Problems

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Solution: For \(h(x)\) to be written in this form \(b^2=4ac\). Suppose \(f(x) = 3x^2+4x\), \(g(x) = x^2-2\). then, \begin{align*} && f(x) + \lambda g(x) &= (3+\lambda)x^2+4x - 2 \lambda \\ \Rightarrow && 0 &= 16 + 8(3+\lambda) \lambda \\ \Rightarrow && 0 &= 2+ 3 \lambda + \lambda^2 \\ &&&= (\lambda +1)(\lambda + 2) \\ \Rightarrow && \lambda &= -1 , -2 \\ \end{align*} \begin{align*} && f(x) - g(x) &= 2(x+1)^2 \\ && f(x) -2g(x) &= (x+2)^2 \\ \Rightarrow && g(x) &= 2(x+1)^2 - (x+2)^2 \\ && f(x) &= 4(x+1)^2 - (x+2)^2 \end{align*} Suppose \(f(x) = 3x^2+4x, g(x) = x^2 + \alpha\), then \begin{align*} && f(x) + \lambda g(x) &= (3+\lambda)x^2+4x+\lambda \alpha \\ \Rightarrow && 0 &= 16 -2\lambda \alpha(\lambda + 3) \\ && 0 &= \alpha \lambda^2 +3\lambda-8 \\ \Rightarrow && 0 &= 9 +32 \alpha \\ \Rightarrow && \alpha &= -\frac{9}{32} \end{align*}

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Solution: \begin{align*} && 0 &= 3y^{2}-10xy+3x^{2}+16y-16x+15 \\ \Rightarrow && 0 &= 6y \frac{\d y}{\d x} - 10x \frac{\d y}{\d x} - 10y + 6x+ 16 \frac{\d y}{\d x } - 16 \\ &&&= \frac{\d y}{\d x} \left (6y - 10x +16 \right) - (10y-6x+16) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{5y-3x+8}{3y-5x+8} \\ \Rightarrow && \frac{y-t}{x-s} &= \frac{5t-3s+8}{3t-5s+8} \\ && y &= \left(\frac{5t-3s+8}{3t-5s+8}\right)x -\left(\frac{5t-3s+8}{3t-5s+8}\right)s + t \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(3s^2+3t^2-10st)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(16s-16t-15)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8t-8s+15}{3t-5s+8} \\ \end{align*} While \(x \to \infty\) we still have \(3 \frac{y^2}{x^2} - 10 \frac{y}{x} + 3 + 16 \frac{y}{x^2} - 16\frac{1}{x} + 15 \frac{1}{x^2} = 0\), ie if \(\frac{y}{x} = k\), then \(3k^2 - 10k + 3 \to 0 \Rightarrow k \to 3, \frac13\). Therefore, as \(s \to \infty\) we can write \begin{align*} && y &= \left(\frac{5\frac{t}{s}-3+8\frac{1}{s}}{3\frac{t}{s}-5+8\frac1{s}}\right)x - \frac{8\frac{t}s-8+15\frac{1}{s}}{3\frac{t}{s}-5+8\frac{1}{s}} \\ k \to 3: &&& \to \left(\frac{15-3}{9-5}\right)x - \frac{24-8}{9-5} \\ &&&= 3x - 4 \\ k \to \frac13: && &\to \left(\frac{\frac53-3}{1-5}\right)x - \frac{\frac83-8}{1-5} \\ &&&= \frac13 x - \frac43 \end{align*} Therefore the equations are \(y = 3x-4\) and \(3y=x-4\) The lines are parallel to \(y = 3x\) and \(y = \frac13x\), so by considering the triangles formed with the origin and a point \(1\) along the \(x\) or \(y\) axis we can see the angles are identical. This means the line \(y = x\) is parallel to one axis and \(y = -x\) is parallel to the other. They must meet where our two lines meet which is \((1,-1)\), so our lines are \(y = x-2\) and \(y = -x\)

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Solution: \begin{align*} && x-y &= \frac1z - \frac1y \\ && x-z &= \frac1x - \frac1y \\ && y-z &= \frac1x - \frac1z \\ \Rightarrow && (x-y)(x-z)(y-z) &= \frac{(y-z)(y-x)(z-x)}{x^2y^2z^2} \\ \Rightarrow && x^2y^2 z^2 &= 1 \\ \end{align*} Suppose \(x + \frac1{y} =k \Rightarrow xy + 1 = ky\) Therefore \(y + \frac{1}{z} = y \pm xy = k\) Therefore \(1 \mp y = k(y \mp 1) \Rightarrow k = \pm 1\)

