Problems

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Solution: \begin{align*} && b &= c - ma \\ \Rightarrow && c &= b+ma \\ \Rightarrow && y &= m(a-x)+b \\ \Rightarrow && q &= ma+b \\ && p &= \frac{ma+b}{m} \\ \\ && d &= p+q \\ &&&= a + \frac{b}{m} + ma + b \\ \Rightarrow && d' &= -bm^{-2}+a \\ \Rightarrow && m &= \sqrt{b/a} \\ \\ \Rightarrow &&d &= a + \sqrt{ba}+\sqrt{ba} + b \\ &&&= (\sqrt{a}+\sqrt{b})^2 \\ \\ && |PQ|^2 &= p^2 + q^2 \\ &&&= a^2 + \frac{2ab}{m} + \frac{b^2}{m^2} + m^2a^2 + 2mab + b^2 \\ &&&= a^2+b^2 + \frac{b^2}{m^2} + \frac{2ab}{m}+ 2abm + a^2m^2 \\ && \frac{\d}{\d m}&= -2b^2m^{-3}-2abm^{-2}+2ab + 2a^2m \\ && 0 &=2a^2m^4+2abm^3-2abm-2b^2 \\ &&&= 2(am^3-b)(am+b) \\ \Rightarrow && m &= \sqrt[3]{\frac{b}{a}} \\ \\ &&|PQ|^2 &= \left[ a^{1/3}(a^{2/3} + b^{2/3}) \right]^2 + \left[ b^{1/3}(a^{2/3} + b^{2/3}) \right]^2 \\ &&&= a^{2/3}(a^{2/3} + b^{2/3})^2 + b^{2/3}(a^{2/3} + b^{2/3})^2 \\ &&&= (a^{2/3} + b^{2/3})^2 \cdot (a^{2/3} + b^{2/3}) \\ &&&= (a^{2/3} + b^{2/3})^3 \\ \Rightarrow && |PQ| &= (a^{2/3} + b^{2/3})^{3/2} \end{align*} We can also do this with AM-GM instead: \begin{align*} && d &= a + b + \frac{b}{m} + am \\ &&&\geq a+b + 2 \sqrt{\frac{b}{m} \cdot am} \\ &&&= a+2\sqrt{ab}+b \\ \\ && |PQ|^2 &= a^2+b^2 + \frac{b^2}{m^2} + \frac{2ab}{m}+ 2abm + a^2m^2 \\ &&&= a^2+b^2 + \frac{b^2}{m} + abm + abm + a^2m^2 + \frac{ab}{m} + \frac{ab}{m} \\ &&&= a^2+b^2 + 3\sqrt[3]{ \frac{b^2}{m} \cdot abm \cdot abm} + 3 \sqrt[3]{ a^2m^2 \cdot \frac{ab}{m} \cdot \frac{ab}{m} } \\ &&&= a^2 + 3b^{4/3}a^{2/3}+3b^{2/3}a^{4/3}+b^2 \\ &&&= (a^{2/3}+b^{2/3})^3 \end{align*}

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Solution:

1. \(\,\)
   

   + - ↺
   1. \(n = 0\) if \(b > 9\)
   2. \(n = 1\) is not possible, since by symmetry if \(x\) is a root, so is \(-x\), and \(0\) can never be the only root.
   3. \(n = 2\) if \(b < 0\) or \(b = 9\)
   4. \(n = 3\) if \(b = 0\)
   5. \(n = 4\) if \(0 < b < 9\)
2. \(\,\) \begin{align*} && y' &= 4x^3-12x+a \\ && y'' &= 12x^2-12 \\ \Rightarrow && x &= \pm 1 \\ \Rightarrow && 0 &= 4(\pm 1) - 12 (\pm 1) + a \\ &&&= a \mp 8 \\ \Rightarrow && a &= \pm 8 \end{align*} When \(a = 8\), we have \(y = x^4-6x^2+8x\) and \begin{align*} &&y' &= 4x^3-12x+8 \\ &&&= 4(x^3-3x+2) \\ &&&= 4(x-1)^2(x+2) \\ \Rightarrow && y(1) &= 3\\ && y(-2) &= -24 \end{align*}
   

