Problems

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Solution:

1. \(\,\)
   

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   Note that we have a right angled triangle, with the sides in a ratio of \(m\). So if our target length is \(x\) we have \(x^2 + (mx)^2 = c^2\) and so \(x = c(m^2+1)^{-\frac12}\)
2. The distance from the origin to \(L\) is \(a = c(m^2+1)^{-\frac12}\) so \begin{align*} && a^2(m^2+1) &= c^2 \\ && \frac{c-y(x)}{0-x} &= y' \\ \Rightarrow && c-y &= -xy' \\ \Rightarrow && a^2((y')^2+1) &= (y-xy')^2 \\ \\ && 2a^2y'y'' &= 2(y-xy')(y'-xy''-y') \\ &&&= 2(xy'-y)xy'' \\ \Rightarrow && y'' &= 0 \\ \text{ or } && 2a^2y' &= 2(xy'-y)x \end{align*} If \(y'' = 0\) then \(y = mx + c\) and the result follows immediately. \begin{align*} && 0 &= (a^2-x^2)y' + yx \\ \Rightarrow &&\frac1{y} y' &= -\frac{x}{a^2-x^2} \\ \Rightarrow && \ln y &= \frac12\ln (a^2-x^2) + K \\ \Rightarrow && y^2 &= M(a^2-x^2) \\ \Rightarrow && x^2 + y^2 &= a^2 \end{align*} Where in the last step we know the tangents from an ellipse are not all equidistant to the origin.
3. 

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Solution: \(c_r = \binom{n}{r}a^{n-r}b^r\) \begin{align*} && c_m &\geq c_{m+1} \\ \Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m+1} a^{n-m-1}b^{m+1} \\ \Rightarrow && \frac{1}{(n-m)}a &\geq \frac{1}{m+1}b \\ \Rightarrow && (m+1)a &\geq (n-m)b \\ \Rightarrow && m(a+b) &\geq nb -a \\ \Rightarrow && m &\geq \frac{n(b+1)-a-b}{a+b}=\frac{n(b+1)}{a+b} - 1 \\ \\ && c_m &\geq c_{m-1} \\ \Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m-1} a^{n-m+1}b^{m-1} \\ \Rightarrow && \frac{1}m b &\geq \frac{a}{(n-m+1)} \\ \Rightarrow && (n-m+1)b &\geq ma \\ \Rightarrow && (n+1)b &\geq m(a+b) \\ \Rightarrow && m &\leq \frac{(n+1)b}{a+b} \end{align*} Since \(m\) lies between two values \(1\) apart is is either equal to one of those two values or is the unique integer between them. Let \(\displaystyle G(n,a,b) = \left \lfloor \frac{b(n+1)}{a+b} \right \rfloor\), so

1. \(\,\) \begin{align*} && G(9,1,3) &= \left \lfloor \frac{3(9+1)}{1+3} \right \rfloor \\ &&&= \left \lfloor \frac{30}{4} \right \rfloor \\ &&&= 7 \\ \\ && G(9,2,3) &= \left \lfloor \frac{3(9+1)}{2+3} \right \rfloor \\ &&&= \left \lfloor \frac{30}{5} \right \rfloor \\ &&&= 6 \end{align*}
2. \(\,\) \begin{align*} && G(2k, a, a) &= \left \lfloor \frac{a(2k+1)}{a+a} \right \rfloor \\ && &= \left \lfloor \frac{2k+1}{2} \right \rfloor \\ &&&= k \\ \\ && G(2k-1, a, a) &= \left \lfloor \frac{a(2k-1+1)}{a+a} \right \rfloor \\ && &= \left \lfloor k\right \rfloor \\ &&&= k \\ \end{align*}
3. \(G(n,a,b)\) is decreasing in \(a\), therefore take \(a = 1\).
4. For fixed \(n\), we are looking at \(\frac{a(n+1)}{a+b}\) and we want this to be as large as possible. By considering \((n+1) - \frac{b(n+1)}{a+b}\) we can see this is increasing in \(b\) and the largest value possible is \(n\). This is achieved when \begin{align*} && \frac{b(n+1)}{a+b} & \geq n \\ \Leftrightarrow && bn + b &\geq na + bn \\ \Leftrightarrow && b& \geq na \end{align*}

