Problems

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Solution:

If \(n^m = m^n \Rightarrow \frac{\ln m}{m} = \frac{\ln n}{n}\) so if \(n < m\) we must have that \(n < e \Rightarrow n = 1,2\). \(n = 1\) has no solutions, so consider \(n = 2\) which obviously has the corresponding solution \(m=4\). Since \(h(x)\) is decreasing for \(x > e\) we know this is the only solution. Hence the only solution for distinct \(m,n\) are \((m,n) = (2,4), (4,2)\)

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Solution: \(g'(x) = f'(x) - \frac{f'(x)}{(f(x))^2} = f'(x) \l 1 - \frac{1}{(f(x))^2} \r\) \(g''(x) = f''(x) - \frac{f''(x)f(x)^2-f'(x)\cdot 2f(x) f'(x)}{(f(x))^4} = f''(x) + \frac{f''(x)f(x)-2(f'(x))^2}{(f(x))^3}\) \begin{align*} f(x) &=4+\cos2x+2\sin x \\ f'(x) &=-2\sin2x+2\cos x \\ f''(x) &= -4\cos2x-2\sin x \end{align*} Therefore, since the stationary points of \(g\), ie points where \(g'(x) = 0\) are where \(f'(x) = 0\) or \(f(x) = \pm 1\) we should look at \begin{align*} && 0 &= f'(x) \\ && 0 &= 2 \cos x - 2 \sin 2x \\ &&&= 2 \cos x - 4 \sin x \cos x \\ &&&= 2\cos x (1 - 2 \sin x) \\ \Rightarrow && x &= \frac{\pi}2, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6} \end{align*} \begin{align*} && 1 &= f(x) \\ && 1 &= 4 + \cos 2x + 2 \sin x \\ \Rightarrow && \cos 2x = -1,& \sin x = -1 \\ \Rightarrow && x &= \frac{3\pi}{2} \end{align*} which we were already checking. For each of these points we have: \begin{array}{c|c|c|c||c} x & f(x) & f'(x) & f''(x) & g''(x) \\ \hline \frac{\pi}{2} & 5 & 0 & 2 & > 0\\ \frac{3\pi}{2} & 1 & 0 & 6 &> 0\\ \frac{\pi}{6} & 5.5 & 0 & -3 & < 0 \\ \frac{5\pi}{6} & 5.5 & 0 & -3 & < 0\\ \end{array} Therefore \(\frac{\pi}{2}, \frac{3\pi}{2}\) are minimums and \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) are maxima.

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Solution: First note that \((\sqrt{6} - \sqrt{2})^2 = 8 - 2\sqrt{12} = 8 - 4\sqrt{3}\) and since \(49 > 16 \times 3\) \(\sqrt{6}-\sqrt{2} > 1\). Therefore we can assume without loss of generality that \(P\) and \(Q\) both do not lie on the same side as each other, a side containing \(O\), otherwise one of those lengths would be \(1 \text{ cm} < (\sqrt{6}-\sqrt{2}) \text{ cm} \). Let \(O = (0,0)\), \(P = (1,x)\), \(Q = (y,1)\), then our lengths squared are: \(1 + x^2, 1 + y^2, (1-x)^2+(1-y)^2\). To maximise the length of the smallest side, each side should be equal in length (otherwise we could increase the length of the smallest side by moving the point between the shortest side and the longest side (without affecting the other side). Therefore \(x = y\) and \(1+x^2 = 2(1-x)^2 \Rightarrow x^2-4x+1 = 0 \Rightarrow x = 2 - \sqrt{3} \). Therefore the distances are all \(\sqrt{1+7-4\sqrt{3}} = \sqrt{8-4\sqrt{3}} = (\sqrt{6}-\sqrt{2}) \text{ cm}\)

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Solution:

1. There are \(m \cdot n\) line segments, since each of the \(m\) points on the \(y\)-axis correspond to \(n\) points of the \(x\)-axis.
2. Every pair of points on the \(x\)-axis and pair of points on the \(y\) axis will generate \(4\) lines, but only \(1\) of those points will meet in the middle, therefore there will be \(\binom{m}{2} \binom{n}{2}\) intersections.

