^{9} & x^{15} & 3x^{24} \\ 3x^{6} & x^{18} & 3x^{24} \\ x^{3} & x^{21} & x^{24} \\ x^{0} & x^{24}& x^{24} \end{array} So the total is \(55\). Similarly for \(25\) we can only obtain this in the same ways but also taking an extra power of \(x\) from the geometric series, ie \(55\) For \(66\) we obtain by similar reasoning that it is: \(\frac{13\cdot12}{2} + 2 \left (1 + 3 + \cdots + \frac{13 \cdot 12}{2} \right) = \frac{13\cdot12}{2} + 2 \binom{14}{3} = \frac{13 \cdot 12}2 ( 1 + \frac{30}{3}) = 11 \cdot 6 \cdot 13 = 858\)

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Solution: If \(\p(x)\) and \(\q(x)\) are polynomials of degree \(m\) and \(n\), \(\p(\q(x))\) has degree \(mn\).

1. Suppose \(\p(\p(\p(x)))- 3 \p(x)= -2x\), and suppose \(p(x)\) has degree \(n = \geq 2\), then \(\p(\p(\p(x)))\) has degree \(n^3\) and so the left hand side has degree higher than \(1\) and the right hand side is degree \(1\). Therefore \(\p(x)\) is degree \(1\) or \(0\). If \(p(x) = c\) then \(c^3-3c = -2x\) but the LHS doesn't depend on \(x\) which is also a contradiction. Therefore \(\p(x)\) is degree \(1\). Suppose \(\p(x) = ax+b\) then: \begin{align*} && -2x &= \p(\p(\p(x))) - 3\p(x) \\ &&&= \p(\p(ax+b)) - 3(ax+b) \\ &&&= \p(a(ax+b)+b) - 3ax -3b \\ &&&= a(a^2x+ab+b) + b - 3ax - 3b \\ &&&= (a^3-3a)x + b(a^2+a-2) \\ \Rightarrow &&& \begin{cases} a^3-3a&=-2 \\ b(a^2+a-2) &= 0\end{cases} \\ \Rightarrow &&& \begin{cases} a^3-3a+2 = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow &&& \begin{cases} (a-1)(a^2+a-2) = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow && (a,b) &= (1, b), (-2,b) \end{align*}
2. Suppose \(2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4\) and let \(\deg \p(x) = n\), then LHS has degree \(\max(n^2,2n,n)\) and the right hand side has degree \(4\). Therefore \(\p(x)\) must have degree \(2\). Let \(\p(x) = ax^2 + bx + c\), then, considering the coefficient of \(x^4\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we will have \(2a^3+3a^2=1 \Rightarrow 2a^3+3a^2-1 = (a+1)^2(2a-1) \Rightarrow a = -1, a=\frac12\). Consider the coefficient of \(x^3\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we have \(4a^2b+6ab = 0 \Rightarrow 2ab(2a+3) = 0\) Since \(a = -1, \frac12\) this means \(b = 0\). Consider the constant coefficient in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) (using \(b = 0\)). \(2ac^2+c+3c^2-4c = 0 \Rightarrow c(2ac+3c-3) = 0\). Therefore \(c = 0\) or \(a = -1, c = 3, a = \frac12, c = \frac34\), so our possible polynomials are: \(\p(x) = -x^2, \frac12x^2, -x^2+3, \frac12x^2+\frac34\)