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Solution:

1. Suppose we have the shortest distance between the two curves, and the path between the points is not a normal to both curves. Then we could shift the endpoints to reduce the distance. (Assuming we're not at a point of intersection). Therefore, the normal to the curves must be the same (or in other words) the gradients of the curves must be the same. ie we are at a point where \(2y y' = 4a\) we must have \(y' = m\), so \(y = \frac{2a}{m}\) and \(x = \frac{a}{m^2}\) and the distance from this point to the line \(y=mx+c\) is \(\frac{|m \frac{a}{m^2} - \frac{2a}{m}+c|}{\sqrt{m^2+1}} = \frac{|mc-a|}{m\sqrt{m^2+1}} = \frac{mc-a}{m\sqrt{m^2+1}}\). If \(mc \leq a\) then we find \(\frac{a-mc}{m\sqrt{m^2+1}}\) However, we must check that the two curves do not intersect (otherwise the closest distace is \(0\)). ie we need to check if \((mx+c)^2 = 4ax\) has any solutions, this quadratic has discriminant \((2mc-4a)^2 - 4 \cdot m^2 \cdot c^2 = 16a^2-16amc = 16a(a-mc)\) which is clearly greater than \(0\) when \(a \geq mc\). Therefore the shortest distance in this case is \(0\).
2. The distance squared between the point \((p,0)\) and a point of the form \((at^2,2at)\) is \(D^2 = (at^2-p)^2+4a^2t^2 = a^2t^4+(4a^2-2ap)t^2+p^2\) \begin{align*} && \frac{D^2}{a^2} &= t^4 + 2\left(2-\frac{p}{a}\right)t^2 + \frac{p^2}{a^2} \\ &&&= \left (t^2 - \left (\frac{p}{a}-2 \right)\right)^2 + \frac{p^2}{a^2} - \left (2-\frac{p}{a} \right)^2 \\ &&&= \left (t^2 - \left (\frac{p}{a}-2 \right)\right)^2 +\frac{4p}{a} -4 \\ \end{align*} Therefore if \(2 \leq \frac{p}{a}\) then we can find a \(t\) such that we attain the minimum for \(D^2/a^2\) of \(\frac{4p}{a}-4\) and so \(D = \sqrt{4pa-4a^2} = 2\sqrt{a(p-a)}\) . If not the smallest value will be when \(t = 0\) and we will have \(|p|\) Now consider all the lines joining points on the parabola to the centre of the circle. The shortest distance from the parabola to the circle will be normal to the circle and therefore will also be a line through the center. Therefore we need only consider the shortest distance from \((p,0)\) to the parabola \(-b\). Case 1: If \(p \geq 2a\) we have \(2\sqrt{a(p-a)} - b\) or \(0\) if \(b \geq 2\sqrt{a(p-a)}\) Case 2: If \(p < 2a\) we have \(p-b\) or \(0\) if \(b \geq p\)

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Solution:

1. \begin{align*} && I - I_1 &= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) \d x - \int_0^1 \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) - \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left (-y^2-2yy' \tan x - y^2 \tan^2 x \right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2(1+ \tan^2 x )\right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2 \sec^2 x\right) \d x\\ &&&= \int_0^1 -\frac{\d}{\d x} \left (y^2 \tan x \right) \d x\\ &&&= \left [-y^2 \tan x \right]_0^1 \\ &&&= 0 \\ \\ \Rightarrow && I &= I_1 = \int_0^1 \left ( y' + y \tan x \right)^2 d x \geq 0 \end{align*} The only way \(I_0 = 0\) is is \(y' + y \tan x =0\), so \begin{align*} && \frac{\d y}{\d x} &= - y \tan x \\ \Rightarrow && \int \frac{1}{y} &= \int -\tan x \d x \\ \Rightarrow && \ln |y| &= \ln |\cos x| + C \\ \Rightarrow && y &= A \cos x \\ \Rightarrow && A &= 0 \Rightarrow y = 0 \end{align*}
2. Let \(J_1 = \int_0^1 (y'+ay\tan ax)^2 \d x\), then \begin{align*} && J-J_1 &= \int_0^1 \left ( \left ( y' \right)^2 - a^2y^2 \right) - \left ( y' + ya \tan ax \right)^2 \d x\\ &&&= \int_0^1 \left (-a^2y^2-2yy' a \tan a x-y^2a^2 \tan^2 ax \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2(1+\tan^2 ax) \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2\sec^2 ax \right) \d x \\ &&&= \left [ - a y^2 \tan a x \right]_0^1 = 0 \end{align*} This is true if \(a < \frac{\pi}{2}\), since otherwise we might care about the order of the zero for \(y\) at \(x = 1\). Consider \(y = \cos \frac{\pi}{2} x\), then \(y' = -\frac{\pi}{2} \sin^2\frac{\pi}{2} x\) and \begin{align*} && \int_0^1 \frac{\pi^2}{4} \left (\sin^2 \frac{\pi}{2}x - \cos^2 \frac{\pi}{2} x \right) \d x &= -\frac{\pi^2}{4} \int_0^1 \cos(\pi x) \d x \\ &&&= 0 \end{align*}

