Problems

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Solution: Consider \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\), then we know that \begin{align*} && \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= \sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2} \cos \theta \\ \Rightarrow && (ax+by+cz)^2 &= (a^2+b^2+c^2)(x^2+y^2+z^2) \cos^2 \theta \\ &&&\leq (a^2+b^2+c^2)(x^2+y^2+z^2) \end{align*} For equality to hold, we must have that the vectors are parallel, ie \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)

1. By applying our inequality from the first part with \(a=1, b = 2, c=2\) we have \((x+2y+2z)^2 \leq (1+2^2+2^2)(x^2+y^2+z^2) = 9(x^2+y^2+z^2)\)
2. Since \begin{align*} && (p^2+(2q)^2+(3r)^2)\left (8^2 +4^2+1^2 \right) &\geq (8p+8q+3r)^2 \\ \Leftrightarrow && 729 \cdot 81 &\geq 243^2 \\ &&3^6 \cdot 3^4 &\geq 3^{10} \end{align*} Therefore we must be in the equality case, ie \(p = 8\lambda, 2q = 4\lambda, 3r = \lambda\) as well as \(64\lambda + 16\lambda +\lambda = 243 \Rightarrow 81\lambda = 243 \Rightarrow \lambda = 3\) so we have \[ (p,q,r) = \left (24, 6, 1 \right) \]

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Solution:

1. After the first collision, it takes \(B\), \(\frac{d}{v_B} = \frac{d}{u} \frac{2(k+1)}{3} = \frac{4d}{3u}\) to collide with \(C\). During which time \(B\) and \(A\) have been moving apart with speed \(\frac12u\) and so are a distance \(\frac{2d}{3}\) apart. After the second collision, \(w_B = \frac{3(4\cdot 1 - 3)}{8(1+1)(1+3)}u = \frac{3}{64}u\) and \(v_A = \frac{1}{4}u\) so they are moving together at speed \(\frac{16-3}{64}u = \frac{13}{64}u\). It will take them \(\frac{2d}{3} \div \frac{13}{64}u = \frac{128d}{3 \times 13u}\) to do this for a total time of \(\frac{128d}{3 \times 13u} + \frac{4d}{3u} = \frac{(128+52)d}{3 \times 13 u} = \frac{60d}{13u}\)

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Solution:

1. \(\,\) \begin{align*} && 1 &= \frac{x}{x-a} + \frac{x}{x-b} \\ \Leftrightarrow && (x-a)(x-b) &= x(2x-a-b) \\ \Leftrightarrow && 0 &= x^2-ab \end{align*} Therefore if \(a,b\) are both positive or both negative, \(ab > 0\) and there are two distinct solutions \(x = \pm \sqrt{ab}\)
2. \(\,\) \begin{align*} && 1+c &= \frac{x}{x-a} + \frac{x}{x-b} \\ \Leftrightarrow && (1+c)(x-a)(x-b) &= x(2x-a-b) \\ \Leftrightarrow && 0 &= (c-1)x^2-c(a+b)x+ab(1+c) \\ \\ && 0 &= \Delta = c^2(a+b)^2 - 4 \cdot(c-1)\cdot ab(1+c) \\ &&&= c^2(a+b)^2-4ab(c^2-1) \\ &&&= c^2 ((a+b)^2-4ab)+4ab \\ &&&= c^2(a-b)^2+4ab \\ \Rightarrow && c^2 &= -\frac{4ab}{(a-b)^2} \\ &&&= -\frac{(a+b)^2-(a-b)^2}{(a-b)^2} \\ &&&= 1 - \left ( \frac{a+b}{a-b} \right)^2 \end{align*} Note that \(c^2 \geq 0\) and \(1-x^2 \leq 1\) so \(0 \leq c^2 \leq 1\). \(c^2 = 0 \Rightarrow ab = 0\), but this is not possible since \(a,b \neq 0\), therefore \(0 < c^2 \leq 1\)

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Solution:

1. \(\,\)
   

   + - ↺
   The overlap can be at most 0.4, which would mean \(p =0.7-0.4 = 0.3\) It must be at least 0.1, which would mean \(p =0.7-0.1 = 0.6\) so \(0.3 \leq p \leq 0.6\)
2. 