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Solution: \begin{align*} && t &= x + \sqrt{1+x^2} \\ &&\frac1t &= \frac{1}{x+\sqrt{1+x^2}} \\ &&&= \frac{\sqrt{1+x^2}-x}{1+x^2-1} \\ &&&= \sqrt{1+x^2}-x \\ \Rightarrow && x &=\frac12 \left ( t - \frac1t\right) \\ \Rightarrow && \d x &=\frac12 \left (1 + \frac1{t^2} \right)\d t \\ \\ \Rightarrow && \int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x &= \int_{t=1}^{t = \infty}f(t) \frac12\left (1 + \frac1{t^2} \right)\d t \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right)f(t) \d t \end{align*} \begin{align*} && I &= \int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \d x \\ &&&= \int_0^\infty \frac1 {(x+\sqrt{x^2+1})^2} \d x \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right) \frac{1}{t^2} \d t \\ &&&= \frac12 \left [-\frac1t-\frac13\frac1{t^3} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac43 = \frac23 \end{align*} \begin{align*} && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&x &= \tan \theta\\ && \d x &= \sec^2 \theta = (1+x^2) \d \theta\\ && \tan\theta &= \frac{s}{\sqrt{1-s^2}}\\ \Rightarrow && \tan^2 \theta &= \frac{s^2}{1-s^2} \\ \Rightarrow && \sin \theta &= \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \\ && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&&= \int_0^{\frac12 \pi} \frac{1}{\left (1+ \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \right )^3} \d \theta \\ &&&= \int_{x=0}^{x=\infty} \frac{1}{\left(1 + \frac{x}{\sqrt{1+x^2}} \right)^3} \frac{1}{1+x^2} \d x \\ &&&= \int_0^{\infty} \frac{\sqrt{1+x^2}}{(\sqrt{1+x^2}+x)^3} \d x \\ &&J_a &= \int_0^{\infty} \frac{\sqrt{1+x^2}+x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \frac23 \\ &&J_b &= \int_0^{\infty} \frac{\sqrt{1+x^2}-x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \int_0^{\infty} \frac{1}{(\sqrt{1+x^2}+x)^4} \d x\\ &&&= \frac12\int_1^{\infty} \left (1 +\frac1{t^2} \right)\frac{1}{t^4} \d t \\ &&&= \frac12 \left [-\frac13 t^{-3}-\frac15t^{-5} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac8{15} = \frac4{15} \\ \Rightarrow && J &= \frac12(J_a+J_b) = \frac7{15} \end{align*}

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Solution:

1. Since \(k \geq 1\) we have \(n^{k+1} > n^k\) and \((k+1) > k\), therefore \((k+1)n^{k+1} >kn^k \Rightarrow \frac{1}{(k+1)n^{k+1}} < \frac{1}{kn^k}\) \begin{align*} && \ln \left ( 1 + \frac1n \right) &= \frac1n -\frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \cdots \\ &&&= \frac1n - \underbrace{\left (\frac{1}{2n^2}-\frac{1}{3n^3} \right)}_{>0}- \underbrace{\left (\frac{1}{4n^4}-\frac{1}{5n^5} \right)}_{>0} - \cdot \\ &&&< \frac1n \\ \\ \Rightarrow && n \ln \left ( 1 + \frac1n \right) &< 1 \\ \Rightarrow && \ln \left ( \left ( 1 + \frac1n \right)^n \right) &< 1 \\ \Rightarrow && \left ( 1 + \frac1n \right)^n &< e \end{align*}
2. \(\,\) \begin{align*} &&\ln \left(\frac{2y+1}{2y-1}\right) &= \ln \left (1 + \frac{1}{2y} \right)-\ln \left (1 - \frac{1}{2y} \right) \\ &&&= \frac{1}{2y} - \frac{1}{2(2y)^2} + \frac{1}{3(2y)^3} - \cdots - \left (-\frac{1}{2y} - \frac{1}{2(2y)^2} - \frac{1}{3(2y)^3} - \cdots \right) \\ &&&= \frac{1}{y} + \frac{2}{3(2y)^3} + \frac{2}{5(2y)^5} \\ &&&= \sum_{r=1}^{\infty} \frac{2}{(2r-1)(2y)^{2r-1}} \\ &&&> \frac1y \\ \\ \Rightarrow && \ln \left (1 + \frac{1}{y-\frac12} \right) &> \frac{1}{y} \\\Rightarrow && \ln \left (1 + \frac{1}{n} \right) &> \frac{1}{n+\frac12} \\ \Rightarrow &&(n+\tfrac12) \ln \left (1 + \frac{1}{n} \right) &> 1\\ \Rightarrow && \ln \left ( \left (1 + \frac{1}{n} \right)^{n+\tfrac12} \right) &> 1\\ \Rightarrow && \left (1 + \frac{1}{n} \right)^{n+\tfrac12} & > e \end{align*}

\item Since \(\left (1 + \frac1n \right)^n\) is both bounded above, and increasing, it must tend to some limit \(L\). \begin{align*} && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n+\frac12} \\ \Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \lim_{n \to \infty} \sqrt{1 + \frac1n} \\ \Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \\ \end{align*} And therefore equality must hold.

