Problems

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Solution: \begin{align*} && \var[X+Y] &= \E\left [(X+Y-\E[X+Y])^2 \right] \\ &&&= \E \left [ (X - \E[X] + Y - \E[Y])^2 \right] \\ &&&= \E \left [(X - \E[X])^2 + (Y-\E[Y])^2 + 2(X-\E[X])(Y-\E[Y]) \right] \\ &&&= \E \left [(X - \E[X])^2 \right]+\E \left [(Y-\E[Y])^2 \right]+\E \left [2(X-\E[X])(Y-\E[Y]) \right] \\ &&&= \var[X] + \var[Y] + 2 \mathrm{Cov}(X,Y) \end{align*}

1. \(W+L+D = m\) where \(m\) is the number of games, which has variance \(0\). Therefore \(W+L+D\) has variance \(0\).
2. The probability of a decisive game is \(\frac23\) and \(W+L\) is the number of decisive games. Each game is independent so this meets the criteria for a binomial distribution.

Notice \(W+L \sim B(m, \tfrac23)\) and \(W, L, D \sim B(m, \tfrac13)\), in particular \(\var[W+L] = m \tfrac23 \tfrac13 = \tfrac29m\) and \(\var[W] = \var[D] = \var[D] = m \tfrac13 \tfrac13 = \tfrac29m\) \begin{align*} && \var[W+L] &= \var[W] + \var[L] + 2\mathrm{Cov}(W,L) \\ \Rightarrow && \mathrm{Cov}(W,L) &= -\tfrac19m \\ \Rightarrow && \frac{\mathrm{Cov}(W,L) }{\sqrt{\var[W]\var[L]}} &= -\frac12 \end{align*}

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Solution:

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   \begin{align*} && 1+2x-x^2 = 2/x \\ \Rightarrow && 0 &= x^3-2x^2-x+2 \\ &&&= (x+1)(x^2-3x+2) \\ &&&= (x+1)(x-1)(x-2) \end{align*} Therefore the inequality is satisfied on \((1,2)\) and \((-1,0)\)
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   \begin{align*} && \sqrt{3x+10} &= 2+\sqrt{x+4} \\ && 3x+10 &= x+8 + 4\sqrt{x+4} \\ && 16(x+4) &= 4(x+1)^2 \\ && 4x+16 &= x^2+2x+1 \\ \Rightarrow && 0 &= x^2-2x-15 \\ &&&= (x-5)(x+3) \end{align*} Therefore \(x > 5\)

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Solution:

Since there are two turning points the derivative (a quadratic) has two distinct real roots. \begin{align*} && f'(x) &= 3x^2-2(p+q+r)x+(pq+qr+rp) \\ && 0 &< \Delta = 4(p+q+r)^2 - 4\cdot 3(pq+qr+rp) \\ \Rightarrow && (p+q+r)^2 &> 3(pq+qr+rp) \end{align*} If \(g^2 > 4h\) then \(p(x) = (x^2+gx+h)(x-k)\) has at least 2 real roots (possibly one repeated, and in particular it has two turning point, ie \begin{align*} && p'(x) &= (2x+g)(x-k)+(x^2+gx+h) \\ &&&= 3x^2+(2g-2k)x + (h-kg) \\ && 0 &< \Delta = 4(g-k)^2 - 4\cdot 3 (h-gk) \\ \Rightarrow && (g-k)^2 &> 3(h-gk) \end{align*} Pick \(g = h = 1\) and \(k = 1000\) then \((-999)^2 > 0 > 3(1-1000)\) so it is sufficient but not necessary.

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Solution: Let \(\tan \theta = t\) \begin{align*} \tan 3 \theta &\equiv \tan (2 \theta + \theta) \\ &\equiv \frac{\tan 2 \theta +\tan \theta}{1 - \tan 2 \theta \tan \theta} \\ &\equiv \frac{\frac{2t}{1-t^2}+t}{1-\frac{2t^2}{1-t^2}} \\ &\equiv \frac{2t+t-t^3}{1-t^2-2t^2} \\ &\equiv \frac{3t-t^3}{1-3t^3} \\ &\equiv \frac{3\tan \theta - \tan^3 \theta}{1 - 3 \tan^3 \theta} \end{align*} If \(\theta = \cos^{-1} (2/\sqrt{5})\), then \(\sin \theta = 1/\sqrt{5}\) and \(\tan \theta = 1/2\). Hence \begin{align*} \tan 3 \theta &= \frac{3 \cdot \frac12 - \frac18}{1 - \frac34} \\ &= \frac{11}{2} \end{align*}

