Problems

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Solution: Let \(f(x) = x^4 + ax^3 + bx^2 + ax + 1\), and suppose \(f(k) = 0\). Since \(f(0) = 1\), \(k \neq 0\), therefore we can talk about \(k^{-1}\). \begin{align*} && f(k^{-1}) &= k^{-4} + ak^{-3} + bk^{-2} + ak^{-1} + 1 \\ &&&= k^{-4}(1 + ak + bk^2 + ak^3 + k^4) \\ &&&= k^{-4}(k^4+ak^3+bk^2+ak+1) \\ &&&= k^{-4}f(k) = 0 \end{align*} Therefore \(k^{-1}\) is also a root of \(f\)

1. If \(f\) has only on distinct root, we must have \(f(x) = (x+k)^4\) therefore \(k = k^{-1} \Rightarrow k^2 = 1 \Rightarrow k = \pm1\), or \(a = 4, b = 6\) or \(a = -4, b = 6\)
2. If \(f\) has exactly three distinct roots then one of the roots must be a repeated \(1\) or \(-1\), ie \(0 = f(1) = 1 + a + b + a + 1 = 2 + b +2a \Rightarrow b = -2a-2\) or \(0 = f(-1) = 1 -a + b -a + 1 \Rightarrow b = 2a - 2\)
3. If \(b = 2a-2\), we have \begin{align*} && f(x) &= 1 + ax + (2a-2)x^2 + ax^3 + x^4 \\ &&&= (x^2+2x+1)(1+(a-2)x+x^2) \\ \Rightarrow && x &= \frac{2-a \pm \sqrt{(a-2)^2 - 4}}{2} \\ &&&= \frac{2-a \pm \sqrt{a^2-4a}}{2} \end{align*}

\(f\) has exactly three distinct real roots iff \(b = \pm 2a - 2\) and \(b \neq 6\)

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Solution: \begin{align*} x_{n+2} &= \frac{ax_{n+1}-1}{x_{n+1}+b} \\ &= \frac{a \frac{ax_n - 1}{x_n+b}-1}{\frac{ax_n - 1}{x_n+b}+b} \\ &= \frac{a(ax_n-1)-(x_n+b)}{ax_n-1+b(x_n+b)} \\ &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \end{align*}

1. If \(x_{n+2} = x_n\) then \begin{align*} && x_n &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \\ \Rightarrow && 0 &=(a+b)x_n^2+(b^2-a^2)x_n+(a+b) \\ &&&= (a+b)(x_n^2+(a-b)x_n + 1) \end{align*} If \(x_{n+1} = x_n\) then \(x_n^2+(a-b)x_n + 1\) and since our sequence has period \(2\) rather than \(1\) it must be the case this is non-zero. Therefore \(a+b =0\).
2. \begin{align*} x_{n+4} &= \frac{(a^2-1)x_{n+2}-(a+b)}{(a+b)x_{n+2}+b^2-1} \\ &= \frac{(a^2-1)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} -(a+b)}{(a+b)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} +b^2-1} \\ &= \frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \end{align*} If \(x_{n+4} = x_n\) then \begin{align*} x_n &=\frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \\ 0 &= (a^2+b^2-2)(a+b)x_n^2 + \l (b^2-1)^2-(a^2-1)^2 \r x_n+(a^2+b^2-2)(a+b) \\ &= (a^2+b^2-2)(a+b)x_n^2+(b^2-a^2)(a^2+b^2-2)x_n + (a^2+b^2-2)(a+b) \\ &= (a^2+b^2-2)(a+b)(x_n^2+(b-a)x_n + 1) \end{align*} Since we do not want \(x_n\) to be periodic with period \(1\) we must have the quadratic in \(x_n\) \(\neq 0\). If \(a+b = 0\) then \(x_n\) is periodic with period \(2\) since \(x_{n+2} = \frac{(a^2-1)x_n}{((-a)^2-1)} = x_n\). Therefore it is necessary that \(a^2+b^2-2 = 0\). If \(a^2+b^2-2= 0\) then \begin{align*} x_{n+4} &= \frac{((a^2-1)^2-(a+b)^2)x_n}{(b^2-1)^2-(a+b)^2} \\ &=\frac{((a^2-1)^2-(a+b)^2)x_n}{((2-a^2)-1)^2-(a+b)^2} \\ &=\frac{((a^2-1)^2-(a+b)^2)x_n}{((1-a^2)^2-(a+b)^2} \\ &= x_n \end{align*} Therefore it is sufficient too. So our conditions are \(a+b \neq 0, \, \, x_n^2+(a-b)x_n + 1 \neq 0\) and \(a^2+b^2-2 = 0\)