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Solution:

\begin{align*} \text{N2}(\uparrow): && T_A \cos \alpha - T_B \cos\alpha - mg &= 0 \\ \Rightarrow && T_A \cos \alpha - T_B \cos\beta &= mg \\ \text{N2}(\leftarrow, \text{radially}): && T_A \sin \alpha + T_B \sin \beta &= m x \sin \alpha \omega^2 \\ \Rightarrow && T_A(\cos \alpha \sin \beta+\sin \alpha \cos \beta) &= mg \sin \beta + mx \sin \alpha \omega^2 \cos \beta \\ \Rightarrow && T_A &=\frac{mg\sin \beta + m x \sin \alpha \omega^2 \cos \beta }{\sin(\alpha + \beta)} \\ \Rightarrow && T_B(\sin \beta \cos \alpha- \cos \beta \sin \alpha)&= mx \sin \alpha \omega^2 \cos \alpha -mg \sin \alpha \\ \Rightarrow && T_B &= \frac{m x \sin \alpha \omega^2 \cos \alpha - mg \sin \alpha}{\sin(\beta - \alpha)} \\ &&&= \frac{m \sin \alpha(\omega^2 \cos\alpha - g)}{\sin (\beta - \alpha)} \end{align*} Since \(T_B \geq 0 \Rightarrow \omega^2 \cos\alpha - g \geq 0\) as required. \(\sqrt{h^2-d^2}\) is the length of the final side on the dashed right angle triangle with hypotenuse \(AB\). \(h \cos \alpha\) will be clearly longer as the angle \(\alpha\) will be smaller and so \(\cos \alpha\) will be larger. When \(\omega^2 x \cos \alpha = g\) we must have \(T_B = 0\). \(T_A\cos \alpha = mg \Rightarrow T_A > mg\) since \(\alpha \neq 0\). \(T_A = \frac{mg}{\cos \alpha} \leq \frac{mgh}{\sqrt{h^2-d^2}}\) \(T_A\) will attain it's upper bound when \(\angle APB\) is a right angle.

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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

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Solution:

1. First, not that the point must be ony the curve:
   

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   Since otherwise it's pretty clear we could make the area of the rectangle larger by moving the point onto the curve. Therefore \(A = x(f(x)-f(t))\). To maximise this we need \(xf'(x) + f(x)-f(t) = 0\), ie \(x_0f'(x_0) + f(x_0) = f(t)\)
2. Suppose \(\displaystyle \g(t) =\frac 1t \int_0^t \f(x) \d x\) then \begin{align*} && \g(t) &=\frac 1t \int_0^t \f(x) \d x\\ \Rightarrow && tg(t) &= \int_0^t \f(x) \d x \\ \Rightarrow && tg'(t) +g(t) &= f(t) \\ \Rightarrow && tg'(t) &= f(t) - g(t) \end{align*}
   

