Problems

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Solution: \(AG = r\), therefore the area is: \begin{align*} A &= [AHG] - 2*[ABC] + [CDE] \\ &= \frac12 \pi r^2 - \pi h^2 + \frac12 \pi (r-2h)^2 \\ &= \frac12 \pi \l r^2 - 2h^2 + r^2 -4rh+4h^2 \r \\ &= \frac12 \pi \l 2r^2 -4rh + 2h^2\r \\ &= \pi (r-h)^2 \end{align*} This is the same area as a circle radius \(r-h\) But \(HD = r + (r-2d) = 2(r-d)\), ie the circle with diameter \(HD\) has radius \(r-h\) as required. Suppose \(A = (-h, 0), C = (h, 0), B = (0, h)\) then our parabola is \(y = \frac1{h}(h^2-x^2)\)

The area of \(ABC\) would then be \(\int_{-h}^h \frac{1}{h}(h^2-x^2) \d x = \frac1{h} \left [ h^2x - \frac{x^3}{3} \right] = \frac1{h} \l 2h^3-2\frac{h^3}{3} \r = \frac{4}{3}h^2\) so we have: \begin{align*} A &= [AHG] - 2*[ABC] +[CDE] \\ &= \frac43 r^2-\frac83h^2+\frac43(r-2h)^2 \\ &= \frac43 \l r^2 -2h^2+r^2-4rh+4h^2) \\ &= \frac43 (r-h)^2 \end{align*}

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Solution: Integrating by parts we obtain: \begin{align*} \int x \ln x \, \d x &= [\frac12 x^2 \ln x] - \int \frac12x^2 \cdot \frac1x \d x \\ &= \frac12 x^2 \ln x - \frac14 x^2 + C \end{align*}

We should have: \begin{align*} \int_0^1 x \ln \frac{1}{x} \d x &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ \left [ -\frac12 x^2 \ln x + \frac14 x^2 \right]_0^1 &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ \frac{1}{4} &=\lim_{n \to \infty} \frac{1}{n^2} \sum_{i=1}^n \l i \ln n - i \ln i \r \\ &= \lim_{n \to \infty} \frac{1}{n^2}\l \frac{n(n+1)}{2} \ln n - \sum_{i=1}^n i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{i=1}^n i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{i=1}^n \sum_{k=1}^i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{k=0}^{n-1} \sum_{i=0}^k \ln (n-i) \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{k=0}^{n-1} \ln \frac{n!}{(n-k)!}\r \\ \end{align*}

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Solution:

Denoting \(\overrightarrow{OY}\) by \(\mathbf{y}\) and \(\overrightarrow{OC}\) by \(\mathbf{c}\) etc, we have: \begin{align*} \mathbf{c} &= \frac23 \mathbf{a} + \frac13 \mathbf{b} \\ \mathbf{d} &= \frac13 \mathbf{a} + \frac23 \mathbf{b} \\ \mathbf{x} &= \lambda \mathbf{a} \\ \mathbf{y} &= (1-\lambda) \mathbf{b} \\ 0 &= (\mathbf{d}-\mathbf{x}) \cdot (\mathbf{c} - \mathbf{y}) \\ &=((\frac13 -\lambda)\mathbf{a} + \frac23 \mathbf{b})\cdot(\frac23 \mathbf{a} + (\frac13-1+\lambda) \mathbf{b} ) \\ &= \frac{2}{3} \cdot (\frac13-\lambda) +\frac23 \cdot(\lambda - \frac23)+(\frac{4}{9}+(\frac13-\lambda)(-\frac23+\lambda))\mathbf{a}\cdot\mathbf{b} \\ &= -\frac{2}{9} + (\frac{4}{9} - \frac{2}{9}+\lambda-\lambda^2)\mathbf{a}\cdot \mathbf{b} \\ &= - \frac{2}{9} + (\frac{2}{9} + \lambda - \lambda^2)\mathbf{a}\cdot \mathbf{b} \\ \mathbf{a}\cdot \mathbf{b} &= \frac{2/9}{2/9+\lambda - \lambda^2} \end{align*} Since \(0 \leq \lambda - \lambda^2 \leq \frac14\), \(\frac{\frac29}{\frac29+\frac14} = \frac8{17} \leq \mathbf{a}\cdot\mathbf{b} \leq 1\) If \(\angle AOB\) is as large as possible, \(\mathbf{a}\cdot\mathbf{b}\) is as small as possible, ie \(\lambda = \frac12\) and \(\mathbf{a}\cdot \mathbf{b} = \frac{8}{17}\) First notice that the length \(OM\) to the midpoint of \(AB\) is \(\sqrt{\frac14 (\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})} = \sqrt{\frac14 (2 + 2\mathbf{a}\cdot \mathbf{b})} = \sqrt{\frac12 + \frac4{17}} = \sqrt{\frac{25}{34}} = \frac{5}{\sqrt{34}}\) Notice that \(XYE\) and \(DCE\) are similar triangles, and so the heights satisfy \(\frac{h_1}{h_2} = \frac{\frac12}{\frac13} = \frac32\). Therefore the length \(OE\) is \(\frac12 \frac{5}{\sqrt{34}} + \frac{3}{5} \frac12 \frac{5}{\sqrt{34}} = \frac{8}{10} \frac{5}{\sqrt{34}} = \frac{4}{\sqrt{34}}\)

