Problems

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Solution: Let \(f(x) = e^{ax} \cos bx\) then, \(f'(x) = ae^{ax} \cos bx - be^{ax} \sin bx = e^{ax} \l a\cos bx - b \sin bx \r\). Therefore the stationary points are where \(f'(x) = 0 \Leftrightarrow \tan bx = \frac{b}a\), ie \(x = \tan^{-1} \frac{a}{b} + \frac{n}{b} \pi, n \in \mathbb{Z}\). \begin{align*} f(\tan^{-1} \frac{a}{b} + \frac{n}{b} \pi) &= e^{a \tan^{-1} \frac{a}{b} + \frac{an}{b} \pi} \cos \l b \tan^{-1} \frac{a}{b} +n \pi\r \\ &= e^{a \tan^{-1} \frac{a}{b}} \cos \l b \tan^{-1} \frac{a}{b}\r \cdot e^{\frac{an}{b} \pi}(-1)^n \\ &= e^{a \tan^{-1} \frac{a}{b}} \cos \l b \tan^{-1} \frac{a}{b}\r \cdot (-e^{\frac{a}{b} \pi})^n \\ \end{align*} showing the form the desired geometric progression.

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Solution:

The shown isoceles triangle has base \(2r\), and the two side lengths are \(R\). The angle at the center of the circle is \(\frac{2\pi}{n}\). The height of the triangle (by Pythagoras) is \(\sqrt{R^2-r^2}\) and so the area enclosed in the triangle is \(\frac12 2r \sqrt{R^2-r^2}\). The area in the three sectors are: \(\frac{\pi}{n}(R-r)^2\), two sets of \(\frac12(\frac12 \l \pi - \frac{2\pi}{n}\r)r^2 = \l\frac{1}{2} - \frac{1}{n} \r \frac12 \pi r^2\). Therefore the remaining area \(Y/n\) is \(r \sqrt{R^2-r^2} -\frac{\pi}{n}(R-r)^2 - \l\frac{1}{2} - \frac{1}{n} \r \pi r^2\). Multiplying this by \(n\) we get the desired result. For \(X\) we can look at the larger sector, which we obtain using the extension of this triangle. This has area \(\frac12 \frac{2 \pi}{n} (R+r)^2\). The areas to be removed are the area of the triangle: \(r \sqrt{R^2-r^2}\) and the areas of the two sectors, which have radii \(r\) and angles \(\pi - \l \frac12 -\frac1{n} \r\pi = \l \frac{1}{2} + \frac{1}{n} \r \pi\). Therefore the area for \(X/n\) is \(\frac{\pi}{n} (R+r)^2 - r \sqrt{R^2-r^2} - \l \frac{1}{2} + \frac{1}{n} \r \pi r^2\) and so \(X\) has area: \[ X = \pi (R+r)^2 - nr \sqrt{R^2-r^2} - n\pi r^2\l \frac{1}{2} +\frac{1}{n} \r \] \(Z = n\pi r^2\) \begin{align*} \frac{(X+Y)}{Z} &= \frac{\pi (R+r)^2 - nr \sqrt{R^2-r^2} - n\pi r^2\l \frac{1}{2} +\frac{1}{n} \r}{n \pi r^2 }\\ &= \qquad \frac{nr\sqrt{R^{2}-r^{2}}-\pi(R-r)^{2}-n\pi r^{2}\left(\frac{1}{2}-\frac{1}{n}\right)}{n \pi r^2} \\ &= \frac{4\pi R r - n \pi r^2}{n\pi r^2} \\ &= \frac{4R-nr}{n r} \\ &= \frac{4R}{nr} - 1 \end{align*} Since \(\frac{r}{R} = \sin \frac{\pi}{n}\) we have: \begin{align*} \frac{(X+Y)}{Z} &= \frac{4R}{nr} - 1 \\ &= \frac{4}{n \sin \frac{\pi}n} - 1 \\ & \to \frac{4}{\pi} - 1 \end{align*}