1. Since \(\tan 3 y = 11/2\) has the solution \(y = \cos^{-1} (2/\sqrt{5})\) it will also have the solutions \(y = \cos^{-1}(2/\sqrt{5}) + \frac{\pi}{3}\) and \(y = \cos^{-1}(2/\sqrt{5})+\frac{2\pi}{3}\), therefore \begin{align*} && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5})\\ \Rightarrow && x &= 2/\sqrt{5} \\ && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5}) + \frac{\pi}{3}\\ \Rightarrow && x &= \frac{2}{\sqrt{5}} \frac{1}{2} - \frac{1}{\sqrt{5}} \frac{\sqrt{3}}{2} \\ &&&= \frac{2-\sqrt{3}}{2\sqrt{5}} \\ && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5}) + \frac{2\pi}{3}\\ \Rightarrow && x &= \frac{2}{\sqrt{5}} \left (-\frac{1}{2} \right)- \frac{1}{\sqrt{5}} \frac{\sqrt{3}}{2} \\ &&&= \frac{-\sqrt{3}-2}{2\sqrt{5}} \\ \end{align*}
2. Since \(\cos \frac13 x = \frac{2}{\sqrt{5}}\) has the solution \(x = \tan^{-1} \frac{11}{2}\) it will also have the solutions \(x = \tan^{-1} \frac{11}{2} + 2n \pi\) and \(x = -\tan^{-1} \frac{11}{2} + 2n \pi\). \begin{align*} && \tan^{-1} y &= \tan^{-1} \frac{11}{2} \\ \Rightarrow && y &= \frac{11}{2} \\ && \tan^{-1} y &= \tan^{-1} \frac{11}{2} + 2n \pi \\ \Rightarrow && y &= \frac{\frac{11}{2} + 0}{1-0} \\ &&&= \frac{11}{2} \\ && \tan^{-1} y &= -\tan^{-1} \frac{11}{2} + 2n \pi \\ \Rightarrow && y &= \frac{-\frac{11}{2} + 0}{1-0} \\ &&&= -\frac{11}{2} \\ \end{align*} So our two solutions are \(y = \pm \frac{11}{2}\)

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Solution:

1. For the first one, consider \begin{align*} && \int_0^1 \frac{1}{(1+tx)^2} \d x &= \left [ -\frac{1}{t}(1+tx)^{-1} \right]_0^1 \\ &&&= \frac{1}{t} - \frac{1}{t(1+t)} \\ &&&= \frac{t+1-1}{t(t+1)} = \frac{1}{t+1} \end{align*}
2. Consider \begin{align*} && \int_0^1 \frac{-2x}{(1+tx)^3} \d x &= \int_0^1 \frac{\frac{2}{t}(1+tx) -\frac{2}{t}}{(1+tx)^3} \d x \\ &&&= -\frac{2}{t} \int_0^1 \left (\frac{1}{(1+tx)^2}- \frac{1}{(1+tx)^3} \right) \d x \\ &&&= -\frac{2}{t} \frac{1}{t+1} + \frac{2}{t} \left [ \frac{1}{-2t}(1+tx)^{-2}\right]_0^1 \\ &&&= -\frac{2}{t(t+1)} + \frac2t\left (\frac{1}{2t} - \frac{1}{2t(1+t)^2} \right) \\ &&&= -\frac{2}{t} \left ( \frac{1}{t+1} + \frac{1}{2t(1+t)^2} - \frac{1}{2t}\right) \\ &&&= -\frac{2}{t} \frac{2t(1+t)+1-(1+t)^2}{2t(1+t)^2} \\ &&&= -\frac{2}{t} \frac{2t^2+2t+1-1-2t-t^2}{2t(1+t)^2} \\ &&&= -\frac{1}{(1+t)^2} \end{align*}

I would expect it to be \(\frac{2}{(1+t)^3}\). This is actually an application of differentiating under the integral sign and is completely valid where functions are well behaved.

