Problems

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Solution: \(x = \log_bc\) means that \(b^x = c\) Therefore, we can write \(\frac{\log_ac}{\log_ab} = \frac{\log_ab^{x}}{\log_ab} = \frac{x \log_ab}{\log_ab} = x = \log_bc\), giving us the change of base rule. Rearranging the chance of base rule, we get \(\frac{1}{\log_bc} = \frac{\log_ab}{\log_ac}\)

1. Since \(\pi^2 < 10\), we have \begin{align*} \Leftrightarrow && \pi^2 &< 10 \\ \Leftrightarrow && 2\log_{10} \pi &< 1 \\ \Leftrightarrow && 2 &< \frac{1}{\log_{10} \pi} \\ \Leftrightarrow && 2 &< \frac{\log_{10} 2 + \log_{10} 5}{\log_{10} \pi} \\ \Leftrightarrow && 2 &< \frac{\log_{10} 2}{\log_{10} \pi} + \frac{\log_{10} 5}{\log_{10} \pi} \\ \Leftrightarrow && 2 &< \frac1{\log_2\pi}+ \frac1{\log_5\pi} \\ \end{align*}
2. Since \(e^2 < 8 \Rightarrow 2 < 3 \ln 2 \Rightarrow \ln 2 > \frac23\) \begin{align*} && \log_2 \frac{\pi}{e} &= \frac{\ln \frac{\pi}{e}}{\ln 2} \\ &&&= \frac{\ln \frac{\pi}{e}}{\ln 2} \\ &&&< \frac{3(\ln \pi - 1)}{2} \\ \Rightarrow && \frac{1}{5} &< \frac{3 \ln \pi - 1}{2} \\ \Rightarrow && \frac{2}{15} + 1 = \frac{17}{15}&< \ln \pi \end{align*}
3. From the inequalities given we can set up two linear inequalities in \(\ln 2\) and \(\ln 5\) \begin{align*} && 20 &< e^3 \\ \Rightarrow && 2\ln 2 + \ln 5 &< 3 \tag{*}\\ \\ && \frac{3}{10} &< \log_{10} 2 \\ \Rightarrow && \frac{3}{10} &< \frac{\ln 2}{\ln 10} \\ \Rightarrow && 3 \ln 2 + 3 \ln 5 &< 10 \ln 2 \\ \Rightarrow && -7 \ln 2 + 3 \ln 5 &< 0 \tag{**}\\ \\ \\ && \pi^2 & < 10 \\ \Rightarrow && 2 \ln \pi &< \ln 2 + \ln 5 \tag{***}\\ \end{align*} It would be nice to combine \((*)\) and \((**)\) to bound \(\ln 2+ \ln 5\), so we want to use a linear combination such that \begin{align*} && \begin{cases} 2x -7y &= 1 \\ x + 3y &= 1\end{cases} \\ \Rightarrow && \begin{cases} y &= \frac1{13} \\ x &= \frac{10}{13}\end{cases} \\ \\ \Rightarrow && \ln 2 + \ln 5 < \frac{30}{13} + 0 \\ \Rightarrow && \ln \pi < \frac{15}{13} \end{align*}

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Solution:

1. \begin{align*} &&\tan \beta &= \frac{y}{2a - x} \\ &&\tan \alpha &= \frac{y}{x+a} \\ && \tan \beta &= \tan 2 \alpha \\ && &= \frac{\tan \alpha}{1 - \tan^2 (\alpha)} \\ \Leftrightarrow && \frac{y}{2a-x}&= \frac{\l \frac{y}{x+a} \r}{1 - \l \frac{y}{x+a} \r^2} \\ && &= \frac{2y(x+a)}{(x+a)^2 - y^2} \\ \Leftrightarrow && (x+a)^2 - y^2 &= 2(x+a)(2a-x) \tag{\(y \neq 0\)} \\ \Leftrightarrow && x^2 + 2ax + a^2 - y^2 &= -2x^2 + 2ax - 4a^2 \\ \Leftrightarrow && y^2 &= 3(x^2-a^2) \end{align*}
2. Therefore if \(y^2 = 3(x^2-a^2)\) we know that \(\tan \beta = \tan 2\alpha\), so \(2\alpha = \beta + n \pi\). Since \(0 < \alpha + \beta < \pi\) (since they are angles in a triangle we must have that \(0 < \alpha + 2\alpha - n \pi = 3\alpha - n\pi < \pi\), so \(0 < \alpha - \frac{n\pi}{3} < \frac{\pi}3\), therefore we have \(3\) cases:

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Solution: When \((\ln x)^2 = 1\) we have \(f(x) = \frac{1}{x\ln x}(1 - 1^2)^2 = 0\) \(f'(x) = \frac{2(1 - (\ln x)^2) \cdot (-2 \ln x ) \cdot \frac1x \cdot (x \ln x) - (\ln x +1)(1-(\ln x)^2)^2}{(x\ln x)^2} = \frac{2\cdot 0 \cdot (-2 \ln x ) \cdot \frac1x \cdot (x \ln x) - (\ln x +1) \cdot 0}{(x\ln x)^2} = 0\)

1. First consider \(0 < x < 1\), so \begin{align*} && F(x) &= \int_{1/e}^x f(t) \d t \\ &&&= \int_{1/e}^x \frac{1}{t\ln t} \left(1 - (\ln t)^2 \right)^2 \d t \\ u = \ln t, \d u = \frac1t \d t: &&&= \int_{u=-1}^{u=\ln x} \frac{1}{u}(1-u^2)^2 \d u \\ &&&= \int_{-1}^{\ln x} \left ( u^3 - 2u+\frac1u \right) \d u \\ &&&= \left [ \frac{u^4}{4} - u^2+ \ln |u| \right]_{-1}^{\ln x} \\ &&&= \frac{(\ln x)^4}{4} -(\ln x)^2 + \ln |\ln x| - \frac14+1 \end{align*} Now consider \(x > 1\) \begin{align*} && F(x) &= \int_{e}^x f(t) \d t \\ &&&= \int_{e}^x \frac{1}{t\ln t} \left(1 - (\ln t)^2 \right)^2 \d t \\ u = \ln t, \d u = \frac1t \d t: &&&= \int_{u=1}^{u=\ln x} \frac{1}{u}(1-u^2)^2 \d u \\ &&&= \int_{1}^{\ln x} \left ( u^3 - 2u+\frac1u \right) \d u \\ &&&= \left [ \frac{u^4}{4} - u^2+ \ln |u| \right]_{1}^{\ln x} \\ &&&= \frac{(\ln x)^4}{4} -(\ln x)^2 + \ln| \ln x| - \frac14+1 \end{align*} Notice that \begin{align*} F(x^{-1}) &= \frac{(\ln x^{-1})^4}{4} -(\ln x^{-1})^2 + \ln| \ln x^{-1}| - \frac14+1 \\ &= \frac{(-\ln x)^4}{4} -(-\ln x)^2 + \ln| -\ln x| - \frac14+1 \\ &= \frac{(\ln x)^4}{4} -(\ln x)^2 + \ln| \ln x| - \frac14+1 \\ &= F(x) \end{align*}
2. \(\,\)
   

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Solution:

1. \(p(x) = C(x-1)(x-2)(x-3)(x-4)+1\)
2. Suppose \(P(N+1) = 1\) them we could consider \(f(x) = P(x) - 1\) to be a polynomial of degree \(N\) with at least \(N+1\) roots, which would be a contradiction. Therefore \(P(N+1) \neq 1\). Since \(P(x) = C(x-1)(x-2)\cdots(x-N) + 1\) and \(P(N+1) = 2\) we must have \(C \cdot N! + 1 = 2 \Rightarrow C = \frac{1}{N!}\), hence \(P(x) = \binom{x-1}{N} + 1\) ie \(P(N+r) = \binom{N+r-1}{N}+1\) so \(P(N+2) = \binom{N+1}{N} +1= N+2\), so we can take \(r=2\).
3. 1. Suppose consider \(p(x) = S(x) - 2001\), then \(p(x)\) has roots \(a,b,c,d\) and suppose we can find \(e\) such that \(p(e) = 17\) then we must have \((e-a)(e-b)(e-c)(e-d) = 17\) but the only possible factors of \(17\) are \(-17,-1,1,17\) and we cannot have all \(4\) of them. Hence this is not possible.
   2. Now we have \(abcd = 16\), so we can have factors \(-16,-8,-4, -2, -1, 1, 2, 4,8,16\) (and we need to have \(4\) of them). If we have \(0\) negatives, the smallest product is \(1 \cdot 2 \cdot 4 \cdot 8 > 16\) If we have \(2\) negatives we must have \(1\) and \(-1\) (otherwise we have the same problem of being too large. So \(\{-1,1,-2,8\},\{-1,1,2,-8\},\{-1,1,-4,4\},\) If we have \(4\) negatives that's the same issue as with \(0\) negatives.