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Solution: \begin{align*} && R\cosh(x + \gamma) &=R \cosh x \cosh \gamma + R \sinh x \sinh \gamma \\ \Rightarrow && R \cosh \gamma &= B \\ && R \sinh \gamma &= A \\ \Rightarrow && R^2 &= B^2 - A^2 \\ \Rightarrow && \tanh \gamma &= \frac{A}{B} \\ \end{align*} Therefore it is possible, by writing \(R = \sqrt{B^2-A^2}\) and \(\gamma = \textrm{artanh} \left ( \frac{A}{B} \right)\). This works as long as \(|B| > A > 0\). Supposing \(A >|B| \), try \(S \sinh (x + \delta) = S \sinh x \cosh \delta +S \cosh x \sinh \delta\) \begin{align*} && S \cosh \delta &= A \\ && -S \sinh \delta &= B \\ \Rightarrow && S^2 &= A^2 - B^2 \\ \Rightarrow && \tanh \delta &= \frac{B}{A} \\ \end{align*} Therefore in this case we can write \(\sqrt{A^2-B^2} \sinh \left (x + \tanh^{-1} \left ( \frac{B}{A} \right) \right)\) If \(A = \pm B > 0\) we can we have \(A \sinh x + B \cosh x = \pm Ae^{\pm x}\)

1. Suppose \(y \cosh x = 1\) and \(y \cosh x = a \sinh x +b \cosh x\) so \begin{align*} && 1 & = a \sinh x + b \cosh x \\ &&&= \sqrt{b^2-a^2} \cosh(x + \textrm{artanh} \frac{a}{b} ) \\ \Rightarrow && x + \textrm{artanh} \frac{a}{b} &= \pm \textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) \\ \Rightarrow && x &= \pm \textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) -\textrm{artanh} \frac{a}{b} \end{align*}
2. If \( a > b > 0\) we have \begin{align*} && 1 & = \sqrt{a^2-b^2} \sinh \left ( x - \textrm{artanh} \frac{b}{a} \right) \\ \Rightarrow && x &= \textrm{arsinh} \left ( \frac{1}{\sqrt{a^2-b^2}} \right) + \textrm{artanh} \left ( \frac{b}{a} \right) \end{align*}
3. To intersect at distinct points we must have \(b > a\) and \(\textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) \neq 0\) which is always true.
4. For the curves to touch, we need them to intersect and have matching derivatives, ie \begin{align*} && -\tanh x \cdot \textrm{sech}x &= a\textrm{sech}^2 x \\ \Rightarrow && 0 &= \textrm{sech}^2 x (a + \sinh x) \\ \Rightarrow && x &= -\textrm{arsinh} \, a\\ \Rightarrow && \sinh x &= - a\\ \Rightarrow &&\cosh x &= \sqrt{1 + a^2} \\ \end{align*} So if the curves touch, we must have \(1 = -a^2+b\sqrt{1+a^2} \Rightarrow b = \sqrt{1+a^2}\) and since this does work it is a necessary and sufficient condition. We will also have the \(y\) coordinate is \(\frac{1}{\sqrt{1+a^2}}\)

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Solution:

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2. Notice that there are no solutions when \(x < 0\) since \(f(x) \geq 1\) in that region. Suppose \(x = n + \epsilon, 0 < \epsilon < 1\), then \(f(x) = \frac{n}{n+\epsilon}\), ie \(12n = 7n + 7 \epsilon \Rightarrow 5 n = 7\epsilon \Rightarrow \epsilon = \frac{5}{7}n \Rightarrow n < \frac75\), so \(n = 1 ,\epsilon = \frac57, x = \frac{12}5\). \begin{align*} && \frac{17}{24} &= f(x) \\ \Rightarrow && 17n + 17 \epsilon &= 24 n \\ \Rightarrow && 17 \epsilon &= 7 n \\ \Rightarrow && n &< \frac{17}{7} \\ \Rightarrow && n &= 1, 2 \\ \Rightarrow && x &= \frac{24}{17}, \frac{48}{17} \end{align*}. For \(f(x) = \frac{4}{3}\) we notice that \(x < 0\), so let \(x = -n +\epsilon\), ie \begin{align*} && \frac43 &= f(x) \\ \Rightarrow && \frac43 &= \frac{-n}{-n+\epsilon} \\ \Rightarrow && 4\epsilon &= n \\ \Rightarrow && n &= 1,2,3 \\ \Rightarrow && x &= -\frac{5}{4}, -\frac{3}{2}, -\frac{9}{4} \end{align*}
3. \begin{align*} && \frac9{10} &= f(x) \\ \Rightarrow && 9n + 9 \epsilon &= 10 n \\ \Rightarrow && 9 \epsilon &= n \\ \Rightarrow && n < 9 \end{align} So largest will be when \(n = 8, \epsilon = \frac{8}{9}\), ie \(\frac{80}{9}\)

If \(c < 1\) \begin{align*} && c &= \frac{k}{k + \epsilon} \\ \Rightarrow && \frac{c}{1-c} \epsilon &= k \end{align*} For this to have exactly \(n\) solutions, we need \(n < \frac{c}{1-c} \leq n+1\). If \(c > 1\) \begin{align*} && c &= \frac{-k}{-k+\epsilon} \\ \Rightarrow && c \epsilon &= (c-1) k \\ \Rightarrow && \frac{c}{c-1} \epsilon &= k \end{align*} Therefore for there to be exactly \(n\) solutions we need \(n < \frac{c}{c-1} \leq n+1\)