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   Clearly the blue area + green area is larger than the green area. So \(\displaystyle \int_0^t (f(x) - f(t))\d x > A_0(t)\). Notice that \(f(t) = \frac1{t} \int_0^t f(t) \d x \) so \(-t^2g'(t) = \int_0^t f(x) \d x > A_0(t)\)
3. Not that if \(f(x) = \dfrac{1}{1+x}\), the \(f'(x) = -\frac{1}{(1+x)^2}\) and so \begin{align*} && -\frac{x_0}{(1+x_0)^2} + \frac{1}{1+x_0} &= \frac{1}{1+t} \\ && \frac{1}{(1+x_0)^2} &= \frac{1}{1+t} \\ \Rightarrow && x_0 &= \sqrt{1+t} - 1 \\ && A_0(t) &= (\sqrt{1+t} - 1) \left ( \frac{1}{\sqrt{1+t}} - \frac{1}{t+1} \right) \\ &&&= 1 - \frac{1}{\sqrt{1+t}} - \frac{1}{\sqrt{1+t}} + \frac{1}{1+t} \\ &&&= \frac{2+t}{1+t} - \frac{2}{\sqrt{1+t}} \\ && g(t) &= \frac{1}{t} \int_0^t \frac{1}{1+x} \d x \\ &&&= \frac{\ln(1+t)}{t} \\ \Rightarrow && g'(t) &= \frac{\frac{t}{1+t} - \ln(1+t)}{t^2} \\ \Rightarrow && -t^2g(t) &= \ln(1+t) - \frac{t}{1+t} \\ \Rightarrow && \ln(1+t) - \frac{t}{1+t} &> \frac{2+t}{1+t} - \frac{2}{\sqrt{1+t}} \\ \Rightarrow && \ln \sqrt{1+t} & > 1 - \frac{1}{\sqrt{1+t}} \end{align*}

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Solution: \begin{align*} && b &= c - ma \\ \Rightarrow && c &= b+ma \\ \Rightarrow && y &= m(a-x)+b \\ \Rightarrow && q &= ma+b \\ && p &= \frac{ma+b}{m} \\ \\ && d &= p+q \\ &&&= a + \frac{b}{m} + ma + b \\ \Rightarrow && d' &= -bm^{-2}+a \\ \Rightarrow && m &= \sqrt{b/a} \\ \\ \Rightarrow &&d &= a + \sqrt{ba}+\sqrt{ba} + b \\ &&&= (\sqrt{a}+\sqrt{b})^2 \\ \\ && |PQ|^2 &= p^2 + q^2 \\ &&&= a^2 + \frac{2ab}{m} + \frac{b^2}{m^2} + m^2a^2 + 2mab + b^2 \\ &&&= a^2+b^2 + \frac{b^2}{m^2} + \frac{2ab}{m}+ 2abm + a^2m^2 \\ && \frac{\d}{\d m}&= -2b^2m^{-3}-2abm^{-2}+2ab + 2a^2m \\ && 0 &=2a^2m^4+2abm^3-2abm-2b^2 \\ &&&= 2(am^3-b)(am+b) \\ \Rightarrow && m &= \sqrt[3]{\frac{b}{a}} \\ \\ &&|PQ|^2 &= \left[ a^{1/3}(a^{2/3} + b^{2/3}) \right]^2 + \left[ b^{1/3}(a^{2/3} + b^{2/3}) \right]^2 \\ &&&= a^{2/3}(a^{2/3} + b^{2/3})^2 + b^{2/3}(a^{2/3} + b^{2/3})^2 \\ &&&= (a^{2/3} + b^{2/3})^2 \cdot (a^{2/3} + b^{2/3}) \\ &&&= (a^{2/3} + b^{2/3})^3 \\ \Rightarrow && |PQ| &= (a^{2/3} + b^{2/3})^{3/2} \end{align*} We can also do this with AM-GM instead: \begin{align*} && d &= a + b + \frac{b}{m} + am \\ &&&\geq a+b + 2 \sqrt{\frac{b}{m} \cdot am} \\ &&&= a+2\sqrt{ab}+b \\ \\ && |PQ|^2 &= a^2+b^2 + \frac{b^2}{m^2} + \frac{2ab}{m}+ 2abm + a^2m^2 \\ &&&= a^2+b^2 + \frac{b^2}{m} + abm + abm + a^2m^2 + \frac{ab}{m} + \frac{ab}{m} \\ &&&= a^2+b^2 + 3\sqrt[3]{ \frac{b^2}{m} \cdot abm \cdot abm} + 3 \sqrt[3]{ a^2m^2 \cdot \frac{ab}{m} \cdot \frac{ab}{m} } \\ &&&= a^2 + 3b^{4/3}a^{2/3}+3b^{2/3}a^{4/3}+b^2 \\ &&&= (a^{2/3}+b^{2/3})^3 \end{align*}

