Problems

Solution:

1. The curves meet when they have the same radius for a given \(\theta\) ie \begin{align*} && a + 2 \cos \theta &= 2 + \cos 2 \theta \\ &&&= 2 + 2\cos^2 \theta - 1 \\ \Rightarrow && 0 &= 2 \cos ^2 \theta - 2 \cos \theta + 1 - a \end{align*} The curves touch if this has a repeated root, ie \(0 = \Delta = 4 - 8(1-a) \Rightarrow a = \frac12\). The second way the curves can touch is if there is a single root, but it's at an extreme value of \(\cos \theta = \pm 1\) ie \(0 = 2 - 2\cdot(\pm1) + 1 - a \Rightarrow a = 3 \pm 2 = 1, 5\)
2. Suppose \(a = \frac12\) then the curves touch when \(0 = 2\cos^2 \theta - 2 \cos \theta + \frac12 = (2 \cos \theta-1 )(\cos \theta -\frac12) \Rightarrow \theta = \pm \frac{\pi}{3}\)
   

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3. \(a = 1\)
   

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   \(a = 5\)
   

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Solution:

1. The horizontal position of the particle at time \(t\) is \(u \sin\alpha t\), so \(T = \frac{h \tan \beta}{u \sin \alpha}\) The vertical position of the particle at this time \(T\) satisifes: \begin{align*} && h &= u \cos\alpha \frac{h \tan \beta}{u \sin\alpha} - \frac12 g \left ( \frac{h \tan \beta}{u \sin\alpha} \right)^2 \\ &&&= h\cot \alpha \tan \beta - \frac{gh^2}{2u^2} \tan^2 \beta \cosec^2 \beta \\ \Rightarrow && 1 &= c \tan \beta - \frac{1}{k} \tan^2 \beta (1 + c^2) \\ \Rightarrow && k \cot^2 \beta &= kc\cot \beta -1-c^2 \\ \Rightarrow && 0 &= c^2 -ck \cot \beta + 1 + k \cot^2 \beta \end{align*}
   1. As a quadratic in \(c\) the sum of the roots is \(k \cot \beta\), therefore \(\cot \alpha_1 + \cot \alpha_2 = k \cot \beta\). We also have that \(\cot \alpha_1 \cot \alpha_2 = 1 + k \cot^2 \beta\), so \begin{align*} && \cot (\alpha_1 + \alpha_2) &= \frac{\cot \alpha_1 \cot \alpha_2-1}{\cot \alpha_1 + \cot \alpha_2} \\ &&&= \frac{1 + k \cot^2 \beta - 1}{k \cot \beta} \\ &&&= \cot \beta \\ \Rightarrow && \beta &= \alpha_1 + \alpha_2 \pmod{\pi} \end{align*} but since \(\alpha_i \in (0, \frac{\pi}{2})\) the equation must hold exactly.
   2. Since it has two real roots we must have \begin{align*} && 0 &<\Delta = k^2 \cot^2 \beta - 4 (1 + k \cot^2 \beta) \\ &&&= k^2 \cot^2 \beta-4k \cot^2 \beta -4 \\ &&&= \cot^2 \beta (k^2 - 4k - 4(\sec^2 \beta - 1)) \\ &&&= \cot^2 \beta ( (k-2)^2 -4\sec^2 \beta) \\ \Rightarrow && k &> 2 + 2\sec \beta = 2(1+\sec \beta) \end{align*}
2. The greatest height will satisfy \(v^2 = u^2 + 2as\) so \(0 = u^2 \cos^2 \alpha - 2gh_{max} \Rightarrow 4\sec^2 \alpha = \frac{2u^2}{gh_{max}} = k_{max}\), but this decreases with \(h\), so the smallest \(k\) can be is \(4\sec^2 \alpha\), ie \(k \geq 4 \sec^2 \alpha\)

Solution:

1. Let \(x = \frac{pz+q}{z+1}\) then \begin{align*} && 0 &= x^3-3pqx+pq(p+q) \\ &&&= \left ( \frac{pz+q}{z+1} \right)^3 - 3pq \left ( \frac{pz+q}{z+1} \right) + pq(p+q) \\ &&&= \frac{(pz+q)^3-3pq(pz+q)(z+1)^2+pq(p+q)(z+1)^3}{(z+1)^3} \\ &&&= \frac{1}{(z+1)^3} \Big ((p^3+pq(p+q)-3p^2q)z^3 + (3p^2q-6p^2q+3pq^2+3p^2q+3pq^2)z^2 + \\ &&&\qquad \qquad\quad\quad +(3pq^2-3p^2q-6pq^2+3p^2q+3qp^2)z+(q^3-3pq^2+p^2q+pq^2) \Big ) \\ &&&= \frac{(p^3+pq^2-2p^2q)z^3+(q^3+p^2q-2pq^2)}{(z+1)^3} \\ \Rightarrow && 0 &= (p^3+pq^2-2p^2q)z^3+(q^3+p^2q-2pq^2) \\ &&&= p(p-q)^2z^3 + q(p-q)^2 \\ \Rightarrow && 0 &= pz^3 + q \end{align*}
2. We would like to find \(pq = c\) and \(pq(p+q) = d\), so \(p\) and \(q\) are roots of the quadratic \(x^2-\frac{d}{c}x + c = 0\), which has distinct real roots if \(\Delta = \frac{d^2}{c^2}-4c > 0 \Rightarrow d^2>4c^3\)
3. Note that \(c = -2, d = -2\) so \begin{align*} && 0 &= x^3+6x-2 \\ \text{consider} && 0 &= X^2-X-2 \\ && &= (X+1)(X-2) \\ \Rightarrow && p = -1, &q = 2\\ \Rightarrow && 0 &= x^3-3\cdot 2 \cdot(-1) x + 2\cdot(-1) \cdot(-2+1) \\ \Rightarrow && 0 &= -z^3+2 \\ \Rightarrow && z &= \sqrt[3]{2} \\ \Rightarrow && \frac{-z+2}{z+1} &= \sqrt[3]{2} \\ \Rightarrow && -z+2 &= \sqrt[3]{2} z + \sqrt[3]{2} \\ \Rightarrow && z &= \frac{2-\sqrt[3]{2}}{\sqrt[3]{2}+1} \end{align*}
4. \(\,\) \begin{align*} && 0 &= x^3 - 3p^2x + 2p^3 \\ &&&= (x-p)(x^2+px-2p^2) \\ &&&=(x-p)^2(x+2p)\\ \Rightarrow && x &= p, p, -2p \end{align*} Therefore if we have a repeated root to our associated quadratic we can find a cubic of the form \(x^3-3p^2x+2p^3\), but we know this has roots we can find.

Solution:

1. \(\,\)
   

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   Notice that \(1+r+r^2\) ranges from \(\frac34\) to \(3\) over \((-1,1)\) therefore \(\frac34 \leq p < 3\) attaining its minimum but not its maximum. If \(p > 1\) we know we must be on the right branch and hence we can determine \(r\) and \(S\) uniquely. If \(\frac34 < p < 1\) then we must have two possible values for \(r\), satisfying \(r_i^2 + r+1 = p\) and so our two possible values for \(S_i = \frac{1}{1-r_i}\) or \(r_i = 1-\frac{1}{S_i}\) and so \begin{align*} && p &= \left ( 1 - \frac{1}{S} \right)^2 + 1 - \frac1S + 1 \\ \Rightarrow && pS^2 &= (S-1)^2 + S^2-S + S^2 \\ \Rightarrow && 0 &= (3-p)S^2 -3S + 1 \end{align*}
2. \(\,\)
   

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   So \(\frac23 \leq q < 6\) and \(r\) and hence \(T\) is uniquely determined if \(2 \leq q < 6\). if \(\frac23

Solution:

1. \(\,\) \begin{align*} && y&=\frac{x^2+x\sin\theta+1}{x^2+x\cos\theta+1} \\ \Leftrightarrow && 0 &= x^2(y-1) + x(y \cos \theta - \sin \theta) + y-1 \\ \Leftrightarrow && 0 &\leq \Delta = (y\cos \theta - \sin \theta)^2 - 4(y-1)^2 \\ \Leftrightarrow && (y\cos \theta - \sin \theta)^2 &\geq 4(y-1)^2 \end{align*} [Assuming that \(y \neq 1\), if \(y = 1\) then the RHS is \(0\) and it is automatically satisfied]. Notice that \((y\cos \theta - \sin \theta)^2 \leq (y^2+1)(\cos^2 \theta + \sin^2 \theta)\) by Cauchy-Schwarz, so \(y^2 + 1 \geq 4(y-1)^2\). \begin{align*} && y^2 + 1 &\geq 4(y-1)^2 \\ \Leftrightarrow && 0 &\geq 3y^2-8y+3 \\ \text{c.v.} && y&= \frac{8 \pm \sqrt{64-4\cdot3 \cdot 3}}{6} \\ &&&= \frac{4 \pm \sqrt{16-9}}{3} = \frac{4 \pm \sqrt{7}}3 \end{align*} so \(\frac{4-\sqrt{7}}3 \leq y \leq \frac{4+\sqrt7}3\).
2. If \(y = \frac{4+\sqrt7}3\) then \(y - 1 = \frac{1+\sqrt7}3\) and since \(y^2+1 = 4(y-1)^2\) taking square roots we obtain \(\sqrt{y^2+1} = 2(y-1)\). Since equality must hold in our C-S identity, we must have \(\langle y, -1 \rangle\) parallel to \( \langle \cos \theta , \sin \theta \rangle\), ie \(\tan \theta = -\frac{3}{4+\sqrt{7}}\) and \begin{align*} && x & = \frac{-(y \cos \theta - \sin \theta) \pm \sqrt{\Delta}}{2(y-1)} \\ &&&= \frac{\pm2(y-1)}{2(y-1)} \\ &&&= \pm1 \end{align*}

Solution:

1. \(\,\) \begin{align*} && 0 &= (y')^2 -xy'+y\\ \Rightarrow && 0 &= 2y' y'' -y' - xy'' + y' \\ &&&= 2y'y'' - xy'' \\ &&&= y'' (2y'-x) \end{align*} Therefore \(y'' = 0 \Rightarrow y = mx + c\) or \(y' = \frac12 x \Rightarrow x = \frac14x^2 + C\). Plugging these into the original equation we have \(m^2 - xm+mx+c = 0 \Rightarrow c = -m^2\) \(\frac14 x^2 - \frac12 x^2 + \frac14x^2 + C = 0 \Rightarrow C = 0\). Therefore \(4y = x^2\)
2. \begin{align*} && 0 &= (x^2-1)(y')^2 -2xyy'+y^2-1 \\ \Rightarrow && 0 &= 2x(y')^2 +(x^2-1)2y'y'' - 2yy' - 2x(y')^2-2xyy''+2yy' \\ &&&= (x^2-1)2y'y'' -2xyy'' \\ &&&= 2y'' ((x^2-1)y'-xy) \end{align*} Therefore \(y'' = 0\) so \(y = mx + c\) or \begin{align*} && \frac{\d y}{\d x} &= \frac{xy}{x^2-1} \\ \Rightarrow && \int \frac1y \d y &= \int \frac{x}{x^2-1} \d x \\ \Rightarrow && \ln |y| &= \frac12 \ln |x^2-1| + C \\ \Rightarrow && y^2 &= A(x^2-1) \end{align*} Suppose \(y = mx+c\) then we must have \((x^2-1)m^2-2xm(mx+c)+(mx+c)^2 = -m^2+c^2 \Rightarrow c^2 = m^2\) If \(y^2 = A(x^2-1)\) then \(2yy' = 2xA\) and \begin{align*} && 0 &= \frac{y^2}{A}\left ( \frac{xA}{y} \right)^2 - 2x^2A+A(x^2-1)-1 \\ &&&= x^2A-2x^2A+x^2A-A-1 \\ \Rightarrow && A &= -1 \end{align*} Therefore \(x^2 + y^2 = 1\)

Solution:

1. 