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Solution: \begin{align*} && 2\sin\theta \,\big (\sin\theta + \sin 3\theta + \cdots + \sin (2n-1)\theta\,\big ) &= 2\sin\theta\sin\theta + 2\sin\theta\sin 3\theta + \cdots + 2\sin\theta\sin (2n-1)\theta \\ &&&= \cos((1-1)\theta) - \cos((1+1)\theta)+\cos((3-1)\theta)-\cos((3+1)\theta) + \cdots + \cos (((2n-1)-1)\theta) -\cos(((2n-1)+1)\theta) \\ &&&= \cos 0 - \cos(2n\theta) \\ &&&= 1 - \cos 2n \theta \end{align*}

1. \(\,\)
   

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   Therefore the area is: \begin{align*} A_n &= \frac{\pi}{n} \sin \left ( \frac{\pi}{2n} \right) + \frac{\pi}{n} \sin \left ( \frac{3\pi}{2n} \right) + \frac{\pi}{n} \sin \left ( \frac{5\pi}{2n} \right) + \cdots \frac{\pi}{n} \sin \left ( \frac{(2n-1)\pi}{2n} \right) \\ &= \frac{\pi}{n} \left( \frac{1-\cos \frac{2n \pi}{2n}}{2\sin \frac{\pi}{2n}} \right) \\ &= \frac{\pi}{n} \frac{1}{\sin \frac{\pi}{2n}} \end{align*} as required
2. 

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   Therefore the area is: \begin{align*} && B_n &= \frac{\pi}{n} \frac{\sin(0)+\sin(\frac{\pi}{n})}{2}+\frac{\pi}{n} \frac{\sin(\frac{\pi}n)+\sin(\frac{2\pi}{n})}{2} + \cdots \frac{\pi}{n} \frac{\sin(\frac{(n-1)\pi}{n})+\sin(\frac{n\pi}{n})}{2} \\ &&&= \frac{\pi}{n} \left ( \sin \frac{\pi}{n} + \sin \frac{2\pi}{n} + \cdots+\sin \frac{(n-1)\pi}{n} \right) \\ \Rightarrow && 2\sin \left ( \frac{\pi}{2n} \right)B_n &= \frac{\pi}{2} \left (2\sin \left ( \frac{\pi}{2n} \right)\sin \frac{\pi}{n} + 2\sin \left ( \frac{\pi}{2n} \right)\sin \frac{2\pi}{n} + \cdots+2\sin \left ( \frac{\pi}{2n} \right)\sin \frac{(n-1)\pi}{n} \right) \\ &&&= \frac{\pi}2 \left ( \cos \frac{\pi}{2n} - \cos \frac{3\pi}{n} + \cos\frac{3\pi}{2n} - \cos \frac{5\pi}{2n} + \cos \frac{(2n-3)\pi}{2n} - \cos \frac{(2n-1)\pi}{2n} \right) \\ &&&= \frac{\pi}{n} \left ( \cos \frac{\pi}{2n} - \cos \left ( \pi - \frac{\pi}{2n} \right) \right) \\ &&&= 2 \frac{\pi}{n} \cos \frac{\pi}{2n} \\ \Rightarrow && \sin \left ( \frac{\pi}{2n} \right)B_n &= \frac{\pi}n \cos \frac{\pi}{2n} \end{align*} as required
3. \begin{align*} \frac12(A_n+B_n) &= \frac12 \frac{\pi}{n} \frac{1}{\sin \frac{\pi}{2n}} \left ( 1 + \cos \frac{\pi}{2n} \right) \\ &= \frac{\pi}{n}\frac1{2 \sin \frac{\pi}{2n}} \left (2 \cos^2 \frac{\pi}{4n} \right) \\ &=\frac{\pi}{n} \frac{1}{4 \sin \frac{\pi}{4n} \cos \frac{\pi}{4n}} \left (2 \cos^2 \frac{\pi}{4n} \right) \\ &= \frac{\pi}{2n} \frac{\cos \frac{\pi}{4n}}{\sin \frac{\pi}{4n}} \\ &= B_{2n} \\ \\ A_nB_{2n} &= \frac{\pi}{n\sin \frac{\pi}{2n}} \cdot \frac{\pi}{2n} \frac{\cos \frac{\pi}{4n}}{\sin \frac{\pi}{4n}} \\ &= \frac{\pi^2}{(2n)^2} \frac{\cos \frac{\pi}{4n}}{\sin^2 \frac{\pi}{4n} \cos \frac{\pi}{4n}} \\ &= \left ( \frac{\pi}{2n} \frac{1}{\sin \frac{\pi}{4n}}\right)^2 \\ &= A_{2n}^2 \end{align*}