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Solution:

1. \(\,\)
   

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   1. \(n = 0\) if \(b > 9\)
   2. \(n = 1\) is not possible, since by symmetry if \(x\) is a root, so is \(-x\), and \(0\) can never be the only root.
   3. \(n = 2\) if \(b < 0\) or \(b = 9\)
   4. \(n = 3\) if \(b = 0\)
   5. \(n = 4\) if \(0 < b < 9\)
2. \(\,\) \begin{align*} && y' &= 4x^3-12x+a \\ && y'' &= 12x^2-12 \\ \Rightarrow && x &= \pm 1 \\ \Rightarrow && 0 &= 4(\pm 1) - 12 (\pm 1) + a \\ &&&= a \mp 8 \\ \Rightarrow && a &= \pm 8 \end{align*} When \(a = 8\), we have \(y = x^4-6x^2+8x\) and \begin{align*} &&y' &= 4x^3-12x+8 \\ &&&= 4(x^3-3x+2) \\ &&&= 4(x-1)^2(x+2) \\ \Rightarrow && y(1) &= 3\\ && y(-2) &= -24 \end{align*}
   

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   Therefore there are no solutions if \(b > 24\), one solution if \(b = 24\) and two solutions otherwise. Similarly, if \(a = -8\), we have \(y = x^4 - 6x^2-8x\) \begin{align*} && y' &= 4x^3-12x-8 \\ &&&= 4(x^3-3x-2) \\ &&&= 4(x-2)(x+1)^2 \end{align*} So we have stationary points at \(x = 2\) and \(x = -1\) (which is also a inflection point) and at \(x = 2\) \(y = -24\), so we have the same story: there are no solutions if \(b > 24\), one solution if \(b = 24\) and two solutions otherwise.
3. \(\,\)
   

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Solution: \begin{align*} && s &= ut + \frac12 gt^2 \\ \Rightarrow && -2.5 &= 10 \sin \theta \, T - 5 T^2 \\ \Rightarrow && T &= \frac{10\sin \theta \pm \sqrt{100\sin^2 \theta - 4 \cdot 5 \cdot (-2.5)}}{10} \\ &&&= \sin \theta +\sqrt{\sin^2 \theta + \frac12} \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{2} \sin \theta +\sqrt{2 \sin^2 \theta +1} \right) \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{2 (1-\cos^2 \theta)} + \sqrt{2-\cos 2\theta} \right) \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{1-\cos2 \theta} + \sqrt{2-\cos 2\theta} \right) \\ &&&= \frac{1}{\sqrt{2}}\left ( \sqrt{1-c}+\sqrt{2-c} \right)\\ \\ && s &= 10 \cos \theta T \\ &&&= 10 \sqrt{\frac{\cos 2 \theta +1}{2}}\frac{1}{\sqrt{2}}\left ( \sqrt{1-c}+\sqrt{2-c} \right) \\ &&&= 5 \sqrt{c+1}\left ( \sqrt{1-c}+\sqrt{2-c} \right) \\ \\ && \frac15\frac{\d s}{\d c} &= \frac12(c+1)^{-1/2}((1-c)^{1/2} + (2-c)^{1/2}) - \frac12(c+1)^{1/2}\left ((1-c)^{-1/2}+(2-c)^{-1/2} \right) \\ &&&= \frac{((1-c)(2-c)^{1/2}+(2-c)(1-c)^{1/2})-((c+1)(2-c)^{1/2}+(c+1)(1-c)^{1/2})}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ &&&= \frac{\sqrt{2-c}\left (1-c-c-1 \right)+\sqrt{1-c}\left(2-c-c-1) \right)}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ &&&= \frac{\sqrt{1-c}\left(1-2c\right)-2c\sqrt{2-c}}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ \\ \frac{\d s}{\d c} =0: && \sqrt{1-c}\left(1-2c\right)&=2c\sqrt{2-c} \\ \Rightarrow && (1-c)(1-2c)^2&=4c^2(2-c) \\ \Rightarrow && 1-5c+8c^2-4c^3 &= 8c^2-4c^3 \\ \Rightarrow && 0 &= -5c+1 \\ \Rightarrow && c &= \frac15 \end{align*} When \(\theta = 45^{\circ}, c = 0\), so \(s_{45^{\circ}} = 5(1+\sqrt{2})\) When \(c = \frac15\), \begin{align*} s &= 5 \sqrt{\frac15+1}\left ( \sqrt{1-\frac15}+\sqrt{2-\frac15} \right) \\ &= 5 \sqrt{\frac65} \left ( \sqrt{\frac45} + \sqrt{\frac95} \right) \\ &= 2\sqrt{6}+3\sqrt{6} = 5\sqrt{6} \end{align*} Therefore the additional distance is \(5(\sqrt{6}-\sqrt{2}-1)\)