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Solution: Let \(y = xv\) then \(y' = v + xv'\) and so \(x^2y' = x^2v + x^3v'\) Our differential equation is now: \begin{align*} && x^2 y' &= \frac{1+y^2}{(1+x^2)\frac{y}{x}} \\ \Rightarrow && xy' &= \frac{(1+y^2)}{(1+x^2)y} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x(1+x^2)} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x} - \frac{x}{1+x^2} \\ \Rightarrow && \frac12 \ln(1+y^2) &= \ln x - \frac12 \ln(1+x^2) + C \\ \Rightarrow && \frac12 \ln (1 + y^2) &= \frac12 \ln \l \frac{x^2}{1+x^2}\r + C \\ \Rightarrow && 1+y^2 &= \frac{Dx^2}{1+x^2} \\ \Rightarrow && D &= 4 \quad \quad: (x = 1, v = 1, y = 1) \\ \Rightarrow && 1 + x^2v^2&= \frac{4x^2}{1+x^2}\\ \Rightarrow && v^2 &= \frac{3x^2-1}{x^2(1+x^2)} \\ \Rightarrow && v &= \sqrt{\frac{3x^2-1}{x^2(1+x^2)}} \\ \end{align*} As \(x \to \infty\), \(v \to 0\)

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Solution: Let \(S = \log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots\) then \begin{align*} S &= \sum_{n=0}^{\infty} (-1)^n \log_{2^{2^n}} e \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{\log {2^{2^n}}} \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{2^n\log {2}} \\ &= \frac{\log e}{\log 2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n} \\ &= \frac{1}{\log_e 2} \frac{1}{1+\frac12} \\ &= \frac{1}{\ln (2^{3/2})} \\ &= \frac{1}{\ln (2 \sqrt{2})} \end{align*}

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Solution: Using the suggested substitution, we can find. \begin{align*} && x &=\alpha\cos^{2}\theta+\beta\sin^{2}\theta \\ && x-\alpha &=\alpha(\cos^{2}\theta-1)+\beta\sin^{2}\theta \\ &&& = (\beta - \alpha) \sin^2 \theta \\ && \beta - x &= -\alpha\cos^{2}\theta+\beta(1-\sin^{2}\theta) \\ &&&= (\beta-\alpha)\cos^2 \theta \\ && x &=\alpha\cos^{2}\theta+\beta\sin^{2}\theta \\ \Rightarrow && \frac{dx}{d\theta} &= (\beta - \alpha) 2 \cos \theta \sin\theta \\ \\ &&\int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x &= \int_0^{\pi/2} \frac{1}{(\beta - \alpha)\sin\theta \cos \theta} (\beta - \alpha) 2 \cos \theta \sin \theta \, d \theta \\ &&&= \int_0^{\pi/2} \frac{1}{\bcancel{(\beta - \alpha)}\bcancel{\sin\theta \cos \theta}} \bcancel{(\beta - \alpha)} 2 \bcancel{\cos \theta \sin \theta} \, d \theta \\ &&&= \int_0^{\pi/2} 2 d \theta \\ && &= 2 \frac{\pi}{2} = \boxed{\pi} \end{align*} If \(\alpha > \beta\) we can rewrite the integral as: \begin{align*} \int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x &= \int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\beta)(\alpha-x)}}\,\mathrm{d}x \\ &= -\int_{\beta}^{\alpha}\frac{1}{\sqrt{(x-\beta)(\alpha-x)}}\,\mathrm{d}x \\ &= -\pi \end{align*} Where the last step we are directly using the first integral with the use of \(\alpha\) and \(\beta\) reversed. Finally, using the substitution \(xt = 1\), we fortunately lose the \(\frac1{x}\) term: \begin{align*} && x &= \frac{1}{t} \\ && \frac{dx}{dt} &= -\frac1{t^2} \\ \\ && \int_{\alpha}^{\beta}\frac{1}{x\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x &= \int_{\alpha}^{\beta}\frac{t}{\sqrt{(\frac{1}{t}-\alpha)(\beta-\frac{1}{t})}} \frac{-1}{t^2}\,\mathrm{d}t \\ && &= \int_{\frac1{\alpha}}^{\frac1\beta}\frac{-1}{\sqrt{(1-t\alpha)(t\beta-1)}}\,\mathrm{d}t \\ && &= \int_{\frac1{\alpha}}^{\frac1\beta}\frac{-1}{\sqrt{\alpha\beta}\sqrt{(\frac1{\alpha}-t)(t-\frac1{\beta})}}\,\mathrm{d}t \\ && &= \frac1{\sqrt{\alpha\beta}}\int_{\frac1{\alpha}}^{\frac1\beta}\frac{-1}{\sqrt{(\frac1{\alpha}-t)(t-\frac1{\beta})}}\,\mathrm{d}t \\ &&&= \boxed{\frac{\pi}{\sqrt{\alpha\beta}}} \end{align*} Where again the last step we are using the intermediate integral, with the roles of \(\alpha\) and \(\beta\) replaced with \(\frac{1}{\beta}\) and \(\frac1{\alpha}\)

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Solution:

The blue area is \(F(s)\) the red area is \(G(t)\), the dashed rectangle (which is a subset of the red and blue areas) has area \(st\) therefore \(F(s) + G(t) \geq st\). Equality holds if \(f(s) = t\). \begin{align*} && \int_0^t \sin^{-1} y \d y + \int_0^{\sin^{-1} t} \sin x \d x &= t \sin^{-1} t \\ \Rightarrow && \int_0^t \sin^{-1} y \d y &= t \sin^{-1} t - \left [ -\cos (x) \right]_0^{\sin^{-1} t} \\ &&&= t \sin^{-1} t - (1- \cos (\sin^{-1} t)) \end{align*} Let \(y = t \sin^{-1} t - (1- \cos (\sin^{-1} t))\) then, \begin{align*} \frac{\d y}{\d t} &= \sin^{-1} t +t \frac{\d}{\d t} \l \sin^{-1} (t) \r - \sin ( \sin^{-1} t) \frac{\d}{\d t} \l \sin^{-1} (t) \r \\ &= \sin^{-1} t \end{align*} as required

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Solution: \(\sin x = \frac{e^{ix} - e^{-ix}}{2i}\). Also let \(z = e^{ \frac{2\pi i}{23}}\) \begin{align*} \sum_{k=0}^{10} \sin \l \frac{(4k +2)\pi}{23} \r &= \sum_{k=0}^{10} \textrm{Im} \l \exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l \sum_{k=0}^{10} \exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l e^{ \frac{2\pi i}{23}} \sum_{k=0}^{10} z^{2k} \r \\ &= \textrm{Im} \l z \l \frac{z^{22}-1}{z^2-1} \r \r \\ &= \textrm{Im} \l z \l \frac{z^{11}(z^{11}-z^{-11})}{z(z-z^{-1})} \r \r \\ &= \textrm{Im} \l \frac{z^{11}2i \sin \frac{22 \pi}{23} }{2i \sin \frac{2 \pi}{23}} \r \r \\ &= \frac{\sin \frac{22 \pi}{23}}{\sin \frac{2 \pi}{23}} \textrm{Im} ( z^{11}) \\ &= \frac{\sin^2 \frac{22 \pi}{23}}{\sin \frac{2 \pi}{23}} \\ &= \frac{\sin^2 \frac{\pi}{23}}{2\sin \frac{\pi}{23}\cos \frac{\pi}{23}} \\ &= \frac12 \tan \frac{\pi}{23} \end{align*} Similarly, \begin{align*} \sum_{k=0}^{10} (-1)^k\sin \l \frac{(4k +2)\pi}{23} \r &= \sum_{k=0}^{10} \textrm{Im} \l (-1)^k\exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l \sum_{k=0}^{10} (-1)^k\exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l e^{ \frac{2\pi i}{23}} \sum_{k=0}^{10} (-1)^kz^{2k} \r \\ &= \textrm{Im} \l z \l \frac{z^{22}+1}{z^2+1} \r \r \\ &= \textrm{Im} \l z \l \frac{z^{11}(z^{11}+z^{-11})}{z(z+z^{-1})} \r \r \\ &= \textrm{Im} \l \frac{z^{11}2 \cos \frac{22 \pi}{23} }{2 \cos\frac{2 \pi}{23}} \r \r \\ &= \frac{\cos\frac{22 \pi}{23}}{\cos \frac{2 \pi}{23}} \textrm{Im} ( z^{11}) \\ &= \frac{\cos \frac{22 \pi}{23}\sin \frac{22 \pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= \frac12 \frac{\sin \frac{44 \pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= \frac12 \frac{-\sin \frac{2\pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= -\frac12 \tan \frac{2\pi}{23} \end{align*}