   + - ↺
   Notice that: \begin{align*} && 1 &= 0.05 + 0.7 -(hc+hr+0.1) + \\ &&&\quad\quad 0.8 - (hc+cr + 0.1) + \\ &&&\quad \quad \quad 0.4 - (hr+cr+0.1) +\\ &&&\quad \quad \quad \quad hc+hr+cr+0.1 \\ && &= 0.05 +0.7+0.8+0.4 - (hc+hr+cr)-2\cdot 0.1 \\ \Rightarrow && hc+hr+cr &=0.05 +0.7 + 0.8 + 0.4 - 0.2-1 \\ \Rightarrow && \mathbb{P}(\text{exactly 2}) &= 0.75 \end{align*} Notice \(q = \frac{hc}{0.8}\) Notice that we must have: \(hc, hr cr \geq 0\) as well as \(hc+hr+cr = 0.75\) \begin{align*} && \mathbb{P}(\text{only hat}) &= 0.7 -(hc+hr+0.1) \geq 0 \\ \Rightarrow && hc+hr & \leq 0.6 \\ && \mathbb{P}(\text{only cloak}) &= 0.8 - (hc+cr + 0.1)\geq 0 \\ \Rightarrow &&hc+cr & \leq 0.7 \\ && \mathbb{P}(\text{only ring}) &= 0.4 - (hr+cr+0.1) \geq 0 \\ \Rightarrow && hc+hr & \leq 0.3 \\ \end{align*} To find the minimum for \(hc\) we want to maximise \(hr+cr = 0.3\), so \(hc = 0.75 - 0.3 = 0.45\). To find the maximum for \(hc\) we want to minimise \(hr\) and \(cr\) \(cr \leq 0.7 - hc\) and \(hr \leq 0.6 - hc\) so \(0.75 \leq hc + (0.6 - hc) + (0.7 - hc) = 1.3-hc\) so \(hc \leq 1.3 - 0.75 = 0.55\) Therefore the range for \(q\) is \(\frac{.45}{.8}\) to \(\frac{.55}{.8}\) or \(\frac9{16} \leq q \leq \frac{11}{16}\)

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Solution:

Since \(y' = \cos x - \cos x + x \sin x = x \sin x > 0\) which is positive on \((0, \frac{\pi}{2})\), so \(y\) is increasing, and therefore will achieve it's highest value at \(\frac{\pi}{2}\) which is \(y(\frac{\pi}{2}) = 1\) and it's smallest value at \(y(0) = 0\). Therefore \(0 \leq y \leq 1\)

1. \(\,\) \begin{align*} \int_0^{\frac{1}{2}\pi}\,y\;\d x &= \int_0^{\frac{1}{2}\pi} (\sin x - x \cos x) \d x \\ &= \left [-\cos x \right]_0^{\frac{1}{2}\pi} +\left [ -x \sin x \right]_0^{\frac{1}{2}\pi} + \int_0^{\frac{1}{2}\pi} \sin x \d x \\ &= 1-\frac{\pi}{2} + 1 = 2 - \frac{\pi}{2} \end{align*}
2. \(\,\) \begin{align*} \int_0^{\frac{1}{2}\pi}y^2\d x &= \int_0^{\frac{1}{2}\pi} (\sin x - x \cos x)^2 \d x \\ &= \int_0^{\frac{1}{2}\pi} (\sin^2x - 2x\sin x \cos x+x^2\cos^2 x) \d x\\ &= \int_0^{\frac{1}{2}\pi} (\sin^2x -x \sin 2x+\tfrac12x^2(\cos 2 x + 1)) \d x\\ &= \frac{\pi}{4} + \frac{\pi^3}{48} + \int_0^{\frac{1}{2}\pi} (-x \sin 2x+\tfrac12x^2\cos 2 x) \d x \\ &= \frac{\pi}{4} + \frac{\pi^3}{48} + \left [\frac12 x \cos 2x +\frac14 x^2 \sin2x\right]_0^{\frac{1}{2}\pi}-\int_0^{\frac{1}{2}\pi}(\tfrac12 \cos 2x +\tfrac12 x \sin 2x) \d x\\ &= \frac{\pi}{4} + \frac{\pi^3}{48} - \frac{\pi}{4} - \left [ \frac14 \sin 2x \right]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} \tfrac12 x \sin 2x \d x\\ &= \frac{\pi^3}{48} - \left( \left[ -\frac14 x \cos 2x \right]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} -\frac14 \cos 2x \d x \right)\\ &= \frac{\pi^3}{48} - \left( \frac{\pi}{8} + \left[ \frac18 \sin 2x \right]_0^{\frac{1}{2}\pi} \right)\\ &= \frac{\pi^3}{48} - \frac{\pi}{8} \end{align*}

