Problems

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Solution:

The distance to the top of the house (viewed from above) from the long side will be \(p\). The distance from the short side will also be the same, since the roof sections are climbing at the same angle, so they will take just as far to reach the top. Therefore the length of the roof ridge will be \(2q - 2p\) which is independent of \(\theta\). \vspace{1em} The height of the roof will be \(h = p \tan \theta\). The attic can be split into a prism (along the roof ridge) and a pyramid (along the sloping sides). The pyramid will have volume \(\frac13 p \tan\theta (2p)^2 = \frac83 \tan\theta p^3\). The prism will have volume \(2(q-p)p^2 \tan\theta\). Therefore the total volume will be \(\l \frac{2}{3}p + 2q \r p^2\tan\theta \) The distance (along the plane) to the roof of the house will be \(\frac{p}{\cos \theta}\) and therefore the two end roof-sections will be triangles of area \(\frac{p^2}{\cos \theta}\). The two side roof-sections will be trapiziums will area \(\frac{1}{2} \l 2q + 2(q-p) \r \frac{p}{\cos \theta}\) Therefore the total area will be \(\frac{1}{\cos \theta} \l 2p^2 + 4pq - 2p^2 \r = \frac{4pq}{\cos \theta}\)

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Solution: \begin{align*} && y &= x^a \\ && \frac{\d y}{\d x} &= \begin{cases} ax^{a-1} & a \neq 0 \\ 0 & a = 0 \end{cases} \\ \\ && y &= a^x \\ &&&= e^{(\ln a) \cdot x} \\ && \frac{\d y}{\d x} &= \ln a e^{(\ln a) x} \\ &&&= \ln a \cdot a^ x \\ \\ && y &= x^x \\ &&&= e^{x \ln x}\\ && \frac{\d y}{\d x} &= e^{x \ln x} \cdot \left ( \ln x + x \cdot \frac1x \right) \\ &&&= x^x \left (1 + \ln x \right) \\ \\ && y&= x^{(x^x)} \\ &&&= e^{x^ x \cdot \ln x} \\ && \frac{\d y}{\d x} &= e^{x^x \cdot \ln x} \left ( x^x \left (1 + \ln x \right) \cdot \ln x + x^x \cdot \frac1x\right) \\ &&&= x^{x^x} \left (x^x (1+ \ln x) \ln x +x^{x-1} \right) \\ &&&= x^{x^x+x-1} \left (1 + x \ln x + x (\ln x)^2 \right) \\ \\ && y &= (x^x)^x \\ &&&= x^{2x} \\ &&&= e^{2x \ln x} \\ && \frac{\d y}{\d x} &= e^{2 x \ln x} \left (2 \ln x + 2 \right) \\ &&&= 2(x^x)^x(1 + \ln x) \end{align*}

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Solution: The coefficient of \(x^n\) on the LHS is \begin{align*} && (1-x^2)^n &= (1-x)^n(1+x)^n \\ [x^n]: && \begin{cases} (-1)^{\lfloor \frac{n}2 \rfloor}\binom{n}{\lfloor \frac{n}2 \rfloor} &\text{if } n\text{ even} \\ 0 & \text{otherwise} \end{cases} &= \sum_{i=0}^n \underbrace{(-1)^i\binom{n}{i}}_{\text{take }(-x)^i\text{ from first bracket}} \cdot \underbrace{\binom{n}{n-i}}_{\text{take }x^{n-i}\text{ from second bracket}} \\ &&&= \sum_{i=0}^n (-1)^i\binom{n}{i}\binom{n}{i} \\ &&&= \sum_{i=0}^n (-1)^i\binom{n}{i}^2\\ \end{align*}

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Solution:

1. \begin{align*} \frac{1-\cos \alpha}{\sin \alpha} &= \frac{1-(1-2\sin^2 \frac{\alpha}{2})}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\ &= \frac{2 \sin^2 \frac \alpha2}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\ &= \frac{\sin \frac \alpha2}{ \cos \frac\alpha2} \\ &= \tan \tfrac{\alpha}{2} \end{align*}
2. \begin{align*} \int\frac{\mathrm{d}x}{1-2kx+x^{2}} &= \int \frac{\d x}{(x-k)^2+1-k^2} \\ &= \frac{1}{1-k^2}\int \frac{\d x}{\left (\frac{x-k}{\sqrt{1-k^2}} \right)^2+1} \\ &= \frac{1}{\sqrt{1-k^2}} \tan^{-1} \left (\frac{x-k}{\sqrt{1-k^2}} \right)+C \end{align*}

\begin{align*} \int_{0}^{1}\frac{\sin\alpha}{1-2x\cos\alpha+x^{2}}\,\mathrm{d}x &= \sin \alpha \left [\frac{1}{\sqrt{1-\cos ^2\alpha}} \tan^{-1} \left ( \frac{x - \cos \alpha}{\sqrt{1-\cos^2\alpha}} \right) \right]_0^1 \\ &= \tan^{-1} \left ( \frac{1 - \cos \alpha}{\sin \alpha} \right) -\tan^{-1} \left ( \frac{- \cos \alpha}{\sin \alpha} \right) \\ &= \tan^{-1} \tan \tfrac{\alpha}{2} + \tan^{-1} \cot \alpha \\ &= \frac{\alpha}{2} + \frac{\pi}{2} - \alpha \\ &= \frac{\pi-\alpha}{2} \end{align*}

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Solution:

1. The line \(PO\) has gradient \(\frac{p^2}{p} = p\) and teh line \(QO\) has gradient \(q\), therefore we must have that \(pq = -1\). Therefore, \(R\) is the point \begin{align*} && R &= \left ( \frac{p-\frac{1}{p}}{2}, \frac{p^2+\frac{1}{p^2}}{2} \right) \\ &&&= \left ( \frac12\left ( p - \frac{1}{p} \right),2\left (\frac12 \left(p-\frac{1}{p}\right) \right)^2+1 \right) \\ &&&= \left ( t, 2t^2+1\right) \end{align*} So we are looking at another parabola.
   

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2. The tangents are \(y = 2px+c\), ie \(p^2 = 2p^2+c\), ie \(y = 2px -p^2\) so we have \begin{align*} && y - 2px &= -p^2 \\ && y - 2qx &= -q^2 \\ \Rightarrow && (2p-2q)x &= p^2-q^2 \\ \Rightarrow && x &= \frac12 (p+q)\\ && y &= p(p+q)-p^2 \\ && y &= pq = -1 \end{align*} Therefore \(x = \frac12(p - \frac1p), y= -1\), so we have the line \(y = -1\) (the directrix)
   

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Solution:

1. \begin{align*} && f(x) &= \frac{ix-1}{ix+1} \\ &&&= \frac{ix-1}{ix+1} \frac{1-ix}{1-ix} \\ &&&= \frac{ix-1+x^2+ix}{1^2+x^2} \\ &&&= \frac{x^2-1}{x^2+1} + i \frac{2x}{x^2+1} \\ \Rightarrow && \textrm{Re}(f(x)) &= \frac{x^2-1}{x^2+1} \\ && \textrm{Im}(f(x)) &= \frac{2x}{x^2+1} \end{align*}
2. \begin{align*} && f(x)f(x)^* &= \frac{ix-1}{ix+1} \frac{(ix-1)^*}{(ix+1)^*} \\ &&&= \frac{ix-1}{ix+1} \frac{-ix-1}{-ix+1} \\ &&&= \frac{ix-1}{ix+1} \frac{-(ix+1)}{-(ix-1)} \\ &&&= 1
3. \begin{align*} && \begin{pmatrix} i & -1 \\ i & 1 \end{pmatrix}^2 &= \begin{pmatrix} -1-i & -i-1 \\ -1+i & -i+1 \end{pmatrix} \\ \Rightarrow && f(f(z)) &= \frac{-(1+i)(z+1)}{(-1+i)(z-1)} \\ &&&= \frac{2i}{2} \frac{z+1}{z-1} \\ &&&= i \frac{z+1}{z-1} \\ \Rightarrow && \textrm{Re}(f(f(x))) &= 0 \\ && \textrm{Im}(f(f(x))) &= \frac{x+1}{x-1} \end{align*}
4. \begin{align*} && \begin{pmatrix} i & -1 \\ i & 1 \end{pmatrix} \begin{pmatrix} i & i \\ 1 & -1 \end{pmatrix} &= \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \\ \Rightarrow && f(f(f(z))) &= z \end{align*}