The segment joining the smallest \(x\) to the smallest \(y\) and the segment joining the largest \(x\) to the largest \(y\) will not meet any other segments. The segment joining the smallest \(x\) to the largest \(y\) and the segment joining the largest \(x\) to the smallest \(y\) will have the largest number of intersections. They each intersect all the other segment coming out of other points (except the ones headed for the same point) ie \((m-1)(n-1)\) each. They also intersect each other, so \(2(m-1)(n-1)-1\) points. Therefore we are interested in: \begin{align*} \frac{\binom{n}{2}\binom{n}{2} - 2(n-1)^2+1}{n^2-4} &= \frac{\frac{n^2(n-1)^2}{4} - 2n^2+4n-1}{n^2-4} \\ &= \frac{n^4-2n^3+n^2-8n^2+16n-4}{4(n^2-4)} \\ &= \frac{(n - 2) (n^3 - 7 n + 2)}{4(n^2-4)} \\ &= \frac{n^3-7n+3}{4(n+2)} \end{align*}

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Solution: \begin{align*} \frac{1}{(1-ax)(1-bx)} &=\frac{1}{b-a} \l \frac{b}{1-bx}-\frac{a}{1-ax}\r \\ &= \frac{1}{b-a} \l \sum_{k=0}^{\infty} b(bx)^k-\sum_{k=0}^{\infty} a(ax)^k \r \\ &= \frac{1}{b-a} \sum_{k=0}^{\infty} \l b^{k+1} - a^{k+1} \r x^k \end{align*} Therefore \(\displaystyle c_m = \frac{b^{k+1}-a^{k+1}}{b-a}\). \begin{align*} c_m^2 &= \frac{(b^{m+1}-a^{m+1})^2}{(b-a)^2} \\ &= \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \end{align*} \begin{align*} \sum_{m=0}^{\infty} c_m x^m &= \sum_{m=0}^{\infty} \l \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \r x^m \\ &= \frac{1}{(b-a)^2} \l \sum_{m=0}^{\infty} a^{2m+2} x^m-2\sum_{m=0}^{\infty} (ab)^{m+1} x^m+\sum_{m=0}^{\infty} b^{2m+2} x^m \r \\ &= \frac{1}{(b-a)^2} \l a^2\sum_{m=0}^{\infty} a^{2m} x^m-2ab\sum_{m=0}^{\infty} (ab)^{m} x^m+b^2\sum_{m=0}^{\infty} b^{2m} x^m \r \\ &= \frac{1}{(b-a)^2} \l \frac{a^2}{1-a^2x^2} - \frac{2ab}{1-abx} + \frac{b^2}{1-b^2x^2}\r \\ &= \frac{1+ab}{(1-a^2x)(1-abx)(1-b^2x)} \end{align*} Where geometric series will converge if \(|a^2x| < 1, |b^2x| < 1, |abx| < 1\), ie \(|x| < \min (\frac{1}{a^2}, \frac{1}{b^2} )\)

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Solution:

1. \(\overrightarrow{P_{1}P_{2}}=\overrightarrow{P_{5}P_{4}}\) is equivalent to \(z_2 - z_1 = z_4 - z_5\). \(\overrightarrow{P_{2}P_{3}}=\overrightarrow{P_{6}P_{5}}\) is equivalent to \(z_3-z_2 = z_5 - z_6\).
2. \(\overrightarrow{P_{2}P_{4}}\) is perpendicular to \(\overrightarrow{P_{3}P_{6}}\,\) is equivalent to \(\frac{z_4 - z_2}{z_6-z_3} \in i\mathbb{R}\)