Since \(y^2 < y\) on this interval, we must have \( \frac{\pi^3}{48} - \frac{\pi}{8} < 2 - \frac{\pi}{2} \Rightarrow \pi^3 +18\pi < 96\) as required.

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Solution:

1. \(P\) is travelling in a unit circle about the origin in the \(\mathbf{i}-\mathbf{j}\) plane. \(Q\) is travelling in a circle (also about the origin, but in a different plane with radius \(3\)).
2. \(\,\) \begin{align*} && \mathbf{p}\cdot \mathbf{q} &= |\mathbf{p}||\mathbf{q}| \cos \theta \\ \Rightarrow && \cos \theta &= \frac{\tfrac32\cos t \cos(t + \tfrac{\pi}4)+3\sin t \sin (t + \tfrac{\pi}{4})}{3} \\ &&&= \tfrac12\cos t \cos(t + \tfrac{\pi}4)+\sin t \sin (t + \tfrac{\pi}{4}) \\ &&&= \tfrac14 (\cos (2t + \tfrac{\pi}{4}) + \cos(\tfrac{\pi}{4} ))+\tfrac12(\cos(\tfrac{\pi}{4})-\cos(2t + \tfrac{\pi}{4})) \\ &&&= \tfrac{3\sqrt{2}}8 - \tfrac14 \cos ( 2t +\tfrac{\pi}{4}) \end{align*}
3. If \(\theta \geq \frac14\pi\), then \(\cos \theta \leq \frac{\sqrt{2}}2\) \begin{align*} && \frac{\sqrt{2}}2 & \geq \frac{3\sqrt{2}}8 - \frac14 \cos ( 2t +\tfrac{\pi}{4}) \\ \Rightarrow && \frac{\sqrt{2}}2 &\geq -\cos(2t + \tfrac{\pi}{4}) \\ \Rightarrow && \cos(2t + \tfrac{\pi}{4}) &\geq -\frac{1}{\sqrt{2}} \\ \Rightarrow && 2t + \tfrac{\pi}{4} &\not\in (\tfrac{3\pi}{4},\tfrac{5\pi}{4}) \cup (\tfrac{11\pi}{4},\tfrac{13\pi}{4}) \\ \Rightarrow && t &\not\in (\tfrac{\pi}{4}, \tfrac{\pi}{2})\cup (\tfrac{5\pi}{4}, \tfrac{3\pi}{2}) \end{align*} which is is a time of \(\frac{\pi}{2}\), therefore the left over time is \(\frac32\pi\)