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Solution: Suppose \begin{align*} && \frac{1}{a+b} &= \frac{1}{a} + \frac{1}{b} \\ \Rightarrow && ab &= b(a+b)+a(a+b) \\ &&&= (a+b)^2 \\ \Rightarrow && 0 &= a^2+ab + b^2 \\ &&&= \tfrac12 (a^2+(a+b)^2+b^2) \end{align*} Which clearly has no solution for non-zero \(a,b\). Suppose \begin{align*} && \frac{1}{a+b+c} &= \frac1a + \frac1b+\frac1c \\ \Leftrightarrow && abc &= (a+b+c)(bc+ca+ab) \\ \Leftrightarrow && 0 &= (a+b+c)(bc+ca+ab) - abc \\ &&&= (a+b)(b+c)(c+a) \end{align*} Therefore it is true iff \(a+b = 0\) or \(b+c=0\) or \(c+a =0\) as required.

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Solution: \begin{align*} y &= x^2-2bx+c \\ &= (x-b)^2+c-b^2 \end{align*} Therefore the turning point is at \((b,c-b^2)\)

Therefore there will be two distinct roots if \(c -b^2 < 0 \Rightarrow c < b^2\). For the equation to have two distinct positive roots it needs to have two distinct roots (ie the condition above) and \(y(0) = c\) needs to be positive, ie \(c > 0\). Therefore the curve has two distinct positive roots if \(0 < c < b^2\). The turning points on \(y = x^3-3b^2x+c\) will have \(0 = y' = 3x^2-3b^2 \Rightarrow x = \pm b\) Therefore for the cubic to have three distinct real root we must have a root between the turning points, \(y(-b) > 0 > y(b)\) \(b^3-3b^3+c = c-2b^3 < 0\) \((-b)^3+3b^3+c = c+2b^3 > 0\) ie \(-2b^3 < c < 2b^3\). The equation \(\l x - a \r^3 - 3b^2 \l x - a \r + c = 0\) will have 3 distinct roots if \(-2b^3 < c < 2b^3\), they will all be positive if the \(y(0) < 0\) and \(a+b > 0\) (ie the first turning point is in the first quadrant, ie \(-a^3+3b^2a+c < 0, a+b>0\). \begin{align*} 2x^3 - 9x^2 + 7x - 1 &= 2(x^3-\frac92x^2+\frac72 x-\frac12) \\ &= 2((x-\frac{3}{2})^3-\frac{27}{4}x+\frac{27}{8}+\frac72x-\frac12) \\ &= 2((x-\frac{3}{2})^3-\frac{13}{4}x+\frac{23}{8}) \\ &= 2((x-\frac{3}{2})^3-\frac{13}{4}(x-\frac{3}{2})-\frac{39}{8}+\frac{23}{8}) \\ &= 2((x-\frac{3}{2})^3-\frac{13}{4}(x-\frac{3}{2})-2) \\ \end{align*} Therefore in our notation \(a = \frac32, b = \sqrt{13/12}, c = 2\). Clearly \(a+b > 0\), so we need to check \(|c| < 2b^3\) which is clearly true as \(b > 1\). Finally we need to check \(y(0) = -1\), so all conditions are satisfied and there are 3 distinct roots.

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Solution: The point \(ze^{i\theta}\) is obtained by rotating the point \(z\) about \(0\) by an angle \(\theta\) anticlockwise. The complex numbers \(\alpha, \beta\) and \(\gamma\) will form an equilateral triangle iff the angles between each side are \(\frac{\pi}{3}\), ie \begin{align*} \begin{cases}{\gamma - \beta} &= e^{i \frac{\pi}{3}}({\beta - \alpha}) \\ {\alpha- \gamma} &= e^{i \frac{\pi}{3}}({\gamma- \beta}) \\ {\beta- \alpha} &= e^{i \frac{\pi}{3}}({\alpha- \gamma})\end{cases} \end{align*} We don't need all these equations, since the first two are equivalent to the third. Combining the first two equations, we have \begin{align*} && \frac{\gamma - \beta}{\beta-\alpha} &= \frac{\alpha-\gamma}{\gamma - \beta} \\ \Leftrightarrow && (\gamma - \beta)^2 &= (\alpha-\gamma)(\beta-\alpha) \\ \Leftrightarrow && \gamma^2 +\beta^2 - 2\gamma \beta &= \alpha\beta-\alpha^2-\gamma\beta+\gamma\alpha \\ \Leftrightarrow && \alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta&=0 \end{align*} as required. If the roots of \(z^{3}+az^{2}+bz+c=0\) are \(\alpha, \beta, \gamma\) then \(\alpha+\beta+\gamma = -a\) and \(\beta\gamma+\gamma\alpha+\alpha\beta = b\). We also have that \(a^2 - 2b = \alpha^2+\beta^2+\gamma^2\). Therefore there roots will lie at the vertices of an equilateral triangle iff \(a^2-3b = 0\)