Problems

Solution: The tangent to the curve \(y = f(x)\) at \(x = a\) has the equation \(\frac{y-f(a)}{x-a} = f'(a) = g(a)+(a-p)g'(a)\). This passes through \((p,0)\) iff \begin{align*} && \frac{-f(a)}{p-a} &= g(a)+(a-p)g'(a) \\ \Leftrightarrow && -f(a) &= (p-a)g(a) -(a-p)^2g'(a) \\ \Leftrightarrow && -f(a) &= -f(a) -(a-p)^2g'(a) \\ \Leftrightarrow && 0 &= g'(a) \\ \end{align*}

1. In this case \(g(x) = A(x-q)(x-r) = A(x^2-(q+r)x+qr)\) and so we must have that \(g'(a) = 0\), ie \(A(2a-(q+r)) = 0 \Rightarrow 2a = q+r\) The gradient is \(g(a) +(a-p)g'(a) = g(a) = A(a-q)(a-r)\)
2. By the same reasoning, but with \(g(x) = A(x-p)(x-q)\) we have the gradient is \(A(c-p)(c-r)\). This is parallel iff \begin{align*} && (c-p)(c-r) &= (a-q)(a-r) \end{align*} The tangent at \(x = q\) is \(\frac{y-0}{x-q} = A(q-p)(q-r)\) or \( y = A(q-p)(q-r)(x-q)\)

Solution:

1. \(\,\)
   

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   Notice that \begin{align*} && (a+b)^2 &= PQ^2 + (a-b)^2 \\ \Rightarrow && PQ^2 &= 4ab \\ && (b+c)^2 &= QR^2 + (c-b)^2 \\ \Rightarrow && QR^2 &= 4bc \\ && (a+c)^2 &= PR^2 + (a-c)^2 \\ \Rightarrow && PR^2 &= 4ac \\ \Rightarrow && 2\sqrt{ac} &= 2\sqrt{ab}+2\sqrt{bc} \\ \Rightarrow && \frac{1}{\sqrt{b}} &= \frac{1}{\sqrt{c}} + \frac1{\sqrt{a}} \\ \end{align*} Let \(x, y, z = \frac{1}{\sqrt{a}}, \frac1{\sqrt{b}}, \frac{1}{\sqrt{z}}\) so we would like to prove that \(2(x^4+y^4+z^4) = (x^2+y^2+z^2)^2\) or \(x^4+y^4+z^4 = 2x^2y^2+2y^2z^2+2z^2x^2\). We also have \begin{align*} && y &= x+z \\ \Rightarrow &&y^2 &= x^2+z^2+2xz \\ \Rightarrow && (y^2-x^2-z^2)^2 &= 4x^2z^2 \\ \Rightarrow && y^4+x^4+z^4 - 2x^2y^2-2y^2z^2+2x^2z^2 &= 4x^2z^2\\ \Rightarrow && y^4+x^4+z^4 &= 2x^2y^2+2y^2z^2+2z^2x^2 \end{align*}
2. Notice that subject to \(y > z > x\) all these steps are reversible, so we must have the equality we desire

Solution: At the point \((2t, t^2)\) the gradient is \(t\). Suppose \(\tan \theta = t\), then the point \(b\) away in each direction is \(\binom{2t}{t^2} \pm b \binom{\cos \theta}{\sin \theta}\), ie one end can be written in the form \((x,y) = (2\tan \theta - b \cos \theta, \tan^2 \theta - b \sin \theta)\). Notice we must have \(2\tan \alpha- b \cos \alpha= 0 \Rightarrow b = 2 \frac{\sin \alpha}{\cos ^2 \alpha}\), therefore the coordinates are \((2 \tan \alpha - 2 \tan \alpha, \tan^2 \alpha - 2\tan^2 \alpha) = (0, -\tan^2 \alpha)\) and \((4 \tan \alpha, 3\tan^2 \alpha)\)

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Solution:

1. Notice that \(P\) lies on \(C_1C_2\), and that the triangles formed from \(C_iPT_i\) where \(T_i\) are the tangent points are similar, with ratios \(\frac{r_1}{r_2}\). Therefore \(\frac{C_1P}{r_1} = \frac{C_2P}{r_2}\), and hence \(\frac{C_1P}{C_1C_2} = \frac{C_1P}{C_1P-C_2P} = \frac{1}{1-\frac{r_2}{r_1}} = \frac{r_1}{r_1-r_2}\) So we have \(\mathbf{p} = \mathbf{x_1} + (\mathbf{x}_2 - \mathbf{x}_1)\cdot\frac{r_1}{r_1-r_2} = \frac{r_1\mathbf{x}_2 - r_2\mathbf{x}_1}{r_1-r_2}\)
2. Suppose \(\mathbf{x}_3 = \binom{\alpha}{\beta}\) in the basis of \(\{ \mathbf{x}_1, \mathbf{x}_2 \}\), then we can see that \begin{align*} && \mathbf{p} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} \\ && \mathbf{q} &= \frac{r_1(\alpha \mathbf{x}_1 +\beta \mathbf{x}_2) - r_3\mathbf{x}_1}{r_1-r_3} \\ &&&= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\ && \mathbf{r} &=\frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\ && \mathbf{p}-\mathbf{q} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} - \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\ &&&= \frac{1}{(r_1-r_2)(r_1-r_3)} \binom{(r_1-r_3)(-r_2)-(r_1-r_2)(r_1\alpha-r_3)}{(r_1-r_3)r_1 - (r_1-r_2)r_1\beta} \\ &&&= \frac{r_1}{(r_1-r_2)(r_1-r_3)} \binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \\ && \mathbf{q} - \mathbf{r} &= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} - \frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\ &&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(r_2-r_3)(r_1\alpha-r_3) - (r_1-r_3)r_2\alpha)}{(r_2-r_3)r_1\beta - (r_1-r_3)(r_2\beta - r_3)} \\ &&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(-r_2r_3+r_3^2) - \alpha(r_1r_3-r_3r_2)}{r_3(r_1-r_3)-\beta(r_1-r_2)} \\ &&&= \frac{r_3}{(r_1-r_3)(r_2-r_3)}\binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \end{align*} Therefore they are clearly parallel, and hence lie on a line.
3. \(Q\) is halfway between \(P\) and \(R\) if \begin{align*} && \frac{r_1}{(r_1-r_2)(r_1-r_3)} &= \frac{r_3}{(r_1-r_3)(r_2-r_3)} \\ \Leftrightarrow && r_1(r_2-r_3) &= r_3(r_1-r_2) \\ \Leftrightarrow && r_1r_2 - r_1r_3 &= r_1r_3 - r_2r_3 \\ \Leftrightarrow && r_2 &= \frac{2r_1r_3}{r_1+r_3} \end{align*}

Solution: \begin{questionparts} \item The tangent to \(y = \ln x\) is \begin{align*} && \frac{y - \ln x_1}{x - x_1} &= \frac{1}{x_1} \\ \Rightarrow && \frac{x_1y -x_1 \ln x_1}{ x- x_1} &= 1 \\ \Rightarrow && x_1 y - x_1 \ln x_1 &= x - x_1 \end{align*} So to run through the origin, we need \(\ln x_1 = 1 \Rightarrow x_1 = e\) so the line will be \(y = \frac1{e} x\) If \(ma = \ln a \Rightarrow m = \frac{\ln a}{a} = \frac{\ln b}{b} \Rightarrow b \ln a = a \ln b \Rightarrow a^b = b^a\). \item

Solution: The condition is that we match the first and second derivative (as well as passing through the point in question, which is \((\frac{\pi}{4}, 2 - \frac{\pi}{4})\) The gradient is \(y' = -1 + \sec^2 x\), so the value is \(1\). The second derivative is \(y'' = 2 \sec^2 x \tan x\), which is \(4\) If we have a circle, radius \(r\), so \((x-a)^2 + (y-b)^2 = r^2\) then \(2(x-a) + 2(y-b) \frac{\d y}{\d x} = 0\) and \(2 + 2 \left ( \frac{\d y}{\d x} \right)^2 + 2(y-b) \frac{\d^2y}{\d x^2} = 0\). Therefore we must have \(1+1+(2-\frac{\pi}{4}-b)4 = 0 \Rightarrow b =\frac52-\frac{\pi}{4}\) We know that the centre lies on the line \(y = 2-x\), so we must have \(a = \frac{\pi}{4}-\frac12\) and so the centre is \(( \frac{\pi}{4} - \frac12,\frac52 - \frac{\pi}{4})\) and the radius is \(\sqrt{\frac14 + \frac14} = \frac{\sqrt{2}}{2}\)

Solution:

1. This is a circle centre \((2t,t)\) with radius \(t\). Therefore it is exactly \(t\) away from the line \(y = 0\) so just touches that line. Not that \(\tan \alpha = \frac{t}{2t} = \frac12\) so \(\tan 2\alpha = \frac{2\tan \alpha}{1-\tan^2\alpha} = \frac{1}{1-\frac14} = \frac43\). Therefore the line \(y = \frac43x\) or \(3y = 4x\) is tangent to \(C\).
2. Note that \(3y=4x\) and \(4y+3x=15\) are perpendicular, so this is a right-angled triangle with incenter \((2t,t)\) for some \(t\) and hypotenuse \(15\) We can find the third coordinate when \(3y-4x = 0\) and \(4y+3x = 15\) meet, ie \((\frac{9}{5}, \frac{12}5)\) The incentre lies on the bisector of the right angle at this point, which is the line through \((\frac{9}{5}, \frac{12}5)\) and \((\frac{15}{2}, 0)\), so \begin{align*} && \frac{2t-\frac{12}{5}}{t - \frac{9}{5}} &= \frac{-\frac{12}{5}}{\frac{15}2-\frac95} \\ \Rightarrow && \frac{10t-12}{5t-9} &= \frac{-24}{57} = -\frac{8}{19} \\ \Rightarrow && 190t - 12 \cdot 19 &= -40t + 72 \\ \Rightarrow && t &= 2 \end{align*} Therefore the center is \((4, 2)\) and the equation is \((x-4)^2+(y-2)^2=2^2\)