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Solution:

1. \(\,\)
   

   + - ↺
2. \(\,\) \begin{align*} && \frac{\d}{\d x} \left ( \frac{1}{g(x)} \right) &= \frac{\d }{\d x} \left ( \big( (x-a)^2-1 \big) \big( (x-b)^2 -1\big)\right) \\ &&&= ((x-a)^2-1)(2(x-b))+((x-b)^2-1)(2(x-a)) \\ &&&= 2(2x-a-b)(x^2-(a+b)x+ab-1) \\ \Rightarrow && x &= \frac{a+b}{2}, \frac{a+b \pm \sqrt{(a+b)^2-4ab+4}}{2} \\ &&&= \frac{a+b}{2}, \frac{a+b \pm \sqrt{(a-b)^2+4}}{2} \end{align*} If \(b > a+2\):
   

   + - ↺
   If \(b = a+2\):
   

   + - ↺

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Solution:

\begin{align*} && BD^2 &= a^2+d^2 - 2ad \cos \theta \\ && BD^2 &= b^2+c^2-2bc \cos (\pi - \theta) \\ \Rightarrow && a^2+d^2 - 2ad \cos \theta &= b^2+c^2+2bc \cos \theta \\ \Rightarrow && 2(ad+bc)\cos \theta &= a^2+d^2-b^2-c^2 \\ \Rightarrow && \cos \theta &= \frac{a^2+d^2-b^2-c^2}{2(ad+bc)} \\ \\ && Q &= \frac12 ad \sin \theta + \frac12 bc \sin (\pi - \theta) \\ &&&= \frac12 (ad+bc) \sin \theta \\ \Rightarrow && \sin \theta &= \frac{2Q}{ad+bc} \\ \\ && 1 &= \sin^2 \theta + \cos^2 \theta \\ &&&= \frac{4Q^2}{(ad+bc)^2} + \frac{(a^2+d^2-b^2-c^2)^2}{4(ad+bc)^2} \\ \Rightarrow && 4(ad+bc)^2 &= 16Q^2 + (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= 4(ad+bc)^2- (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= (2ad+2bc - a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2) \\ &&&= ((b+c)^2-(a-d)^2)((a+d)^2-(b-c)^2) \\ &&&= (b+c-a+d)(b+c+a-d)(a+d+b-c)(a+d-b+c) \\ \Rightarrow && Q^2 &= (s-a)(s-b)(s-c)(s-d) \end{align*} Since all triangles are cyclic, we can place \(D\) at the same point as \(A\) to obtain Heron's formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac12(a+b+c)\)

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Solution:

\begin{align*} && \mathbf{y}_1 &= \overrightarrow{OG}+\overrightarrow{GY_1} \\ &&&= \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3) -\lambda_1 \left (\mathbf{x}_1 - \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ && 1 &= \mathbf{y}_1 \cdot \mathbf{y}_1 \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \cdot \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac19\left ( (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\mathbf{x}_1 \cdot \mathbf{x}_2+\mathbf{x}_1 \cdot \mathbf{x}_3) + 2(1+\lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3 \right) \\ \Rightarrow && 9 &= (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\beta+\gamma) + 2(1+\lambda_1)^2 \alpha \\ &&&= 3+6\lambda_1^2+2(\beta+\gamma)-2(\beta+\gamma)\lambda_1 - 4\lambda_1^2(\beta+\gamma) + 2\alpha+4\lambda_1\alpha + 2\lambda_1^2 \alpha \\ && 0 &= (-6+2(\alpha+\beta+\gamma))+2(2\alpha-(\beta+\gamma))\lambda_1 + (6+2(\alpha-2(\beta+\gamma)))\lambda_1^2 \\ \Rightarrow && 0 &= ((\alpha+\beta+\gamma)-3)+(2\alpha-(\beta+\gamma))\lambda_1 + (3+\alpha-2(\beta+\gamma))\lambda_1^2 \\ &&&= (\lambda_1+1)((3+\alpha-2(\beta+\gamma))\lambda_1+ ((\alpha+\beta+\gamma)-3)) \\ \Rightarrow && \lambda_1 &= \frac{3-(\alpha+\beta+\gamma)}{3+\alpha-2(\beta+\gamma)} \end{align*} as required. Since \(\dfrac {GX_1}{GY_1} = \frac1{\lambda_1}\) we must have, \begin{align*} && \frac {GX_1}{GY_1} + \frac {GX_2}{GY_2} + \frac {GX_3}{GY_3} &= \frac1{\lambda_1}+\frac1{\lambda_2}+\frac1{\lambda_3} \\ &&&= \frac{(3+\alpha-2\beta-2\gamma)+(3+\beta-2\gamma-2\alpha)+3+\gamma-2\alpha-2\beta)}{3-\alpha-\beta-\gamma} \\ &&&= \frac{9-3(\alpha+\beta+\gamma)}{3-(\alpha+\beta+\gamma)} \\ &&&= 3 \end{align*}