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Solution: Consider the \(5\) segements radiating from \(A\). By the pigeonhole principle, at least \(3\) of them must be the same colour (say gold and say reaching \(B,C,D\)). If any of the segments joining any of \(B,C,D\) are gold then we have found a monochromatic gold triangle. But if none of them are gold, they are all silver, therefore \(BCD\) is a monochromatic silver triangle.

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Solution: \[ (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k \]

1. \begin{align*} (1+1)^n &= \sum_{k=0}^n \binom{n}{k} \\ (1-1)^n &= \sum_{k=0}^n (-1)^n\binom{n}{k} \\ &= \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} -\sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} \end{align*} Therefore \(\displaystyle \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} = \sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} = \frac{2^n}{2} = 2^{n-1}\)
2. \begin{align*} && (1+x)^n &= \sum_{k=0}^n \binom{n}{k}x^k \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{k=0}^n k\binom{n}{k} x^{k-1} \\ x = 1: && n2^{n-1} &= \sum_{k=1}^n k\binom{n}{k} \end{align*} as required

\begin{align*} \sum_{r=0}^n (r + (-1)^r) \binom{n}{r} &= n2^{n-1}+0 = n2^{n-1} \end{align*}

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Solution: The normal to the curve \(y = f(x)\) has gradient \(-\frac{1}{f'(x)}\) and so has equation: \begin{align*} && \frac{Y - f(x)}{X - x} &= -\frac{1}{f'(x)} \\ \Rightarrow && Y &= -\frac{1}{f'(x)}X + \frac{x}{f'(x)}+f(x) \end{align*} Hence the \(Q\) is \(\displaystyle \left (0, f(x) + \frac{x}{f'(x)} \right)\). \begin{align*} && |PQ|^2 &= x^2 + \frac{x^2}{(f'(x))^2} \\ \Rightarrow && x^2 + e^{x^2} &= x^2 + \frac{x^2}{(f'(x))^2} \\ \Rightarrow && (f'(x))^2 &=x^2 e^{-x^2} \end{align*} Therefore \(f'(x) = \pm x e^{-x^2/2}\). If \(f(x)\) has a minimum at \((0,-2)\) then \(f''(0) > 0\), and \(f''(x) = \pm (e^{-x^2/2} - x^2e^{-x^2/2}) = \pm e^{-x^2/2}(1-x^2)\) so we should take the positive branch of the solution, ie \(f'(x) = xe^{-x^2/2}\). Therefore \(f(x) = - e^{-x^2/2}+C\). Since \(f(0) = -2\) we must have \(-2 = -1 + C\), ie \(C = -1\). Therefore \(f(x) = -1 - e^{-x^2/2}\)