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   Therefore there are no solutions if \(b > 24\), one solution if \(b = 24\) and two solutions otherwise. Similarly, if \(a = -8\), we have \(y = x^4 - 6x^2-8x\) \begin{align*} && y' &= 4x^3-12x-8 \\ &&&= 4(x^3-3x-2) \\ &&&= 4(x-2)(x+1)^2 \end{align*} So we have stationary points at \(x = 2\) and \(x = -1\) (which is also a inflection point) and at \(x = 2\) \(y = -24\), so we have the same story: there are no solutions if \(b > 24\), one solution if \(b = 24\) and two solutions otherwise.
3. \(\,\)
   

   + - ↺

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Solution:

1. \(\,\)
   

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   The area under the blue curve is \(1-\cos b\). The area under the green line is \(\frac12 b \sin b\) The area under the red line is \(\frac12 b^2\) Therefore \(\frac12 b \sin b < 1- \cos b < \frac12 b^2 \Rightarrow 1- \frac12 b^2 < \cos b < 1 - \frac12 b \sin b\)
2. \(\,\)
   

   + - ↺
   \begin{align*} &&\text{Area under blue curve}: &= \int_0^1 a^x \d x\\ &&&= \left [ \frac{1}{\ln a}e^{x \ln a} \right]_0^1 \\ &&&= \frac{a-1}{\ln a} \\ \\ &&\text{Area under green line}: &=\frac12 \cdot 1 \cdot (a + 1)\\ &&&= \frac{a+1}{2} \\ \\ &&\text{Area under tangent}: &=\frac12 \cdot 1 \cdot (1+\ln a + 1)\\ &&&= \frac{\ln a+2}{2} \\ \\ \Rightarrow && \frac{a+1}{2} & > \frac{a-1}{\ln a} \\ \Rightarrow && \ln a& > \frac{2(a-1)}{a+1} \\ \\ \Rightarrow && \frac{a-1}{\ln a} &> \frac{\ln a +2}{2} \\ \Rightarrow && 2(a-1) -2\ln a - (\ln a)^2 &> 0 \\ \Rightarrow && \ln a & < -1 + \sqrt{2a-1} \end{align*}

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Solution: \begin{align*} && \frac{\d y}{\d x} &= -\frac1{2x^2} \\ \Rightarrow && \frac{y - \frac{1}{2p}}{x - p} &= - \frac{1}{2p^2} \\ \Rightarrow && y - \frac1{2p} &= -\frac{1}{2p^2}x +\frac1{2p} \\ \Rightarrow && y +\frac{1}{2p^2}x &= \frac1p \\ \Rightarrow && 2p^2 y + x &= 2p\\ \Rightarrow && 2q^2 y + x &= 2q \\ \Rightarrow && (p^2-q^2)y &= p-q \\ \Rightarrow && y &= \frac{1}{p+q} \\ && x &= \frac{2pq}{p+q} \end{align*} \begin{align*} \text{normal}: && \frac{y-\frac1{2p}}{x-p} &= 2p^2 \\ \Rightarrow && y - \frac1{2p} &= 2p^2x - 2p^3 \\ \Rightarrow && 2py -4p^3x &= 1-4p^4 \\ \Rightarrow && 2qy -4q^3x &= 1-4q^4 \\ pq = \tfrac12: && y - 2p^2 x &= q-2p^3 \\ && y - 2q^2 x &= p-2q^3 \\ \Rightarrow && (2q^2-2p^2)x &= q-p +2q^3-2p^3 \\ &&&= (q-p)(q+p+2q^2+1+2p^2) \\ \Rightarrow && x &= \frac{1+2(p^2+q^2)+1}{2(p+q)} \\ &&&= \frac{1+2(p^2+q^2+2pq-1)+1}{2(p+q)} \\ &&&= p+q\\ && y &= 2p^2 \left ( p+q \right) + q - 2p^3 \\ &&&= p+q \end{align*} So \(N(p+q, p+q)\) and \(T\left (\frac{1}{p+q}, \frac{1}{p+q} \right)\), so both points lie on \(y = x\). \[ OT \cdot ON = \frac{\sqrt{2}}{p+q} \cdot (p+q)\sqrt{2} = 2 \] which is clearly constant.