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Solution:

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   Notice the tangent must hit the \(y\)-axis above the origin, ie \begin{align*} && 0 &< R'(t)(0-t) + R(t) \\ \Rightarrow && R'(t) t &< R(t) \\ \Rightarrow && t &< \frac{R(t)}{R'(t)} = \frac{1}{H(t)} \end{align*}
2. Suppose \(H(t) = a/t\) then \begin{align*} && \frac{R'}{R} &= \frac{a}{t} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{a}{t} \d t \\ \Rightarrow && \ln R &= a \ln t + C \tag{t, R > 0} \\ \Rightarrow && R &= Kt^a \end{align*} Since we need \(R(t) > 0\), \(K > 0\), since \(R'(t) > 0\) we need \(a > 0\), since \(R''(t) < 0\) we need \(a(a-1) < 0\) ie \(0 < a < 1\)
3. Suppose instead \(H(t) = bt^{-2}\) then \begin{align*} && \frac{R'}{R} &= \frac{b}{t^2} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{b}{t^2} \d t \\ \Rightarrow && \ln R &= -bt^{-1} + C \tag{R > 0} \\ \Rightarrow && R &= Ke^{-b/t} \end{align*} Since \(R > 0\) we must have \(K > 0\). \begin{align*} R' > 0: && R' &= K(b/t^2)e^{-b/t} > 0 \\ \Rightarrow && b &> 0 \\ R'' < 0: && R'' &= K(b^2/t^4)e^{-b/t} -K2b/t^3 e^{-b/t} \\ &&&= Kb/t^4 (b-2t)e^{-b/t} < 0 \\ \Rightarrow && b &< 2t\\ \Rightarrow && b &< 2t \end{align*} which cannot be true for all \(t\), ie there is no \(b\) which satisfies this.

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Solution: \begin{align*} && y &= x \\ && y' &= 1 \\ && y'' &= 0 \\ \Rightarrow && 0 &= 0 + p(x) + xq(x) \tag{1} \\ \\ && y &= 1-x^2 \\ && y' &= -2x \\ && y'' &= -2 \\ \Rightarrow && 0 &= -2 -2x p(x)+(1-x^2)q(x) \tag{2}\\ \\ 2x*(1) +(2): && 2 &= (2x^2+1-x^2) q(x) \\ \Rightarrow && q(x) &= 2(1+x^2)^{-1} \\ \Rightarrow && p(x) &= -2x(1+x^2)^{-1} \tag{by (1)} \end{align*} \begin{align*} && \frac{\d^2}{\d x^2} \left (a x + b(1-x^2) \right) + p(x) \frac{\d}{\d x} \left (a x + b(1-x^2) \right)+q(x) \left (a x + b(1-x^2) \right) \\ &&= a \frac{\d^2 x}{\d x^2} + b \frac{\d^2}{\d x^2} \left ( 1- x^2 \right) + ap(x) \frac{\d x}{ \d x} + bp(x) \frac{\d }{\d x} \left ( 1- x^2 \right) + aq(x) x + bq(x)(1-x^2) \\ &&= a \left (\frac{\d^2 x}{\d x^2}+ p(x) \frac{\d x}{ \d x} +q(x)x\right)+b \left ( \frac{\d^2}{\d x^2} \left ( 1- x^2 \right)+ p(x) \frac{\d }{\d x} \left ( 1- x^2 \right)+q(x)(1-x^2)\right) &= 0 \end{align*} \begin{align*} && y &= \cos^2(\tfrac12 x^2) = \frac12 \left (1 + \cos(x^2) \right) \\ && y' &= -x \sin(x^2) \\ && y'' &= -2x^2 \cos(x^2)-\sin(x^2) \\ \Rightarrow && 0 &= -2x^2 \cos(x^2)-\sin(x^2)+p(x)(-x \sin(x^2)) +\frac12 \left (1 + \cos(x^2) \right)q(x) \\ \Rightarrow && 2x^2\cos(x^2)+\sin(x^2) &= -x \sin(x^2) p(x) + \frac12(1 + \cos(x^2)) q(x) \tag{3}\\ \\ && y &= \sin^2(\tfrac12 x^2) = \frac12 \left ( 1 - \cos (x^2) \right) \\ && y' &= x\sin(x^2) \\ && y'' &= 2x^2 \cos(x^2)+\sin(x^2) \\ \Rightarrow && 0 &= 2x^2 \cos(x^2)+\sin(x^2) +p(x) x \sin(x^2) + \frac12 \left ( 1 - \cos (x^2) \right)q(x)\\ \Rightarrow && -2x^2 \cos(x^2)-\sin(x^2) &= p(x) x \sin(x^2) + \frac12 \left ( 1 - \cos (x^2) \right)q(x) \tag{4}\\ (3)+(4): && 0 &= q(x) \\ \Rightarrow && p(x) &= -\frac{2x^2 \cos(x^2)+\sin(x^2)}{x \sin(x^2)} \end{align*}