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Solution:

Suppose one of the \(x\) coordinates is \(t > 0\), then the coordinates are \(y = \pm \sqrt{1-t}, x = \pm t\). The area will be \(A = 2t \cdot 2 \sqrt{1-t}\). To maximise this, \begin{align*} && \frac{\d A}{\d t} &= 4 \sqrt{1-t} - 2t(1-t)^{-\frac12} \\ &&&= \frac{4(1-t) - 2t}{\sqrt{1-t}} \\ &&&= \frac{4-6t}{\sqrt{1-t}} \end{align*} Therefore there is a stationary point at \(t = \frac23\). Since we know the area is \(0\) when \(t = 0, 1\) we can see this must be a maximum for the area. Therefore the area is \(\displaystyle 4 \frac23 \sqrt{1-\frac23} = \frac{8}{3\sqrt{3}} = \frac{8}{\sqrt{27}}\). For this similar problem, using a similar approach we find \(y = \pm (1- t)^{n/2m}, x = \pm t\) and so the area is \(A = 4 t \cdot (1-t)^{n/2m}\). \begin{align*} && \frac{\d A}{\d t} &= 4(1-t)^{n/2m} - 4t \frac{n}{2m} (1-t)^{\frac{n}{2m} - 1} \\ &&&= (1-t)^{\frac{n}{2m}-1} \left ( 4(1-t) - \frac{2n}{m} t\right) \\ &&&= (1-t)^{\frac{n}{2m}-1} \left ( 4 - (4 + \frac{2n}{m})t \right) \\ \end{align*} Therefore \(\displaystyle t = \frac{2m}{2m+n}\) and \(\displaystyle A = 4\cdot \frac{2m}{2m+n} \cdot (1 - \frac{2m}{2m+n})^{n/2m} = \frac{8m}{2m+n} \cdot \left ( \frac{n}{2m+n}\right)^{n/2m}\)

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Solution: Given that \(e^{i \theta} = \cos \theta + i \sin \theta\) we must have that

1. \begin{align*} \cos 4 \theta &= \textrm{Re} \l e^{i 4 \theta} \r \\ &= \textrm{Re} \l (\cos \theta + i \sin \theta)^4 \r \\ &= \cos^4 \theta - \binom{4}{2}\cos^2 \theta \sin^2 \theta +\sin^4 \theta \\ &= \cos^4 \theta - 6\cos^2 \theta (1-\cos^2 \theta) +(1-\cos^2 \theta)^2 \\ &= 8\cos^4 \theta - 8\cos^2 \theta + 1 \end{align*}
2. Similarly, \begin{align*} \cos 6 \theta &= \textrm{Re} \l e^{i 6 \theta} \r \\ &= \textrm{Re} \l (\cos \theta + i \sin \theta)^6 \r \\ &= \cos^6 \theta -\binom{6}{2}\cos^4 \theta \sin^2 \theta +\binom{6}{4} \cos^2\theta \sin^4 \theta - \sin^6 \theta \\ &= \cos^6 \theta - 15 \cos^4 \theta (1-\cos^2 \theta) + 15\cos^2 \theta (1-\cos^2\theta)^2 - (1-\cos^2 \theta)^3\\ &= 31\cos^6 \theta-45\cos^4\theta+15\cos^2\theta-1+3\cos^2 \theta-3\cos^4 \theta+\cos^6 \theta \\ &= 32 \cos^6 \theta-48\cos^4 \theta+18\cos^2 \theta-1 \end{align*}