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Solution: Note that \(\cos 4 \phi = \cos \phi, \cos 3 \phi = \cos 2 \phi\) \begin{align*} && LHS & = 2y^2 \\ &&&= 2 \left ( \cos \phi + \cos 2 \phi \right)^2 \\ &&&= 2 \cos ^2 \phi + 2 \cos^2 2 \phi + 4 \cos \phi \cos 2 \phi \\ &&&= \cos 2 \phi+1+ \cos4 \phi+1+2 \left ( \cos \phi + \cos 3 \phi \right) \\ &&&= \cos 2 \phi + 2 + \cos \phi + 2 \cos \phi + 2 \cos 2 \phi \\ &&&= 3(\cos \phi + \cos 2 \phi) + 2 \\ &&&= 3 y + 2 \\ &&&= RHS \end{align*} Therefore \(y\) satisfies \(2y^2 = 3y+2\), which we can solve: \begin{align*} && 0 &= 2y^2-3y-2 \\ &&&= (2y+1)(y-2) \\ \Rightarrow && y &= -\frac12,2 \end{align*} Since \(\cos \phi \neq 1\), \(y \neq 2\), therefore \(y = -\frac12\). \begin{align*} && \sum_{k=0}^{16} \cos k \theta &= \sum_{k=0}^{17} \textrm{Re} \left ( e^{ k \theta i} \right ) \\ &&&= \textrm{Re} \left ( \sum_{k=0}^{16}e^{ k \theta i} \right ) \\ &&&= \textrm{Re} \left ( \frac{1-e^{17 \theta i}}{1-e^{i \theta}} \right ) \\ &&&= 0 \end{align*} Suppose \(z = \cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta\) \begin{align*} z^2 &= \left (\cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta \right)^2 \\ &= \cos^2 \theta + \cos^2 2 \theta + \cos^2 4 \theta + \cos^2 8 \theta \\ & \quad \quad 2( \cos \theta \cos 2 \theta + \cos \theta \cos 4 \theta + \cos \theta \cos 8 \theta + \\ & \quad \quad \quad \cos 2 \theta \cos 4 \theta + \cos 2 \theta \cos 8 \theta + \cos 4 \theta \cos 8 \theta) \\ &= \frac12 \left (\cos 2 \theta + 1+ \cos 4 \theta + 1 + \cos 8 \theta + 1 + \cos 16 \theta + 1 \right ) + \\ &\quad \quad ( \cos \theta + \cos 3 \theta + \cos 3 \theta + \cos 5 \theta + \cos 7 \theta + \cos 9 \theta + \\ & \quad \quad \quad \cos 2 \theta + \cos 6 \theta + \cos 6 \theta + \cos 10 \theta +\cos 4 \theta + \cos 12 \theta ) \\ &= \frac12 z + 2 + \\ & \quad \quad ( \cos 3 \theta + \cos 6 \theta - \cos 8 \theta - \cos 11 \theta \\ & \quad \quad \quad - \cos 13 \theta - \cos 14 \theta - \cos 15 \theta - \cos 16 \theta - 1) \\ &= \frac12 z + 1 - z \\ &= -\frac12 z +1 \end{align*} Therefore \(z\) satisfies \(z^2=-\frac12 z+1 \Rightarrow z = \frac{-\frac12 \pm \sqrt{\frac14+4}}{2} = \frac{-1 \pm \sqrt{17}}{4}\) Therefore \(z = \frac{\sqrt{17}-1}{4}\) since \(z > 0\)

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Solution:

Let \(\displaystyle I_n = \int_0^{\pi/4} \tan^{n} \theta \, \d \theta\), then \begin{align*} I_0 &= \int_0^{\pi/4} \tan \theta \d \theta \\ &= \left [ -\ln \cos \theta \right]_0^{\pi/4} \\ &= -\ln \frac{1}{\sqrt{2}} - 0 \\ &= \frac12 \ln 2 \\ \\ \\ I_{2n+1} &= \int_0^{\pi/4} \tan^{2n+1} \theta \, \d \theta \\&= \int_0^{\pi/4} \tan^{2n-1} \theta \tan ^2 \theta \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta (\sec^2 \theta - 1) \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta \sec^2 \theta - \tan^{2n-1} \theta \, \d \theta \\ &= \left[ \frac{1}{2n} \tan^{2n} \theta \right]_0^{\pi/4} - I_{2n-1} \\ &= \frac1{2n} - I_{2n-1} \end{align*} Therefore we can see that \(\displaystyle I_{2n+1} = (-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right)\). As we can see as \(n \to \infty\), \(I_n \to 0\) Therefore \begin{align*} && 0 &= \tfrac{1}{2}\ln2+\lim_{n \to \infty} \sum_{m=1}^{n}\frac{(-1)^{m}}{2m} \\ \Rightarrow && \tfrac{1}{2}\ln2 &= \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m} \end{align*} \begin{align*} && I_{-1} &= \int_0^{\pi/4} 1 \d \theta \\ &&&= \frac{\pi}{4} \end{align*} Therefore \(\displaystyle I_{2n} = (-1)^n \left ( \frac{\pi}{4} + \sum_{m=1}^n \frac{(-1)^m}{2m-1} \right)\) and since \(I_{2m} \to 0\) the same result follows.