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Solution:

1. Let \(x = \frac{pz+q}{z+1}\) then \begin{align*} && 0 &= x^3-3pqx+pq(p+q) \\ &&&= \left ( \frac{pz+q}{z+1} \right)^3 - 3pq \left ( \frac{pz+q}{z+1} \right) + pq(p+q) \\ &&&= \frac{(pz+q)^3-3pq(pz+q)(z+1)^2+pq(p+q)(z+1)^3}{(z+1)^3} \\ &&&= \frac{1}{(z+1)^3} \Big ((p^3+pq(p+q)-3p^2q)z^3 + (3p^2q-6p^2q+3pq^2+3p^2q+3pq^2)z^2 + \\ &&&\qquad \qquad\quad\quad +(3pq^2-3p^2q-6pq^2+3p^2q+3qp^2)z+(q^3-3pq^2+p^2q+pq^2) \Big ) \\ &&&= \frac{(p^3+pq^2-2p^2q)z^3+(q^3+p^2q-2pq^2)}{(z+1)^3} \\ \Rightarrow && 0 &= (p^3+pq^2-2p^2q)z^3+(q^3+p^2q-2pq^2) \\ &&&= p(p-q)^2z^3 + q(p-q)^2 \\ \Rightarrow && 0 &= pz^3 + q \end{align*}
2. We would like to find \(pq = c\) and \(pq(p+q) = d\), so \(p\) and \(q\) are roots of the quadratic \(x^2-\frac{d}{c}x + c = 0\), which has distinct real roots if \(\Delta = \frac{d^2}{c^2}-4c > 0 \Rightarrow d^2>4c^3\)
3. Note that \(c = -2, d = -2\) so \begin{align*} && 0 &= x^3+6x-2 \\ \text{consider} && 0 &= X^2-X-2 \\ && &= (X+1)(X-2) \\ \Rightarrow && p = -1, &q = 2\\ \Rightarrow && 0 &= x^3-3\cdot 2 \cdot(-1) x + 2\cdot(-1) \cdot(-2+1) \\ \Rightarrow && 0 &= -z^3+2 \\ \Rightarrow && z &= \sqrt[3]{2} \\ \Rightarrow && \frac{-z+2}{z+1} &= \sqrt[3]{2} \\ \Rightarrow && -z+2 &= \sqrt[3]{2} z + \sqrt[3]{2} \\ \Rightarrow && z &= \frac{2-\sqrt[3]{2}}{\sqrt[3]{2}+1} \end{align*}
4. \(\,\) \begin{align*} && 0 &= x^3 - 3p^2x + 2p^3 \\ &&&= (x-p)(x^2+px-2p^2) \\ &&&=(x-p)^2(x+2p)\\ \Rightarrow && x &= p, p, -2p \end{align*} Therefore if we have a repeated root to our associated quadratic we can find a cubic of the form \(x^3-3p^2x+2p^3\), but we know this has roots we can find.