\begin{align*} 0 &= 16x^{6}-28x^{4}+13x^{2}-1\\ &= \frac12 (32x^6-56x^4+26x^2-1) \\ &= \frac12(32x^6-48x^4+18x^2-1-(8x^4-8x^2+1)) \end{align*} Therefore if \(x = \cos \theta\) then we are looking at solving \(\cos 6 \theta = \cos 4 \theta\). \(\cos 6 \theta - \cos 4 \theta = -2 \sin 5\theta \sin \theta = 0\). So we should be looking at \(\sin 5 \theta = 0\) and \(\sin \theta = 0\). \(\sin \theta = 0 \Rightarrow x = \cos \theta = \pm 1\) both of which are roots. The other roots will be \(\cos \frac{\pi}{5}, \cos \frac{2\pi}{5}\) etc but it's unclear this is an acceptable form. Alternatively, given our two roots, we can factorize \begin{align*} 0 &= 16x^{6}-28x^{4}+13x^{2}-1 \\ &= (x^2-1)(16x^4-12x^2+1) \end{align*} We can solve \(16y^2-12y+1=0\) to see that \(x^2 = \frac{3 \pm \sqrt{5}}{8}\) so our roots are: \(x = -1, 1, \pm \sqrt{\frac{3 + \sqrt{5}}{8}}, \pm \sqrt{\frac{3 -\sqrt{5}}{8}}\) (We might notice that \(3+\sqrt{5} =\l \frac{1+\sqrt{5}}{\sqrt{2}} \r^2\) so our final answer could be: \(x = -1, 1, \pm \frac{1+\sqrt{5}}{4}, \pm \frac{\sqrt{5}-1}{4}\))

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Solution: \(8\frac{\d y}{\d x} = 3(x^2-4)\) so the turning points are at \((\pm 2, \mp 2)\)

We need either \begin{align*} && \begin{cases} -2a+b &= 0 \\ 2c+d &= 0 \\ 2a+b &= 1 \\ -2c+d &= 1 \end{cases} && \text{ or } && \begin{cases} -2a+b &= 1 \\ 2c+d &= 1 \\ 2a+b &= 0 \\ -2c+d &= 0 \end{cases} \\ \Rightarrow && \begin{cases} -2a+b &= 0 \\ 2a+b &= 1 \\ 2c+d &= 0 \\ -2c+d &= 1 \end{cases} && \text{ or } && \begin{cases} -2a+b &= 1 \\ 2a+b &= 0 \\ 2c+d &= 1 \\ -2c+d &= 0 \end{cases}\\ \Rightarrow && \begin{cases} (a,b) = (\frac14,\frac12) \\ (c,d) = (-\frac14, \frac12)\end{cases} && \text{ or } && \begin{cases} (a,b) = (-\frac14,\frac12) \\ (c,d) = (\frac14, \frac12)\end{cases} \end{align*} So either \(X = \frac14 x + \frac12, Y = -\frac14 y + \frac12\) or \(X = -\frac14x + \frac12, Y = \frac14y + \frac12\)

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Solution: Using Newton's second law, we have: \begin{align*} && -M\omega(v^2+V^2)/v &= M v \frac{\d v}{\d x} \\ \Rightarrow && \frac{v^2}{v^2+V^2} \frac{\d v}{\d x} &= -\omega \\ \Rightarrow && \omega X &= \int_{V/2}^V \frac{v^2}{v^2+V^2} \d v \\ &&&= \int_{V/2}^V \l 1 - \frac{V^2}{v^2+V^2} \r \d v \\ &&&= \left [v - V\tan^{-1} \frac{v}{V} \right]_{V/2}^V \\ &&&= V \l \frac12 - \tan^{-1} 1 + \tan^{-1} \frac12 \r \\ \Rightarrow X &= \frac{V}{\omega} \l \tan^{-1} \frac12 + \frac12 - \frac{\pi}{4} \r \end{align*}. The resistive force just before the field changes is \(M \omega (\frac{V^2}{4} + V^2)/\frac{V}{2} = \frac52MV\omega\). Therefor the constant resistive force is between \(\frac{11}4MV\omega\) and \(\frac{9}{4}MV \omega\) and acceleration is \(\frac{11}{4}V\omega, \frac{9}{4}V\omega\). Since \(v^2 = u^2 + 2as \Rightarrow s = \frac{v^2-u^2}{2a} = \frac{\frac{V^2}{4}}{2kV\omega} = \frac{V}{8k\omega}\) therefore \(z \in \left [ \frac{V}{22\omega},\frac{V}{18 \omega} \right]\)