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Solution: \begin{align*} && 0 &= u^2 + 2u \sinh x -1 \\ &&&= u^2 + u(e^x-e^{-x})-e^{x}e^{-x} \\ &&&= (u-e^{-x})(u+e^x) \\ \Rightarrow && u &= e^{-x}, -e^x \end{align*} \begin{align*} && 0 &= \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}\sinh x-1 \\ \Rightarrow && \frac{\d y}{\d x} &= e^{-x}, -e^x \\ \Rightarrow && y &= -e^{-x}+C, -e^x+C \\ y(0) = 0: && C &= 1\text{ both cases } \\ y'(0) > 0: && y &= 1-e^{-x} \end{align*} \begin{align*} && 0 &= \sinh x u^2 + 2u -\sinh x \\ \Rightarrow && u &= \frac{-2 \pm \sqrt{4+4\sinh^2 x}}{2\sinh x} \\ &&&= \frac{-1 \pm \cosh x}{\sinh x} = - \textrm{cosech }x \pm \textrm{coth}x \\ \\ && 0 &= \sinh x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}-\sinh x \\ \Rightarrow && \frac{\d y}{\d x} &= - \textrm{cosech }x \pm \textrm{coth}x \\ \Rightarrow && y &= -\ln \left ( \tanh \frac{x}{2} \right) \pm \ln \sinh x+C \end{align*} For \(x \to 0\) to be defined, we need \(+\), so \begin{align*} && y &= \ln \left (\frac{\sinh x}{\tanh \frac{x}{2}} \right) + C \\ && y &= \ln \left (\frac{2\sinh \frac{x}{2} \cosh \frac{x}{2}}{\tanh \frac{x}{2}} \right)+C \\ &&&= \ln \left (2 \cosh^2 x \right) + C \\ y(0) = 0: && 0 &= \ln 2+C \\ \Rightarrow && y &= \ln(2 \cosh^2 x) -\ln 2 \\ && y &= 2 \ln (\cosh x) \end{align*}

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Solution: Claim \(G\) is a group under matrix multiplication

• (Closure) Suppose \(\mathbf{A}\) and \(\mathbf{B}\) are matrices of that form, then \(\begin{pmatrix} a_1 & b_1 \\ 0 & c_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ 0 & c_2 \end{pmatrix} = \begin{pmatrix} a_1a_2 & a_1b_2 + b_1c_2 \\ 0 & c_1c_2 \end{pmatrix}\), this is clearly of the required form since if \(a_1, a_2, c_1, c_2 \neq 0\) then \(a_1a_2, c_1c_2 \neq 0\)
• (Associative) By inheritance from matrix multiplication
• (Identity) Consider \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) also clearly of the required form.
• (Inverse) Consider \((ac)^{-1}\begin{pmatrix} c & -b \\ 0 & a \end{pmatrix}\), since \(ac \neq 0\) we can assume it has an inverse mod \(5\). therefore we have another matrix of the required form.

There are \(4\) possible values for \(a\) and \(c\) and \(5\) possible values for \(b\), so \(4 \times 4 \times 5 = 80\) elements, so the group is order \(80\). \(G\) is not commutative, consider \(\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix}\) \(\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 0 & 2 \end{pmatrix}\) The elements of order \(1\) or \(2\) satisfy \(\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} = \begin{pmatrix} a^{-1} & -ba^{-1}c^{-1} \\ 0 & c^{-1} \end{pmatrix}\) Therefore \(a^2 = 1, c^2 = 1 \Rightarrow a, c = 1, 4\) and \(b = -ba^{-1}c^{-1} \Rightarrow b = 0\) or , \(ac = -1\), so we have \((a,b,c) = (1,0,1), (4,0,4), (1, *, 4), (4, *, 1)\) So there are \(12\) elements of order \(1\) or \(2\). But this can't be a subgroup since \(12 \not \mid 80\)