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Solution: \begin{align*} && s &= ut \\ \Rightarrow && a &= u \cos \alpha t\\ \Rightarrow && t &= \frac{a}{u \cos \alpha}\\ && s &= ut+ \frac12at^2 \\ \Rightarrow && -h &< -u\sin \alpha \frac{a}{u \cos \alpha}-\frac12 g \left (\frac{a}{u \cos \alpha} \right)^2 \\ &&&= -a \tan \alpha-\frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha \\ \Rightarrow && \frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha &< h -a\tan \alpha \\ \Rightarrow &&\frac{1}{u^2} &< \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \end{align*} \begin{align*} && s &= ut + \frac12a t^2 \\ \Rightarrow && 2h &= u\sin \alpha t + \frac12 gt^2 \\ \Rightarrow && t &= \frac{-u\sin \alpha \pm \sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && t &= \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && s &= ut \\ \Rightarrow && b &> u \cos \alpha t \\ \Rightarrow && \frac{b}{u \cos \alpha} &> \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g} \\ \Rightarrow && \sqrt{u^2 \sin^2 \alpha+4hg} &< \frac{bg}{u \cos \alpha} + u \sin \alpha \\ \end{align*} \begin{align*} \Rightarrow && u^2 \sin^2 \alpha+4hg &< \frac{b^2g^2}{u^2 \cos^2 \alpha} +u^2 \sin^2 \alpha + 2bg \tan \alpha \\ \Rightarrow && 4hg - 2bg \tan \alpha &< \frac{b^2g^2}{u^2 \cos^2 \alpha} \\ &&&< \frac{b^2g^2}{\cos^2 \alpha} \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \\ &&&= \frac{2b^2g(h-a\tan \alpha)}{a^2} \\ \Rightarrow && \tan \alpha \left (\frac{2b^2g}{a} - 2bg \right) &< \frac{2b^2gh}{a^2} - 4hg \\ \Leftrightarrow && \tan \alpha \left (\frac{2b^2g- 2abg}{a} \right) &< \frac{2b^2gh- 4hga^2}{a^2} \\ \Leftrightarrow && \tan \alpha \left (\frac{2bg(b- a)}{a} \right) &< \frac{2hg(b^2- 2a^2)}{a^2} \\ \Rightarrow && \tan \alpha &< \frac{h(b^2-2a^2)}{ab(b-a)} \end{align*}