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Solution: \begin{questionparts} \item \begin{align*} && \dfrac{\d }{\d x} \left( \s(x)^3 + \c(x)^3 \right) &= 3\s(x)^2\s'(x) + 3\c(x)^2 \c'(x) \\ &&&= 3\s(x)^2\c(x)^2 - 3\c(x)^2\s(x)^2 \\ &&&= 0 \\ \\ \Rightarrow && \s(x)^3 + \c(x)^3 &= \text{constant} \\ &&&= \s(0)^3 + \c(0)^3 \\ &&&= 1 \end{align*} \item \begin{align*} \frac{\d }{\d x} \, \Big( \s(x) \c(x) \Big) &= \s'(x) \c(x) + \s(x)\c'(x) \\ &= \c(x)^3 - \s(x)^3 \\ &= \c(x)^3 - (1-\c(x)^3) \\ &= 2\c(x)^3 - 1 \\ \\ \dfrac{\d }{\d x} \left( \dfrac{\s(x)}{\c(x)} \right) &= \frac{\s'(x)\c(x) - \s(x)\c'(x)}{\c(x)^2} \\ &= \frac{\c(x)^3 + \s(x)^3}{\c(x)^2} \\ &= \frac{1}{\c(x)^2} \\ \end{align*} \item \begin{align*} \int \s(x)^2 \d x &= -\int -\s(x)^2 \d x \\ &= -\int \c'(x) \d x \\ &= - \s(x) +C \\ \\ \int \s(x)^5 \, \d x &= \int \s(x)^2 \s(x)^3 \d x \\ &= \int \s(x)^2 (1 - \c(x)^3) \d x \\ &= -\int \c'(x) (1 - \c(x)^3) \d x \\ &= - c(x) + \frac{\c(x)^4}{4} + C \end{align*} \item If \(u = \s(x), \frac{\d u}{\d x} = \c(x)^2\) \begin{align*} \int \frac{1}{(1-u^3)^{\frac{2}{3}}} \, \d u &= \int \frac{1}{(1-\s(x)^3)^{\frac{2}{3}}} \c(x)^2 \d x \\ &= \int 1 \d x \\ &= x + C \\ &= \s^{-1}(u) + C \\ \\ \int \frac{1}{{(1-u^3)^{\frac{4}{3}}}} \d u &= \int \frac1{(1-\s(x)^3)^{\frac43} }\c(x)^2 \d x \\ &= \int \frac1{(\c(x)^3)^{\frac43}} \c(x)^2 \d x \\ &= \int \frac1{\c(x)^2} \d x \\ &= \frac{\s(x)}{\c(x)} + C \\ &= \frac{u}{(1-u^3)^{\frac13}} + C \\ \end{align*} \begin{align*} && \int {(1-u^3)}^{\frac{1}{3}} \, \d u &= \int (1-s(x)^3)^{\frac13} c(x)^2 \d x \\ &&&= \int \c(x)^3 \d x = I\\ &&&= \int \c(x) s'(x) \d x \\ &&&= \left [\c(x) \s(x) \right] + \int \s(x)^2 s(x) \d x \\ &&&= \c(x) \s(x) + \int (1 - \c(x)^3) \d x + C \\ &&&= \c(x) \s(x) + x - I + C \\ \Rightarrow && I &= \frac{x + \c(x) \s(x)}{2} + k \\ \Rightarrow && &= \frac12 \l \s^{-1}(u) + u \sqrt[3](1-u^3)\r + k \end{align*}

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Solution: Applying conservation of energy, since there are no external forces (other than gravity) the condition to reach the house (with any speed) is the initial GPE is larger than the final GPE, ie: \begin{align*} && m g x \sin \alpha &\geq m g d \sin \beta \\ \Rightarrow && x \sin \alpha &\geq d \sin \beta \end{align*} Let \(T_1\) be the time taken on the downward section, and \(T_2\) the time taken on the upward section, then: \begin{align*} && s &= ut + \frac12 a t^2 \\ \Rightarrow && x &= \frac12 g \sin \alpha T_1^2 \\ \Rightarrow && T_1^2 &= \frac{2x}{g \sin \alpha} \\ && v &= u + at \\ \Rightarrow && v &= T_1 g \sin \alpha \\ && mg x \sin \alpha &= mg d \sin \beta + \frac12 m w^2 \\ \Rightarrow && w &= \sqrt{2(x \sin \alpha - d \sin \beta)} \\ && w &= v - g \sin \beta T_2 \\ \Rightarrow && T_2 &= \frac{v - w}{g \sin \beta} \\ \Rightarrow && T &= T_1 + T_2 \\ &&&= \sqrt{\frac{2x}{g \sin \alpha}} + \frac{\sqrt{\frac{2x}{g \sin \alpha}} g \sin \alpha- \sqrt{2(x \sin \alpha - d \sin \beta)}}{g \sin \beta} \\ &&&= \left ( \frac{2}{g \sin \alpha} \right)^{\tfrac12} \left ( \sqrt{x} + \sqrt{x}k - \sqrt{k^2x-kd}\right) \end{align*} Differentiating wrt to \(x\), we obtain: \begin{align*} && \frac{\d T}{\d x} &= C(-(1+k)x^{-1/2}+k^2(k^2 x - kd)^{-1/2}) \\ \text{set to }0: && 0 &= k^2(k^2 x - kd)^{-1/2} - (1+k)x^{-1/2} \\ \Rightarrow && \sqrt{x} k^2 &= \sqrt{k^2x - kd} (1+k) \\ \Rightarrow && x k^4 &= (k^2x-kd)(1+k)^2 \\ \Rightarrow && x(k^4-k^2(1+k)^2) &= -kd(1+k)^2 \\ \Rightarrow && x(2k^2+k) &= d \\ \Rightarrow && x &= \frac{d}{(2k^2+k)} \end{align*}