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Solution: \begin{align*} && (n^2+1) + (n^2+2) + \cdots + (n+1)^2 &= n^3+(n+1)^3 \\ \Leftrightarrow && \sum_{i=n^2+1}^{(n+1)^2} i &= n^3 + (n+1)^3 \\ && \sum_{i=n^2+1}^{(n+1)^2} i &= \sum_{i=1}^{(n+1)^2} i- \sum_{i=1}^{n^2} i \\ &&&= \frac{(n+1)^2((n+1)^2+1)}{2} - \frac{n^2(n^2+1)}{2} \\ &&&= \frac{(n+1)^2(n^2+2n+2) - n^2(n^2+1)}{2} \\ &&&= \frac{2(n+1)^3+n^2(n^2+2n+1) - n^2(n^2+1)}{2}\\ &&&= \frac{2(n+1)^3+2n^3 + n^2(n^2+1) - n^2(n^2+1)}{2}\\ &&&= (n+1)^3+n^3 \end{align*} \begin{align*} && \sum_{i=1}^{N^2} i &=(0^3+1^3)+ (1^3+2^3)+(2^3+3^3) + \cdots + ((N-1)^3+N^3) \\ &&&= 2 \left (1^3+2^3 + 3^3 + \cdots + (N-1)^3 \right) + N^3 \\ \Rightarrow && \sum_{i=1}^N i^3 &= \frac12 \left ( N^3+ \sum_{i=1}^{N^2} i \right) \\ &&&= \frac12 \left ( N^3 + \frac{N^2(N^2+1)}{2} \right) \\ &&&= \frac{N^2(N^2+1)+2N^3}{4} \\ &&&= \frac{N^2(N^2+2N+1)}{4} \\ &&&= \frac{N^2(N+1)^2}{4} \\ \end{align*}

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Solution: \begin{align*} && I &= \int_0^{\frac14\pi} \ln (1 + \tan \theta) \d \theta \\ \theta = \tfrac14\pi - \phi, \d \theta = -\d\phi: &&&= \int_0^{\frac14 \pi} \ln ( 1 + \tan (\tfrac14\pi - \phi)) \d \phi \\ &&&= \int_0^{\frac14 \pi} \ln \left ( 1 + \frac{1 - \tan \phi}{1+\tan \phi} \right) \d \phi \\ &&&= \int_0^{\frac14 \pi} \ln \left ( \frac{2}{1+\tan \phi} \right) \d \phi \\ &&&= \tfrac14 \pi \ln 2 - I \\ \Rightarrow && I &= \tfrac18\pi \ln 2 \end{align*} \begin{align*} && J &= \int_0^1 \frac{\ln(1+x)}{1+x^2} \d x \\ x= \tan \theta \d \theta, \d \theta = \frac{\d x}{1+x^2} &&&= \int_0^{\frac14 \pi} \ln(1 + \tan \theta) \d \theta \\ &&&= \tfrac18 \pi \ln 2 \end{align*} \begin{align*} && K &= \int_0^{\frac12 \pi} \ln \left ( \frac{1 + \sin x}{1 + \cos x} \right) \d x \\ y = \tfrac12\pi - x, \d y = -\d x: &&&= \int_0^{\frac12\pi} \ln \left ( \frac{1+\cos y}{1+\sin y}\right) \d y \\ &&&= -K \\ \Rightarrow && K &= 0 \end{align*}