Solution:

1. Clearly \(x = 0\) and \(x = 4\) are vertical asymptotes. Notice that \(\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r}\) tends to \(0\) as \(x \to \infty\). Therefore the oblique asymptote is \(y = x-4\).
2. \begin{align*} && 0 &= \frac{x^2(x-4)^2-4^2(2x+1)^2}{x^2(x-4)} \\ &&&= \frac{(x(x-4)-4(2x+1))(x(x-4)+4(2x+1))}{x^2(x-4)} \\ &&&= \frac{(x^2-12x-4)(x^2+4x+4)}{x^2(x-4)}\\ &&&= \frac{(x^2-12x-4)(x+2)^2}{x^2(x-4)} \end{align*} Therefore \(f(x) = 0\) has a double root at \(x = -2\). Notice it also has roots at \(6 \pm 2\sqrt{10}\)
3. 

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Solution:

1. \(\,\)
   

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   Notice the tangent must hit the \(y\)-axis above the origin, ie \begin{align*} && 0 &< R'(t)(0-t) + R(t) \\ \Rightarrow && R'(t) t &< R(t) \\ \Rightarrow && t &< \frac{R(t)}{R'(t)} = \frac{1}{H(t)} \end{align*}
2. Suppose \(H(t) = a/t\) then \begin{align*} && \frac{R'}{R} &= \frac{a}{t} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{a}{t} \d t \\ \Rightarrow && \ln R &= a \ln t + C \tag{t, R > 0} \\ \Rightarrow && R &= Kt^a \end{align*} Since we need \(R(t) > 0\), \(K > 0\), since \(R'(t) > 0\) we need \(a > 0\), since \(R''(t) < 0\) we need \(a(a-1) < 0\) ie \(0 < a < 1\)
3. Suppose instead \(H(t) = bt^{-2}\) then \begin{align*} && \frac{R'}{R} &= \frac{b}{t^2} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{b}{t^2} \d t \\ \Rightarrow && \ln R &= -bt^{-1} + C \tag{R > 0} \\ \Rightarrow && R &= Ke^{-b/t} \end{align*} Since \(R > 0\) we must have \(K > 0\). \begin{align*} R' > 0: && R' &= K(b/t^2)e^{-b/t} > 0 \\ \Rightarrow && b &> 0 \\ R'' < 0: && R'' &= K(b^2/t^4)e^{-b/t} -K2b/t^3 e^{-b/t} \\ &&&= Kb/t^4 (b-2t)e^{-b/t} < 0 \\ \Rightarrow && b &< 2t\\ \Rightarrow && b &< 2t \end{align*} which cannot be true for all \(t\), ie there is no \(b\) which satisfies this.

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: \begin{align*} && \cot x - \tan x &= 2 \cot 2x \\ \Rightarrow && x\cot x - x\tan x &= 2x\cot 2x \\ \Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n - \sum_{n=0}^{\infty}a_n x^{n+1} &= 1 + \sum_{n=1}^{\infty} b_n (2x)^n \\ \Rightarrow && \sum_{n=1}^{\infty}(1-2^n)b_nx^n &= \sum_{n=1}^{\infty} a_{n-1}x^n \\ \Rightarrow && a_{n-1} &= (1-2^n)b_n \quad \text{if }n \geq 1 \end{align*} \begin{align*} \cot x + \tan x &= 2 \cosec 2x \end{align*} So \begin{align*} && \cot x + \tan x &= 2 \cosec 2x \\ \Rightarrow && x \cot x + x\tan x &= 2x \cosec 2x \\ \Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n + \sum_{n=0}^{\infty} a_n x^{n+1} &= 1+\sum_{n=1}^\infty c_n (2x)^n \\ \Rightarrow && \sum_{n=1}^{\infty} \frac{1}{1-2^n}a_{n-1} +\sum_{n=1}^{\infty}a_{n-1}x^n &= \sum_{n=1}^{\infty} 2^nc_n x^n \\ \Rightarrow && c_n &= \frac{1}{2^n} \left ( 1 + \frac{1}{1-2^n} \right)a_{n-1} \\ &&&= \frac1{2^n} \frac{2^n-2}{2^n-1} a_{n-1}\\ &&&= \frac1{2^{n-1}}\frac{2^{n-1}-1}{2^n-1} a_{n-1} \end{align*}

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*}