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Solution: The position of the ship is \(\mathbf{s} = \binom{20t}{1}\). Suppose the interception is at \(T\), then the ship leaves at \(T-\frac1{12}\underbrace{\sqrt{400T^2+1}}_{\text{distance to intercept}}\). We wish to maximise this, ie \begin{align*} && \frac{\d}{\d T} \left ( T - \frac1{12}\sqrt{400T^2+1}\right) &= 1 - \frac{1}{12} \cdot \frac12 \cdot 400 \cdot 2T \cdot \left (400T^2+1 \right)^{-1/2} \\ &&&= 1 - \frac{100}3 T(400T^2+1)^{-1/2} \\ \Rightarrow && \frac{T}{\sqrt{400T^2+1}} &= \frac{3}{100} \\ \Rightarrow && \frac{T^2}{400T^2+1} &= \frac{9}{10000} \\ \Rightarrow && 10000T^2 &= 3600T^2+9 \\ \Rightarrow && 6400T^2 &= 9 \\ \Rightarrow && T &= \pm \frac{3}{80} \quad \text{(T > 0)} \end{align*} Therefore the distance is \(\sqrt{400 \frac{9}{6400} + 1} = \sqrt{\frac{9}{16}+1} = \frac{5}{4} = 1.25 \text{ km}\)

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Solution: If we fire the gun at an angle to the track, as long as we can travel a horizontal distance \(\geq d\) we can hit the track. Suppose we am at an elevatation \(\alpha\), then \begin{align*} (\uparrow): && s &= ut + \frac12 at^2 \\ && 0 &= v\sin \alpha T - \frac12 g T^2 \\ \Rightarrow &&T &= \frac{2v\sin \alpha}{g}\\ \\ (\rightarrow): && s &= ut \\ && s &= v \cos \alpha T \\ &&&= v\sqrt{1-\sin^2 \alpha} T \\ &&&= vT\sqrt{1 - \frac{g^2T^2}{4v^2}}\\ &&&= \frac{T}{2}\sqrt{4v^2-g^2T^2}\\ \Rightarrow && d & \leq \frac{T}{2}\sqrt{4v^2-g^2T^2} \\ \Rightarrow && 4g^2d^2&\leq g^2T^2(4v^2-g^2T^2) \\ \Rightarrow && 0 &\leq -(g^2T^2)^2 + 4v^2 (g^2T^2)-4g^2d^2 \\ &&&=4v^4-4g^2d^2 -\left (g^2T^2-2v^2 \right)^2 \\ \Rightarrow && \left (g^2T^2-2v^2 \right)^2 & \leq 4v^4-4g^2d^2 \\ \Rightarrow && g^2T^2 &\leq 2v^2+2\sqrt{v^4-g^2d^2} \end{align*} Therefore the maximum value for \(g^2T^2\) is \(2v^2+2\sqrt{v^4-g^2d^2}\) Notice that we are hitting the track directly at \(d\). This is because to maximise the time of flight (for a fixed speed) we want to maximise the angle of elevation. Therefore we want the highest angle where we still hit the track (which is clearly the shortest distance). If we are constraint to \(\alpha \leq 45^\circ\) we know that \(T\) is maximised when \(\alpha = 45^\circ\) (and we will reach the track since the range \(\frac{v^2 \sin 2 \alpha}{g}\) is increasing). Therefore the maximum time is \(T = \frac{\sqrt{2}v}{g}\)