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Solution: \begin{align*} &&\int_0^{\frac14\pi} \sin (2x) \ln(\cos x)\, \d x &= \int_0^{\frac14 \pi} 2 \sin x \cos x \ln (\cos x) \d x \\ u = \cos \theta :&&&= \int_{u=1}^{u=\frac1{\sqrt2}} -2u \ln u \d u \\ &&&= \int_{\frac1{\sqrt{2}}}^1 2u \ln u \d u \\ &&&= \left [u^2 \ln u \right]_{\frac1{\sqrt{2}}}^1-\int_{\frac1{\sqrt{2}}}^1 u \d u \\ &&&= -\frac12 \ln \frac{1}{\sqrt{2}} - \l\frac12 - \frac14 \r \\ &&&= \frac14 (\ln 2 - 1) \end{align*} \begin{align*} && \int_0^{\frac14\pi} \cos (2x) \ln(\cos x)\, \d x &= \left [ \frac12 \sin 2x \ln (\cos x) \right]_0^{\frac14\pi}- \int_0^{\frac14\pi} \frac12 \sin 2x \frac{-\sin x}{\cos x} \d x \\ &&&=\frac12 \ln \frac{1}{\sqrt{2}}+\int_0^{\frac14\pi} \sin^2 x \d x \\ &&&= -\frac14 \ln 2 + \int_0^{\frac14\pi} \frac{1-\cos 2x }{2} \d x \\ &&&= -\frac14 \ln 2 +\frac{\pi}{8} -\frac{1}{4} \\ &&&= \frac18 (\pi - 2\ln 2 - 2) \\ &&&= \frac18 (\pi - \ln 4 - 2) \\ \end{align*} Notice that \(\cos x + \sin x = \sqrt{2} \cos (x -\frac{\pi}{4})\), so: \begin{align*} &&\int_{\frac14\pi}^{\frac12\pi} \big ( \cos(2x) + \sin (2x)\big) \ln \big( \cos x + \sin x\big)\d x &= \int_{\frac14\pi}^{\frac12\pi} \big ( \cos(2x) + \sin (2x)\big) \ln (\sqrt{2} \cos ( x - \frac{\pi}{4}) ) \d x \\ &&&= \int_{u=0}^{u=\frac{\pi}{4}} \l \cos(2u+\frac{\pi}{2})+\sin(2u+\frac{\pi}{2}) \r \cdot \l \frac12 \ln 2 +\ln (\cos u) \r \d x \\ &&&= \int_{0}^{\frac{\pi}{4}} \l -\sin 2u+\cos 2u \r \cdot \l \frac12 \ln 2 +\ln (\cos u) \r \d x \\ &&&= \frac14 \ln 2\left [ \cos 2u + \sin 2u \ \right]_{0}^{\frac{\pi}{4}} - \frac14(\ln2 - 1) + \frac18\pi - \frac14(\ln 2 +1) \\ &&&= \frac{\pi}{8}-\frac12 \ln 2 \end{align*}

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Solution:

\begin{align*} && \cot \alpha &= \frac{AD}{h} \\ && \cot \beta &= \frac{BD}{h} \\ && \cot \phi &= \frac{DC}h \\ && CD^2 &= AB \cdot BD \tag{intersecting chords} \\ \Rightarrow && \cot^2 \phi &= \cot \alpha \cot \beta \end{align*} \begin{align*} && LHS &= \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) \\ &&&= \cos p \cos q \left ( \frac{1-\cos(p+q)}{2} \right) - \sin p\sin q \left (\frac{1+\cos(p+q)}{2} \right) \\ &&&= \frac12 \left (\cos p \cos q(1-\cos(p+q)) - \sin p\sin q (1+\cos(p+q)) \right) \\ &&&= \frac12 \left ((\cos p \cos q- \sin p\sin q) - (\cos p \cos q+ \sin p\sin q)\cos(p+q) \right) \\ &&&= \frac12 \left (\cos(p+q) - \cos (p-q)\cos(p+q) \right) \\ &&&= RHS \end{align*} Therefore \begin{align*} \cot p \cot q -\cot^2 \tfrac12 (p+q) &= \frac{\tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q)}{\sin p \sin q \sin^2 \tfrac12(p+q)} \\ &=\frac{\cos(p+q)(1-\cos(p-q))}{\sin p \sin q \sin^2 \tfrac12(p+q)} \end{align*} Since \(p,q\) are acute, the denominator is positive. Since \(p+q \leq \frac{\pi}{2}\), we have \(\cos(p+q) \geq 0\). Also \((1-\cos(p-q)) \geq 0\). Thus, the expression is \(\geq 0\). So we must have \begin{align*} && \cot^2 \phi &= \cot \alpha \cot \beta \\ &&&\geq \cot^2 \tfrac12(\alpha+\beta) \end{align*} Since \(\cot\) is decreasing on \((0, \tfrac12 \pi)\) we can deduce \(\phi \leq \tfrac12 (\alpha+\beta)\)

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Solution:

1. Suppose \(t_n = x\) for all \(n\), then we must have \begin{align*} && x &= p x + q x \\ \Leftrightarrow && 1 &= p+q \end{align*} and this clearly works for any value of \(x\).
2. Suppose \(t_{2n} = x, t_{2n+1} = y\) for all \(n\), then \begin{align*} && x &= py + q x \\ && y &= px + q y \\ \Rightarrow && 0 &= py + (q-1) x \\ && 0 &= px + (q-1) y \\ \Rightarrow && p &= (q-1) \frac{x}{y} = (q-1) \frac{y}{x} \\ \Rightarrow && \frac{y}{x} = \pm 1 & \text{ or } q = 1, p = 0 \\ \Rightarrow && y = \pm x & \text{ or } (p,q) = (0,1) \\ \end{align*}
3. Suppose \(t_{3n} = x\), \(t_{3n+1}=y\) and \(t_{3n+2}=z\) , so \begin{align*} && x &= pz + qy \\ && y & = px + qz \\ && z &= py + qx \\ \\ && z &= p(px+qz) + q(pz + qy) \\ &&&= p^2x + 2pqz + q^2 y \\ &&&= p^2(pz+qy) + 2pqz + q^2(px+qz) \\ &&&= p^3 z + p^2qy + 2pqz + q^2p x + q^3 z \\ &&&= (p^3+q^3+2pq)z + pq(py+qx) \\ &&&= (p^3 + q^3 + 2pq)z + pq z \\ &&&= (p^3 + q^3 + 3pq)z \\ \Rightarrow && 0 &= p^3 + q^3 + 3pq- 1 \\ &&&= (p+q-1)(p^2+q^2+1+p+q-pq) \\ &&&= \tfrac12(p+q-1)((p-q)^2+(p+1)^2+(q+1)^2) \end{align*} Therefore \(p+q = 1\) or \((p-q)^2+(p+1)^2+(q+1)^2 = 0 \Rightarrow p = q = -1\). If \(p+q = 1\), then \(z = py + (1-p)x\) and \(x = p(py+(1-p)x) + (1-p)y \Rightarrow (1-p+p^2)x = (1-p+p^2)y \Rightarrow x = y \Rightarrow x= y = z\). If \(p = q = -1\) then adding all the equations we get \(x + y + z = -2(x+y+z) \Rightarrow x + y + z = 0\)

Note that what is actually going on here is that solutions must be of the form \(t_n = \lambda^n\) so the only way to be constant is for \(\lambda = 1\) to be a root, the only way for it to be \(2\)-periodic is for \(\lambda = -1\) to be a root, and the only way for it to be \(3\)-periodic is for \(\lambda = 1, \omega, \omega^2\) to be the roots (although we see this via the classic \(x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x + \omega y + \omega^2 z)(x+\omega^2 y +\omega z)\) which is because of the real constraint in the question.