If \(z_2 - z_1 =z_4 - z_5\) and \(z_3-z_2 = z_5 - z_6\) then adding we get \(z_3 - z_1 = z_4 - z_6\) or \(z_4 - z_3 = z_6-z_1\), which is equivalent to \(\overrightarrow{P_{3}P_{4}}=\overrightarrow{P_{1}P_{6}}\,\). \(\textrm{Re}(z) > 1, \textrm{Re}(w) < 0, \textrm{Re}(z) +\textrm{Re}(w)=1\). (We only need one of the first two constraints, since the other is implied by the former). Since \(\overrightarrow{P_{2}P_{4}}\) is perpendicular to \(\overrightarrow{P_{3}P_{6}}\,\) we must have that \(\textrm{Im}(z) = \textrm{Im}(w)\). Combined with the vector logic we must have that \(\textrm{Im}(z) = \frac12\). Let \(z = a + \frac12i\) and \(w = (1-a) + \frac12i\). Since \(w - 1 = k(3-2i)(z-1)\) (the angle constraint) we must have that: \begin{align*} &&-a+\frac12i &= k(3-2i)((a-1) \frac12i) \\ &&&= k( 3 a - 2+(\frac72 - 2 a)i) \\ \Rightarrow && \frac{3a-2}{-a} &= \frac{\frac72-2a}{\frac12} \\ \Rightarrow && 3a-2&= 4a^2-7a \\ \Rightarrow && 0 &= 4a^2-10a+2 \\ \Rightarrow && a &= \frac{5 \pm \sqrt{17}}{4} \end{align*} Since \(a > 1, a = \frac{5 +\sqrt{17}}{4}\) and the distance is: \begin{align*} \left | z - w \right | &= | a+\frac12i - ((1-a) +\frac12i ) | \\ &= |2a-1| \\ &= \frac{5+\sqrt{17}}{2}-1 \\ &= \frac{3+\sqrt{17}}{2} \end{align*}

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Solution: Let \(f(x) = ax^2 + bx + c\) then \begin{align*} f'(x) &= 2ax + b \\ f(0) &= c \\ f(1) &= a+b+c \\ f(-1) &= a-b+c \\ f(1)+f(-1) &= 2(a+c) \\ f(1)-f(-1) &= 2b \\ f'(x) &= x(f(1)+f(-1)) + \frac12 (f(1) - f(-1)) - 2f(0)x \end{align*} as required. Since \(f'(x)\) is a straight line, the maximum value is either at \(1, -1\) or it's constant and either end suffices. \begin{align*} |f'(1)| & \leq |f(1)|\frac{3}{2} + |f(-1)| \frac12 + 2 |f(0)| \\ &\leq \frac{3}{2} + \frac12 + 2 \\ &= 4 \\ \\ |f'(-1)| & \leq |f(1)|\frac{1}{2} + |f(-1)| \frac32 + 2 |f(0)| \\ &\leq \frac{3}{2} + \frac12 + 2 \\ &= 4 \\ \end{align*} Therefore \(|f'(x)| \leq 4\). Suppose \(|f'(x)| = 4\) for some value in \(x \in [-1,1]\), then it must be either \(-1\) or \(1\). If \(f'(1) = 4\) then \(f(1) = 1, f(-1) = 1, f(0) = -1\) so \(f(x) = 1+ k(x^2-1) \Rightarrow f(x) = 1+2(x^2-1) = 2x^2 -1\). If \(f'(-1) = 4\) then \(f(1) = -1, f(-1) = -1, f(0) = 1 \Rightarrow f(x) = -2x^2 + 1\)