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Solution: When we apply the substitution \(x = \frac1{t}\), \(t\) runs from \(-1 \to -\infty\) as \(x\) goes from \(-1 \to 0\). Then it runs from \(\infty \to 1\) as \(x\) runs from \(0 \to 1\). So we would be able to show that: \[ \int_{-1}^{1}\frac{1}{(1+x^{2})^{2}}\,\mathrm{d}x = \int_{-1}^{-\infty}\frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t + \int_{\infty}^1 \frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t \] Let \(x = \tan u, \d x = \sec^2 u \d u\) \begin{align*} \int_{-1}^1 \frac1{(1+x^2)^2} \d x &= \int_{u = -\pi/4}^{u = \pi/4} \frac{\sec^2 u}{(1+\tan^2 u)^2} \d u \\ &= \int_{u = -\pi/4}^{u = \pi/4} \frac{1}{\sec^2 u} \d u \\ &= \int_{-\pi/4}^{\pi/4} \cos^2 u \d u \\ &= \int_{-\pi/4}^{\pi/4} \frac{1 + \cos 2 u}{2} \d u \\ &= \left [ \frac{2u + \sin 2u}{4} \right]_{-\pi/4}^{\pi/4} \\ &= \frac{\pi}{4} + \frac{1}{2} \end{align*} Let \(t = \tan u, \d t = \sec^2 u \d u\) \begin{align*} \int_{-1}^1 \frac{-t^2}{(1+t^2)^2} \d x &= \int_{u = -\pi/4}^{u = \pi/4} \frac{-\tan^2 u \sec^2 u}{(1+\tan^2 u)^2} \d u \\ &= -\int_{u = -\pi/4}^{u = \pi/4} \frac{\tan^2 u}{\sec^2 u} \d u \\ &= -\int_{-\pi/4}^{\pi/4} \sin^2 u \d u \\ &= -\int_{-\pi/4}^{\pi/4} \frac{1 - \cos 2 u}{2} \d u \\ &= -\left [ \frac{2u - \sin 2u}{4} \right]_{-\pi/4}^{\pi/4} \\ &= \frac{1}{2}-\frac{\pi}{4} \end{align*}

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Solution:

Suppose \(P\) is a fraction \(0 \leq k\leq 1\) along \(AB\), then \(\overrightarrow{OP} = \overrightarrow{OA} +\overrightarrow{AP} = \overrightarrow{OA} +k\overrightarrow{AB} = \mathbf{a} + k(\mathbf{b} - \mathbf{a}) = \lambda \mathbf{b} + (1-k) \mathbf{a}\), ie an arbitrary point on \(AB\) has the position vector where \((1-k) = \rho \geq 0\) and \(k= \sigma \geq 0\) and \((1-\lambda) + \lambda = 1\). Any point on the segment \(PC\) will be of the form \(l\mathbf{c} + (1-l) (k \mathbf{b} + (1-k) \mathbf{a})\) which has the form \(\lambda \mathbf{a} + \mu \mathbf{b} + \nu \mathbf{c}\) where \(\lambda + \mu + \nu = (1-l)(1-k) + (1-l)k + l = 1\) and all coefficients are positive.

1. This is equivalent to the area inside the triangle where every point (\(\mathbf{a}, \mathbf{b}, \mathbf{c}\)) has been send to it's negative (\(-\mathbf{a}, -\mathbf{b}, -\mathbf{c}\)), ie
   

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2. Writing points as: \begin{align*} \lambda\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c} &= (1-\mu - \nu)\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c} \\ &= \mathbf{a} + (-\mu)(\mathbf{a}-\mathbf{b}) + (-\nu)(\mathbf{a} - \mathbf{c}) \\ &= \mathbf{a} + (-\mu)(\mathbf{a}-\mathbf{b}) + (-\nu)(\mathbf{a} - \mathbf{c}) + (1+\mu+\nu)\mathbf{0}\\ \end{align*} So this is a translation of \(\mathbf{a}\) of the triangle with vertices at \(\mathbf{0}, \mathbf{a-b}, \mathbf{a-c}\), or a triangle with vertices \(\mathbf{a}, 2\mathbf{a-b}, 2\mathbf{a-c}\).
   

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Solution:

If \(\alpha\) is as labelled then \(\cos \alpha = \frac{1}{x}, \sin \alpha = \frac{\sqrt{x^2-1}}{x}, \tan \alpha = \sqrt{x^2-1}\). We also have the full length of the rubber band is \(2\pi - 2\alpha +2\tan \alpha\) so the extension is \(2 \l \sqrt{x^2-1} - \cos^{-1} \l \frac{1}{x}\r \r\) Therefore \(T = \frac{\l \sqrt{x^2-1} - \cos^{-1} \l \frac{1}{x}\r \r\lambda}{\pi}\). If \(x = 2\), \(T = \frac{\sqrt{3} - \frac{\pi}{3}}{\pi} \lambda, \sin \alpha = \frac{\sqrt{3}}{2}\) \begin{align*} \text{N2}(\uparrow): && 2T\sin \alpha - mg &= 0 \\ \Rightarrow && \frac{\sqrt{3} - \frac{\pi}{3}}{\pi} \lambda \sqrt{3} &= mg \\ \Rightarrow && \lambda &= \frac{\sqrt{3}\pi}{(3\sqrt{3}-\pi)}mg \end{align*}