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Solution: \begin{align*} && \sin A &= \cos B \\ \Leftrightarrow && 0 &= \sin A - \cos B \\ &&&= \sin A - \sin ( \frac{\pi}{2} - B) \\ &&&= 2 \sin \left ( \frac{A + B - \frac{\pi}{2}}{2} \right) \cos \left (\frac{A - B + \frac\pi2}{2} \right) \\ \Leftrightarrow && n \pi &= \frac{A+B - \frac{\pi}{2}}{2}, n\pi + \frac{\pi}{2} = \frac{A-B+\frac{\pi}{2}}{2} \\ \Leftrightarrow && A \pm B &= 2n\pi + \frac{\pi}{2} \\ &&&= (4n+1) \frac{\pi}{2} \end{align*} \begin{align*} |\sin x \pm \cos x| &= | \sqrt{2} \sin(x \pm \frac{\pi}{4} )| \\ & \leq \sqrt{2} \end{align*} Therefore if \(\sin(\sin x) = \cos (\cos x)\) we must have that \(|\sin x \pm \cos x| = |(4n+1) \frac{\pi}{2}| \geq \frac{\pi}{2} > 1.5 > \sqrt{2}\) contradiction.

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Solution: \begin{align*} && f(x) &= \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \\ \Rightarrow && 1 &= \int_0^{\infty} f(x) \d x \\ &&&= \int_0^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\ &&&= Ck^{a+1} \int_0^{\infty} \frac{x^a}{(x+k)^{2a+2} }\d x \\ &&&= Ck^{a+1} \frac{a!(2a-a)!}{(2a+1)!k^{2a-a+1}} \\ &&&= C \frac{a!a!}{(2a+1)!} \\ \Rightarrow && C &= \frac{(2a+1)!}{a!a!} \end{align*} \begin{align*} && I &= \int_0^v \frac{x^a}{(x+k)^{2a+2}} \d x\\ u = k^2/x, \d x = -k^2u^{-2} \d u: &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a}u^{-a}}{(k^2u^{-1} +k)^{2a+2}}(-k^2u^{-2}) \d u \\ &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a-2a-2}u^{2a+2-a}}{(k +u)^{2a+2}}(-k^2u^{-2}) \d u \\ &&&= \int_{ k^2/v}^{\infty} \frac{u^{a}}{(k +u)^{2a+2}} \d u \\ \end{align*} At the median we want a value \(M\) such that \(M = k^2/M\) ie \(M = k\) \begin{align*} && \mathbb{E}(V) &= \int_0^{\infty} x f(x) \d x \\ &&&= \frac{(2a+1)!k^{a+1}}{a!a!} \int_0^{\infty} \frac{x^{a+1}}{(x+k)^{2a+2}} \d x \\ &&&= \frac{(2a+1)!k^{a+1}}{a!a!} \frac{(a+1)!(2a-(a+1))!}{(2a+1)!k^{2a-(a+1)+1}}\\ &&&= \frac{k^{a+1}}{a!} \frac{(a+1)(a-1)!}{k^{a}} \\ &&&= \frac{k(a+1)}{a} = \frac{a+1}a k \end{align*} \begin{align*} && \mathbb{P}(T < t) &= \mathbb{P}(\frac{s}{V} < t) \\ &&&= \mathbb{P}(V > \frac{s}{t}) \\ &&&= \int_{s/t}^{\infty} f(x) \d x \\ &&&= \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\ \\ \Rightarrow && f_T(t) &= \frac{\d}{\d t} \left ( \mathbb{P}(T < t)\right) \\ &&&= \frac{\d}{\d t} \left ( \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \right) \\ &&&= - \frac{C \, k^{a+1} \, \left ( \frac{s}{t} \right)^a}{(\frac{s}{t}+k)^{2a+2}} \cdot \left (-\frac{s}{t^2} \right) \\ &&&= \frac{Ck^{a+1}s^{a+1}t^{2a+2}}{t^{a+2}(s+kt)^{2a+2}} \\ &&&= \frac{C(ks)^{a+1}t^a}{(s+kt)^{2a+2}} \\ &&&= \frac{C(\frac{s}{k})^{a+1}t^a}{(\frac{s}{k}+t)^{2a+2}} \end{align*} Therefore \(T\) follows the same distribution, but with parameter \(s/k\) rather than \(k\). In particular it has median \(s/k\) (and the product of the medians is \(s\)). However, the product of the expected time and expected speed is \(\frac{a+1}{a} k \frac{a+1}{a} \frac{s}{k} = \left ( \frac{a+1}{a} \right)^2s > s\)

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Solution:

1. Suppose \(x_n> 3\) then \begin{align*} && x_{n+1} &= \frac{x_n^2+9-3}{5} \\ &&& \geq \frac{2\sqrt{x_n^2 \cdot 9} - 3}{5} \\ &&&= \frac{6x_n -3}{5} = x_n + \frac{x_n-3}{5} \\ &&&> x_n > 3 \end{align*} Therefore if \(x_i > 3 \Rightarrow x_{i+1} > x_i\) and \(x_{i+1} > 3\) so by induction \(x_n\) strictly increasing for all \(n\).
2. Suppose \(2 < y_n < 3\) then \begin{align*} && y_{n+1} &= 5 - \frac6{y_n} \\ &&&< 5 - \frac63 = 3 \\ \\ && y_{n+1} &= 5 - \frac4{y_n} - \frac{2}{y_n} \\ \\ &&&= y_n + 5 - \frac{2}{y_n} - \left ( y_n + \frac4{y_n} \right) \\ &&&\geq y_n + 5 - \frac{2}{y_n} - 2\sqrt{y_n \frac{4}{y_n}} \\ &&&= y_n + 1 - \frac{2}{y_n} \\ &&&> y_n \end{align*} Therefore if \(y_n \in (2,3)\) we have \(y_{n+1} \in ( y_n, 3)\) and so \(y_n\) is strictly increasing and bounded.

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Solution: \begin{align*} && v^2 &= u^2 + 2as \\ (\uparrow): && 0 &= (u \sin \theta)^2 - 2gh \\ \Rightarrow && h &= \frac{u^2 \sin^2 \theta}{2g} \end{align*} To avoid hitting the ceiling \begin{align*} && \frac9{10}d &>\frac{u^2 \sin^2 \theta}{2g} \\ \Rightarrow && \frac{gd}{u^2}&>\frac{5\sin^2 \theta}{9} \\ \end{align*} In order to pass through \(P\) we need \begin{align*} && d &= u \cos \theta t \\ && \frac12 d &= u \sin \theta t - \frac12 g t^2 \\ \Rightarrow && \frac12 &= \tan \theta - \frac12 \frac{g d}{u^2} \sec^2 \theta \\ &&&<\tan \theta - \frac12 \frac{5 \sin^2 \theta}{9\cos^2 \theta} \\ \Rightarrow && 0 & >5 \tan^2 \theta -18\tan \theta +9 \\ &&&= (5\tan \theta - 3)(\tan \theta - 3) \\ \\ \Rightarrow && \tan \theta &\in \left ( \frac35, 3\right) \\ && \theta &= \left (\arctan \tfrac35, \arctan 3 \right) \end{align*}

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Solution:

1. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} + \sqrt{x} -2x-1 \\ \Rightarrow && 2x+1 &= \sqrt{3x^2+1} + \sqrt{x} \\ \Rightarrow && 4x^2+4x+1 &= 3x^2+1+x+2\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2(x^2+6x+9) &= 4x(3x^2+1) \\ \Rightarrow && 0 &= x^4-6x^3+9x^2-4x \\ &&&= x(x-4)(x-1)^2 \end{align*} So clearly we have \(x = 0, x = 1, x = 4\). \(x = 0\) works, \(x = 1\) works, \(x = 4\) works.
2. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} - 2\sqrt{x} +x-1 \\ \Rightarrow && 1-x &= \sqrt{3x^2+1}-2\sqrt{x} \\ \Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\ \Rightarrow && 0 &= x(x-4)(x-1)^2 \end{align*} Again we must check \(x = 0, x = 1, x = 4\). \(x = 0,1\) work, but \(x = 4\) is not a solution.
3. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} - 2\sqrt{x} -x+1 \\ \Rightarrow && x-1 &= \sqrt{3x^2+1} - 2\sqrt{x} \\ \Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)}\\ \Rightarrow && 0 &= x(x-4)(x-1)^2 \end{align*} So again, we need to check \(x = 0, 1, 4\). \(x = 0, 4\) work, but \(x = 1\) fails.