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Solution: Let \(\mathbf{a}\) denote the vector position of \(A\) and similarly for \(B, C, D\). Then we know that \((\mathbf{b}-\mathbf{a})\cdot(\mathbf{b}-\mathbf{a})=(\mathbf{c}-\mathbf{d})\cdot(\mathbf{c}-\mathbf{d})\) and \((\mathbf{b}-\mathbf{c})\cdot(\mathbf{b}-\mathbf{c})=(\mathbf{a}-\mathbf{d})\cdot(\mathbf{a}-\mathbf{d})\). Subtracting these two equations we see that \(|\mathbf{a}|^2 -2\mathbf{a}\cdot\mathbf{b}+2\mathbf{c}\cdot\mathbf{b} - |\mathbf{c}|^2 = |\mathbf{c}|^2-2\mathbf{c}\cdot\mathbf{d}+2\mathbf{a}\cdot\mathbf{d}-|\mathbf{a}|^2\) or \(2|\mathbf{a}|^2 -2\mathbf{a}\cdot\mathbf{b}+2\mathbf{c}\cdot\mathbf{b} - 2|\mathbf{c}|^2 +2\mathbf{c}\cdot\mathbf{d}-2\mathbf{a}\cdot\mathbf{d}=0\) The midpoints of the diagonals \(AC\) and \(BD\) are \(\frac{\mathbf{a}+\mathbf{c}}{2}\) and \(\frac{\mathbf{b}+\mathbf{d}}{2}\), so the line is parallel to: \(\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}\). The diagonals are parallel to \(\mathbf{a}-\mathbf{c}\) and \(\mathbf{b}-\mathbf{d}\). So it suffices to prove that \((\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}) \cdot (\mathbf{a}-\mathbf{c}) = 0\) (since the other will follow by symmetry, \begin{align*} (\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}) \cdot (\mathbf{a}-\mathbf{c}) &= |\mathbf{a}|^2-\mathbf{a}\cdot\mathbf{b}-\mathbf{a}\cdot \mathbf{d}+\mathbf{b}\cdot \mathbf{c}-|\mathbf{c}|^2+\mathbf{c}\cdot \mathbf{d} \\ \end{align*} But this is exactly half the equation we determined earlier, so we are done.

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Solution:

1. \begin{align*} \int_{1}^{\mathrm{e}}\frac{\ln x}{x^{2}}\,\mathrm{d}x &= \left [-\frac{\ln x}{x} \right]_1^e + \int_1^e \frac{1}{x^2} \, \d x \\ &= -\frac{1}{e} + \left [ -\frac{1}{x} \right]_1^e \\ &= 1 - \frac{2}{e} \end{align*}
2. \begin{align*} \int\frac{\cos x}{\sin x\sqrt{1+\sin x}}\,\mathrm{d}x &= \int \frac{2u}{(u^2-1)u} \d u \tag{\(u^2 = 1+\sin x\)} \\ &= \int \frac{1}{u-1} - \frac{1}{u+1} \d u \\ &= \ln(u-1) - \ln (u+1) + C \\ &= \ln \l \frac{u-1}{u+1} \r + C \\ &= \ln \l \frac{\sqrt{\sin x + 1} + 1}{\sqrt{\sin x + 1} -1} \r + C \end{align*}

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Solution:

The initial velocity is \(\begin{pmatrix} v \cos \theta \\ v \sin \theta \end{pmatrix}\). The trajectory will be: \(\begin{pmatrix} x_0 + (v \cos \theta) t \\ (v \sin \theta)t -\frac12 g t^2 \end{pmatrix}\) we must have that for some time \(t\), this is equal to \(\begin{pmatrix} 0 \\ h \end{pmatrix}\) So \(t = -\frac{x_0}{v \cos \theta}\) and so \begin{align*} &&h &= (v \sin \theta)t -\frac12 g t^2 \\ &&&= -x_0\tan \theta - \frac12 g \frac{x_0^2}{v^2 \cos^2 \theta} \\ &&&= -x_0\tan \theta - \frac{g}{2v^2 \cos^2 \theta}x_0^2 \\ &&&= -x_0\tan \theta - \frac{g}{2v^2} \sec^2 \theta x_0^2 \\ &&&= -x_0\tan \theta - \frac{g}{2v^2} (1+\tan^2 \theta )x_0^2 \\ &&&= -\l \frac{\sqrt{g}x_0}{\sqrt{2}v}\tan \theta +\frac{\sqrt{2}v}{2\sqrt{g}}\r^2+\frac{v^2}{2g}-\frac{g}{2v^2}x_0^2 \\ \Rightarrow && \frac{g}{2v^2}x_0^2 &= \frac{v^2}{2g}-h-\l \frac{\sqrt{g}x_0}{\sqrt{2}v}\tan \theta +\frac{\sqrt{2}v}{2\sqrt{g}}\r^2 \\ \Rightarrow && x_0^2 &= \frac{v^2(v^2-2gh)}{g^2}-K^2 \end{align*} Therefore \(\displaystyle |x_0| \leq \frac{v}{g}\sqrt{v^2-2gh}\)