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   Since \(y(0) < 0\) and \(y(\pm \infty) > 0\) we must cross the axis twice. Therefore there are two distinct real roots. It is not necessary, for example \((x-2)(x-3)\) has distinct real roots by the constant term is \(6 > 0\)
2. For \(x^2+bx+c=0\) to have distinct, positive real roots we need \(\Delta > 0\) and \(\frac{-b -\sqrt{\Delta}}{2a} > 0\) where \(\Delta = b^2-4ac\), ie \(b < 0\) and \(b^2 > \Delta = b^2-4ac\) or \(4ac > 0\). Therefore we need \(b^2-4ac > 0, b < 0, 4ac > 0\)
3. Since \(q < 0\) at least one of the roots is positive. The gradient is \(3x^2+p > 0\) therefore there is exactly one positive root. If \(p < 0\) then there are turning points when \(3x^2+p = 0\) ie \(x = \pm \sqrt{\frac{-p}{3}}\). If the first turning point is above the \(x\)-axis then there will be 3 roots. If it is on the \(x\)-axis then 2, otherwise only 1. \begin{align*} y &= \left (-\sqrt{\frac{-p}{3}}\right)^3 + p\left (-\sqrt{\frac{-p}{3}}\right)+q \\ &= \sqrt{\frac{-p}{3}} \left (p - \frac{p}{3} \right) + q \\ &= \frac{2}{3} \sqrt{\frac{-p}{3}}p +q \\ \end{align*} Therefore it is positive if \(-\frac{4}{27}p^3 >q^2\) ie if \(4p^3+27q^2 < 0\)

Solution:

1. \(\,\) \begin{align*} && 1 &= \frac{x}{x-a} + \frac{x}{x-b} \\ \Leftrightarrow && (x-a)(x-b) &= x(2x-a-b) \\ \Leftrightarrow && 0 &= x^2-ab \end{align*} Therefore if \(a,b\) are both positive or both negative, \(ab > 0\) and there are two distinct solutions \(x = \pm \sqrt{ab}\)
2. \(\,\) \begin{align*} && 1+c &= \frac{x}{x-a} + \frac{x}{x-b} \\ \Leftrightarrow && (1+c)(x-a)(x-b) &= x(2x-a-b) \\ \Leftrightarrow && 0 &= (c-1)x^2-c(a+b)x+ab(1+c) \\ \\ && 0 &= \Delta = c^2(a+b)^2 - 4 \cdot(c-1)\cdot ab(1+c) \\ &&&= c^2(a+b)^2-4ab(c^2-1) \\ &&&= c^2 ((a+b)^2-4ab)+4ab \\ &&&= c^2(a-b)^2+4ab \\ \Rightarrow && c^2 &= -\frac{4ab}{(a-b)^2} \\ &&&= -\frac{(a+b)^2-(a-b)^2}{(a-b)^2} \\ &&&= 1 - \left ( \frac{a+b}{a-b} \right)^2 \end{align*} Note that \(c^2 \geq 0\) and \(1-x^2 \leq 1\) so \(0 \leq c^2 \leq 1\). \(c^2 = 0 \Rightarrow ab = 0\), but this is not possible since \(a,b \neq 0\), therefore \(0 < c^2 \leq 1\)