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(\text{hall switch on}) &= \underbrace{p \cdot \frac34 }_{\text{already on and not flipped}}+ \underbrace{(1-p) \cdot \frac14}_{\text{not on and flipped}} \\ &&&= \frac14 +\frac12 p\\ && \mathbb{P}(\text{kitchen on}) &= \frac12 \\ \Rightarrow && \mathbb{P}(\text{pool is on}) &= \frac18 + \frac14p \end{align*} \begin{align*} && \mathbb{P}(\text{flipped hall switch} | \text{pool on}) &= \frac{\mathbb{P}(\text{flipped hall and pool on})}{\mathbb{P}(\text{pool on})} \\ &&&= \frac{(1-p)\frac14 \cdot \frac 12}{\frac18 + \frac14 p} \\ &&&= \frac{1-p}{1+2p} \end{align*}
2. The number of days the swimming pool light was pressed is \(X = B\left (7, \frac{1-p}{1+2p} \right)\), and we have that \(\mathbb{P}(X = 2) < \mathbb{P}(X = 3) > \mathbb{P}(X=4)\) (since the binomial is unimodal). Let \(q = \frac{1-p}{1+2p} \) \begin{align*} && \mathbb{P}(X = 2) &< \mathbb{P}(X = 3) \\ \Rightarrow && \binom{7}{2} q^2(1-q)^5 &< \binom{7}{3}q^3(1-q)^4 \\ \Rightarrow && 21(1-q) &< 35q \\ \Rightarrow && 21 &< 56q \\ \Rightarrow && \frac{3}{8} &< \frac{1-p}{1+2p} \\ \Rightarrow && 3+6p &< 8-8p \\ \Rightarrow && 14p &< 5\\ \Rightarrow && p &< \frac5{14} \\ \\ && \mathbb{P}(X = 3) &> \mathbb{P}(X = 4) \\ \Rightarrow && \binom{7}{3} q^3(1-q)^4 &> \binom{7}{4}q^4(1-q)^3 \\ \Rightarrow &&(1-q)&> q \\ \Rightarrow && \frac12 &> q \\ \Rightarrow && \frac12 &> \frac{1-p}{1+2p} \\ \Rightarrow && 1+2p &> 2-2p \\ \Rightarrow && 4p &> 1\\ \Rightarrow && p &> \frac1{4} \end{align*} Therefore \(\frac14 < p < \frac{5}{14}\) as required.

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Solution:

Note that the line \(y = x\) is the tangent \((0,0)\) and \(y = \sin x \) is always below it. For any other line through the origin with gradient \(0 < k < 1\) it must start below \(y = \sin x\), but finish above it at \(x = \pi\). It also can only cross once due the the convexity of \(\sin\) in this interval. \begin{align*} && I &= \int_0^\pi | \sin x -kx | \d x \\ &&&= \int_0^\alpha (\sin x -k x) \d x + \int_{\alpha}^\pi (kx - \sin x) \d x \\ &&&= \left [ -\cos x - \frac{kx^2}{2} \right]_0^{\alpha} + \left [ \cos x +\frac{kx^2}{2} \right]_{\alpha}^\pi \\ &&&= -\cos \alpha - \frac{k \alpha^2}{2} +1+(-1)+\frac{k\pi^2}{2} - \cos \alpha - \frac{k\alpha^2}{2} \\ &&&= -2\cos \alpha - k\left (\alpha^2 - \frac{\pi^2}{2} \right) \\ &&&= -2\cos \alpha - \frac{\sin \alpha}{\alpha}\left (\alpha^2 - \frac{\pi^2}{2} \right) \\ &&&= \frac{\pi^2 \sin \alpha}{\alpha} - \alpha \sin \alpha - 2\cos \alpha \end{align*} \begin{align*} && \frac{\d I}{\d \alpha} &= \frac{\pi^2(\alpha \cos \alpha - \sin \alpha)}{2\alpha^2} + 2 \sin \alpha - \sin \alpha - \alpha \cos \alpha \\ &&&= \frac{-2\alpha^3 \cos \alpha + 2\alpha \sin\alpha + \pi^2 \alpha \cos \alpha - \pi^2 \sin \alpha}{2\alpha^2} \\ \\ &&&= \left ( \alpha \cos \alpha - \sin \alpha\right) \left ( \frac{\pi^2}{2\alpha^2}-1 \right)\ \end{align*} Therefore \(I' = 0\) if \(\tan \alpha = \alpha\) or \(\alpha = \frac{\pi}{\sqrt{2}}\). Since \(\tan \alpha = \alpha\) only at \(\alpha = 0\) (between \(0 \leq \alpha < \pi\) (by considering the tangent), we must have a unique turning point when \(\alpha = \frac{\pi}{\sqrt{2}}\). Note that \(I(\frac{\pi}{\sqrt{2}}) = \frac{\pi^2 \sqrt{2} \sin \alpha}{2\pi} - \frac{\pi}{\sqrt{2}} \sin \alpha - 2\cos \frac{\pi}{\sqrt{2}}=- 2\cos \frac{\pi}{\sqrt{2}}\). Notice that \(I(0) = \frac{\pi^2}2 - 2 > 2\) and \(I(\pi) = 2\), but \(-2\cos \frac{\pi}{\sqrt{2}} < 2\) so we must be a at a minimum