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Solution:

1. \(\,\) \begin{align*} && y &= xv \\ && y' &= v + xv' \\ \Rightarrow && 0 &= x^2 v \left ( v + x\frac{\d v}{\d x} \right) +(x^2v^2) - 2x^2 \\ &&&= 2x^2v^2 + x^3 v \frac{\d v}{\d x} - 2x^2 \\ \Rightarrow && 0 &= xv \frac{\d v}{\d x} + 2v^2-2 \\ \\ \Rightarrow && \frac{v}{1-v^2} \frac{\d v}{\d x} &= \frac{2}{x} \\ \Rightarrow && \int \frac{v}{1-v^2} \d v &=2 \ln |x| \\ \Rightarrow && -\frac12\ln |1-v^2| &= 2\ln |x| + C \\ \Rightarrow && 4\ln |x| + \ln |1-v^2| &= K \\ \Rightarrow && x^4(1-v^2) &= K \\ \Rightarrow && x^2(x^2-y^2) &= K \end{align*}
2. \(\,\) \begin{align*} && 0 &= xv \left (v +x \frac{\d v}{\d x} \right) + 6x + 5xv \\ &&&= x^2 v \frac{\d v}{\d x} +xv^2 + 6x+5xv \\ \Rightarrow && 0 &= xv\frac{\d v}{\d x} +v^2 +5v+6 \\ \Rightarrow && -\int \frac{1}{x} \d x &=\int \frac{v}{v^2+5v+6} \d v \\ \Rightarrow && -\ln |x| &= \int \frac{v}{(v+2)(v+3)} \d v \\ &&&= \int \left (\frac{3}{v+3} - \frac{2}{v+2} \right) \d v \\ \Rightarrow && -\ln |x| &= 3\ln |v+3| - 2 \ln |v+2| + C\\ \Rightarrow && -\ln |x| &= \ln \frac{|v+3|^3}{|v+2|^2} + C \\ \Rightarrow && \frac{1}{|x|}|v+2|^2 &= A|v+3|^3 \\ \Rightarrow && \frac{1}{|x|}|\frac{y}{x} + 2|^2&= A|\frac{y}{x} + 3|^3 \\ \Rightarrow && \frac{1}{|x|^3} |y +2x|^2 &= \frac{A}{|x|^3}|y + 3x|^3 \\ \Rightarrow && (y+2x)^2 &= A|y+3x|^3 \end{align*}

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Solution: \begin{align*} && s &= ut + \frac12 gt^2 \\ \Rightarrow && -2.5 &= 10 \sin \theta \, T - 5 T^2 \\ \Rightarrow && T &= \frac{10\sin \theta \pm \sqrt{100\sin^2 \theta - 4 \cdot 5 \cdot (-2.5)}}{10} \\ &&&= \sin \theta +\sqrt{\sin^2 \theta + \frac12} \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{2} \sin \theta +\sqrt{2 \sin^2 \theta +1} \right) \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{2 (1-\cos^2 \theta)} + \sqrt{2-\cos 2\theta} \right) \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{1-\cos2 \theta} + \sqrt{2-\cos 2\theta} \right) \\ &&&= \frac{1}{\sqrt{2}}\left ( \sqrt{1-c}+\sqrt{2-c} \right)\\ \\ && s &= 10 \cos \theta T \\ &&&= 10 \sqrt{\frac{\cos 2 \theta +1}{2}}\frac{1}{\sqrt{2}}\left ( \sqrt{1-c}+\sqrt{2-c} \right) \\ &&&= 5 \sqrt{c+1}\left ( \sqrt{1-c}+\sqrt{2-c} \right) \\ \\ && \frac15\frac{\d s}{\d c} &= \frac12(c+1)^{-1/2}((1-c)^{1/2} + (2-c)^{1/2}) - \frac12(c+1)^{1/2}\left ((1-c)^{-1/2}+(2-c)^{-1/2} \right) \\ &&&= \frac{((1-c)(2-c)^{1/2}+(2-c)(1-c)^{1/2})-((c+1)(2-c)^{1/2}+(c+1)(1-c)^{1/2})}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ &&&= \frac{\sqrt{2-c}\left (1-c-c-1 \right)+\sqrt{1-c}\left(2-c-c-1) \right)}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ &&&= \frac{\sqrt{1-c}\left(1-2c\right)-2c\sqrt{2-c}}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ \\ \frac{\d s}{\d c} =0: && \sqrt{1-c}\left(1-2c\right)&=2c\sqrt{2-c} \\ \Rightarrow && (1-c)(1-2c)^2&=4c^2(2-c) \\ \Rightarrow && 1-5c+8c^2-4c^3 &= 8c^2-4c^3 \\ \Rightarrow && 0 &= -5c+1 \\ \Rightarrow && c &= \frac15 \end{align*} When \(\theta = 45^{\circ}, c = 0\), so \(s_{45^{\circ}} = 5(1+\sqrt{2})\) When \(c = \frac15\), \begin{align*} s &= 5 \sqrt{\frac15+1}\left ( \sqrt{1-\frac15}+\sqrt{2-\frac15} \right) \\ &= 5 \sqrt{\frac65} \left ( \sqrt{\frac45} + \sqrt{\frac95} \right) \\ &= 2\sqrt{6}+3\sqrt{6} = 5\sqrt{6} \end{align*} Therefore the additional distance is \(5(\sqrt{6}-\sqrt{2}-1)\)