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Solution:

1. \(\,\)
   

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   \begin{align*} \text{N2}(\leftarrow, \text{first engine}): && D-T &= Ma \\ \text{N2}(\leftarrow, \text{rest of train}): && T-nR+D &= (M+nm)a \\ \Rightarrow && \frac{D-T}{M} &= \frac{T+D-nR}{M+nm} \\ \Rightarrow && T \left ( \frac{1}{M+nm}+\frac{1}{M} \right) &= \frac{D}{M} + \frac{nR-D}{M+nm} \\ \Rightarrow && T \left ( 2M+nm\right) &= DM +Dnm + nRM - DM \\ &&&= n(mD+MR) \\ \Rightarrow && T &= \frac{n(mD+MR)}{2M+nm} \end{align*}
2. The greatest coupling must occur behind an engine, because each carriage behind an engine acts as a drag. Therefore we need only consider the couple between the second engine and the rest of the carriages:
   

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   \begin{align*} \text{N2}(\leftarrow, \text{up to second engine}): && 2D - T_2 - kR &= (2M+km)a \\ \text{N2}(\leftarrow, \text{everything else}): && T_2 - (n-k)R &= (n-k)ma \\ \Rightarrow && \frac{2D-T_2-kR}{2M+km} &= \frac{T_2-(n-k)R}{(n-k)m} \\ \Rightarrow && T_2 \left (\frac{1}{(n-k)m} + \frac{1}{2M+km} \right) &= \frac{2D-kR}{2M+km} + \frac{R}{m} \\ \Rightarrow && T_2 \left (2M+ nm \right) &= (2D-kR)m(n-k) + R(2M+km)(n-k) \\ \Rightarrow && T_2 &= \frac{(n-k)\left (2Dm+2RM \right)}{2M+nm} \\ &&&= \frac{2(n-k)(mD + MR)}{2M+nm} \end{align*} Therefore \(T > T_2\) provided \(2(n-k) < n \Leftrightarrow k > \frac12n\)
3. If there is a coupling which is in negative tension, it must be between the two engines. In particular, if there is one, there must be one directly in front of the first engine.
   

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   \begin{align*} \text{N2}(\leftarrow, \text{before second engine}): && D - T_3 - kR &= (M+km)a \\ \text{N2}(\leftarrow, \text{everything else}): && T_3 +D- (n-k)R &= (M+(n-k)m)a \\ \Rightarrow && \frac{D-T_3-kR}{M+km} &= \frac{T_3+D-(n-k)R}{M+(n-k)m} \\ \Rightarrow && T_3 \left ( \frac{1}{M+(n-k)m} + \frac{1}{M+km} \right) &= \frac{D-(n-k)R}{M+(n-k)m}+\frac{kR-D}{M+km} \\ \Rightarrow && T_3 (2M+nm) &= (D-(n-k)R)(M+km)+(kR-D)(M+(n-k)m) \\ &&&= D(M+km-M-(n-k)m) + R(kM+k(n-k)m-(n-k)M-k(n-k)m) \\ &&&= D(n-2k)m+RM(2k-n)m \\ &&&= (n-2k)m(D-RM) \end{align*} Therefore \(T_3\) is negative if \(k > \frac12n\) so there are some connections in compression.

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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 \(f'(x) \in [-1, -\tfrac12]\)
   

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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*}