Solution: \begin{align*} && y' &= 2ax+b \\ \Rightarrow && m &= 2ax_t+b \\ \Rightarrow && x_t &= \frac{m-b}{2a} \end{align*} Therefore we must have \begin{align*} mx_t &= 2ax_t^2+bx_t \\ y - mx &= ax_t^2+bx_t+c - mx_t \\ &= ax_t^2+bx_t+c - (2ax_t^2+bx_t) \\ &= c - ax_t^2 \\ &= c-a\left (\frac{m-b}{2a} \right)^2 \\ &= c - \frac{(m-b)^2}{4a} \end{align*} They will have a common tangent if and only if the constant terms are equal, ie \begin{align*} && c_1 - \frac{(m-b_1)^2}{4a_1} &= c_2 - \frac{(m-b_2)^2}{4a_2} \\ \Leftrightarrow && (c_1-c_2) &= \frac{(m-b_1)^2}{4a_1} -\frac{(m-b_2)^2}{4a_2} \\ \Leftrightarrow && 4a_1a_2(c_1-c_2) &= a_2(m-b_1)^2-a_1(m-b_2)^2 \\ &&&= (a_2-a_1)m^2+2(a_1b_2-a_2b_1)m+a_2b_1^2-a_1b_2^2 \end{align*} as required. Treating this as a polynomial in \(m\), we can see that the two curves will have exactly one common tangent iff \(\Delta = 0\), ie: \begin{align*} && 0 &= \Delta \\ &&&= (2(a_1b_2-a_2b_1))^2 - 4 (a_2-a_1)(4a_1 a_2(c_2-c_1)+ a_2 b_1^2-a_1 b_2 ^2) \\ &&&= 4a_1^2b_2^2-8a_1a_2b_1b_2+4a_2b_1^2 - 4a_2^2b_1^2-4a_1^2b_2^2 + 4a_1a_2(b_1^2+b_2^2)-16(a_2-a_1)a_1a_2(c_2-c_1) \\ &&&=-8a_1a_2b_1b_2+4a_1a_2(b_1^2+b_2^2)-16(a_2-a_1)a_1a_2(c_2-c_1) \\ &&&=a_1a_2(4(b_1-b_2)^2-16(a_2-a_1)(c_2-c_1)) \\ &&&= 4a_1a_2((b_2-b_1)^2 - 4(a_2-a_1)(c_2-c_1) \end{align*} But this is just the discriminant of the difference, ie equivalent to the two parabolas just touching. (Assuming \(a_1-a_2 \neq 0\) and we do end up with a quadratic). If \(a_1 = a_2 = a\) then we need exactly one solution to \(2a(b_1-b_2)m +4a^2(c_2-c_1)+a(b_1^2-b_2^2) = 0\), ie \(b_1 \neq b_2\).

Solution: There are two cases to consider by they are equivalent to \(x^2 \pm \alpha x + 2 < 0\), which has no solution solutions if \(\Delta < 0\), ie if \(\alpha^2 - 4\cdot1\cdot2 < 0 \Leftrightarrow |\alpha| < 2\sqrt{2}\). If \(\alpha = 3\), we have \begin{align*} && 0 & > x^2-3x+2 \\ &&&= (x-2)(x-1) \\ \Rightarrow && x & \in (1,2) \\ \\ && 0 &> x^2+3x+2 \\ &&& = (x+2)(x+1) \\ \Rightarrow && x &\in (-2,-1) \end{align*} Both cases work here, so \(x \in (-2, -1) \cup (1,2)\). \begin{align*} && 0 &> x^2 \pm \alpha x + 2 \\ &&&= (x \pm \tfrac{\alpha}{2})^2 -\frac{\alpha^2-8}{4} \end{align*} The potential intervals therefore are \((\frac{\alpha -\sqrt{\alpha^2-8}}{2}, \frac{\alpha +\sqrt{\alpha^2-8}}{2})\) and \((\frac{-\alpha -\sqrt{\alpha^2-8}}{2}, \frac{-\alpha +\sqrt{\alpha^2-8}}{2})\). Neither of these intervals overlap with \(0\), since \(\alpha^2 > \alpha^2-8\), and their lengths are both \(\sqrt{\alpha^2-8}\), therefore \(S = 2\sqrt{\alpha^2-8} < 2\alpha\)

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