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Solution: \begin{align*} && s &= ut + \frac12at^2 \\ && D &= \frac12gt_D^2 \\ \Rightarrow && t_D &= \sqrt{\frac{2D}{g}} \\ \Rightarrow && t_{D-\delta} &= \sqrt{\frac{2(D-\delta}{g}} \end{align*} Therefore the difference in times of what I hear will be: \begin{align*} t &= \underbrace{\sqrt{\frac{2D}{g}}}_{\text{time for first stone to hit water}} + \underbrace{\frac{D}{u}}_{\text{time to hear about it}} - \left (\underbrace{\sqrt{\frac{2(D-\delta)}{g}}}_{\text{time for second stone to hit water}} + \underbrace{\frac{D-\delta}{u}}_{\text{time to hear about it}} \right) \\ &= \sqrt{\frac{2D}g} - \sqrt{\frac{2(D-\delta)}g} + \frac \delta u \end{align*} \begin{align*} && t &= \sqrt{\frac{2D}g} - \sqrt{\frac{2(D-\delta)}g} + \frac \delta u \\ \Rightarrow && \beta &= \sqrt{\frac{2D}g} - \sqrt{\frac{2(D-\delta)}g} \\ && \beta^2 &= \frac{2D}{g} + \frac{2(D-\delta)}{g} - \frac{4}{g}\sqrt{D(D-g)} \\ &&&= \frac{4D}{g} - \frac{2\delta}{g} - \frac{4}{g} \sqrt{D(D-\delta)}\\ \Rightarrow && g\beta^2 &= 4D-2\delta -4\sqrt{D(D-\delta)}\\ \Rightarrow && (g \beta^2-4D+2\delta)^2 &= 16D(D-\delta) \\ \Rightarrow && g^2\beta^4 + 16D^2 + 4\delta^2 -8g\beta^2D+4g\beta^2 \delta -16D\delta &= 16D^2-16D\delta \\ \Rightarrow && 8g\beta^2D &= g\beta^4 +4\delta^2 +4g\beta^2 \delta \\ \Rightarrow && D &= \frac1{8g\beta^2} \left ( g^2\beta^4 +4\delta^2 +4g\beta^2 \delta\right) \\ &&&= \frac1{8g\beta^2} \left ( g\beta^2 +2\delta \right)^2 \\ &&&= \frac12g \left ( \frac{\beta}{2} + \frac{\delta}{g\beta} \right)^2 \end{align*} If \(u = 300, g = 10, t = \frac15, \delta = 10\), then \begin{align*} && \beta &= \frac15-\frac{10}{300}\\ &&&= \frac15 - \frac1{30} \\ &&&= \frac{1}{6}\\ && D &= \frac12 \cdot 10 \left ( \frac1{12} + 6 \right)^2 \\ &&&= 5\cdot (36 + 1 + \frac{1}{12^2}) \\ &&&\approx 37 \cdot 5 = 185 \end{align*}

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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: \(\p(x) = -x^2, \frac12x^2, -x^2+3, \frac12x^2+\frac34\)