Problems

Show that the equation of any circle passing through the points of intersection of the ellipse \((x+2)^2 +2y^2 =18\) and the ellipse \(9(x-1)^2 +16y^2 = 25\) can be written in the form \[ x^2-2ax +y^2 =5-4a\;. \]

Let \(f(x) = x^m(x-1)^n\), where \(m\) and \(n\) are both integers greater than \(1\). Show that the curve \(y=f(x)\) has a stationary point with \(0 < x < 1\). By considering \(f''(x)\), show that this stationary point is a maximum if \(n\) is even and a minimum if \(n\) is odd. Sketch the graphs of \(f(x)\) in the four cases that arise according to the values of \(m\) and \(n\).

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Solution: \begin{align*} && f'(x) &= mx^{m-1}(x-1)^n + nx^m(x-1)^{n-1} \\ &&&= (m(x-1)+nx)x^{m-1}(x-1)^{n-1} \\ &&&= (x(m+n) - m)x^{m-1}(x-1)^{n-1} \\ \end{align*} Therefore when \(x = \frac{m}{m+n}\) there is a stationary point with \(0 < x < 1\). \begin{align*} && f''(x) &= m(m-1)x^{m-2}(x-1)^n + 2mnx^{m-1}(x-1)^{n-1} + n(n-1)x^{m}(x-1)^{n-2} \\ &&&= (m(m-1)(x-1)^2 +2mnx(x-1)+n(n-1)x^2)x^{m-2}(1-x)^{n-2} \\ \Rightarrow && f'' \left ( \frac{m}{m+n} \right) &= \left ( m(m-1) \frac{n^2}{(m+n)^2} - 2mn\frac{mn}{(m+n)^2} + n(n-1) \frac{m^2}{(m+n)^2} \right) \frac{m^{m-2}}{(m+n)^{m-2}} \frac{(-1)^{n-2}n^{n-2}}{(m+n)^{n-2}} \\ &&&= (-1)^{n-2}\frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-2}} \left ( (m-1)n-2mn+(n-1)m\right) \\ &&&= (-1)^{n-2}\frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-2}} \left ( -m-n\right) \\ &&&= (-1)^{n-1} \frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-3}} \end{align*} Therefore this is positive (and a minimum) when \(n\) is odd and negative (and a maximum) when \(n\) is even.

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Show that \((a+b)^2\le 2a^2+2b^2\,\). Find the stationary points on the curve $y=\big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12} + \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}\,$, where \(a\) and \(b\) are constants. State, with brief reasons, which points are maxima and which are minima. Hence prove that \[ \vert a\vert +\vert b \vert \le \big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12} + \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12} \le \big(2a^2+2b^2\big)^{\frac12} \;. \]

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Solution: \begin{align*} && 2a^2+2b^2 &= a^2 + b^2 + (a^2+b^2) \\ &&&\underbrace{\geq}_{AM-GM} a^2+b^2+2\sqrt{a^2b^2} \\ &&&= a^2+b^2 + 2|a||b| \\ &&&\geq a^2+b^2 + 2ab \\ &&&= (a+b)^2 \end{align*} Assume \(a^2 \neq b^2\), otherwise the curve is a constant. \begin{align*} && y & = \big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12} + \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}\\ && \frac{\d y}{\d \theta} &= \tfrac12 \left (a^2\cos^2\theta +b^2\sin^2\theta \right)^{-\frac12} \cdot (2 \sin \theta \cos \theta (b^2 - a^2)) + \tfrac12 (a^2\sin^2\theta +b^2\cos^2\theta)^{-\frac12} \cdot (2 \sin \theta \cos \theta (a^2 - b^2) \\ &&&= \tfrac12\sin2 \theta (b^2 - a^2) \left ( \left (a^2\cos^2\theta +b^2\sin^2\theta \right)^{-\frac12} - (a^2\sin^2\theta +b^2\cos^2\theta)^{-\frac12}\right) \\ \therefore \frac{\d y}{\d x} = 0 \Rightarrow && \sin 2\theta = 0 & \text{ or } a^2\cos^2\theta +b^2\sin^2\theta = a^2\sin^2\theta +b^2\cos^2\theta \\ \Rightarrow && \theta &= 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \\ && (a^2-b^2) \cos ^2\theta &= (a^2-b^2) \sin^2 \theta \\ \Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \end{align*} WLOG \(b^2 - a^2 > 0\), then the two parts of the derivative look like:

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And so \(\frac{\pi}{4}, \frac{3\pi}{4}, \cdots\) are maxima, and the others minima. The maxima are where \(\sin^2 \theta = \cos^2 \theta = \frac12\), so \(y(\frac{\pi}{4}) = 2\left ( \frac{a^2+b^2}{2} \right)^{\frac12} = (2a^2+2b^2)^{\frac12}\) and the maxima are \(\cos^2 \theta = 1, \sin^2 \theta = 0\) and vice versa, ie \(y = |a| + |b|\), therefore we obtain our desired result.

Give a sketch of the curve \( \;\displaystyle y= \frac1 {1+x^2}\;\), for \(x\ge0\). Find the equation of the line that intersects the curve at \(x=0\) and is tangent to the curve at some point with \(x>0\,\). Prove that there are no further intersections between the line and the curve. Draw the line on your sketch. By considering the area under the curve for \(0\le x\le1\), show that \(\pi>3\,\). Show also, by considering the volume formed by rotating the curve about the \(y\) axis, that \(\ln 2 >2/3\,\). [Note: \(\displaystyle \int_0^ 1 \frac1 {1+x^2}\, \d x = \frac\pi 4\,.\;\)]

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Solution:

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\begin{align*} && y &= (1+ x^2)^{-1} \\ \Rightarrow && y' &= -2x(1+x^2)^{-2} \\ \text{eqn of tangent}:&& \frac{y - (1+t^2)^{-1}}{x-t} &= -2t(1+t^2)^{-2} \\ \text{passes thru }(0,1): && \frac{1-(1+t^2)^{-1}}{-t} &= -2t(1+t^2)^{-2} \\ \Rightarrow && (1+t^2)^2-(1+t^2) &= 2t^2 \\ \Rightarrow && t^4-t^2 &= 0 \\ \Rightarrow && t &= 0, \pm 1 \\ \Rightarrow && \frac{y - \frac12}{x - 1} &= -\frac12 \\ && y &=1 -\tfrac12 x \end{align*} There can be no further intersections since the equation is equivalent to the cubic \((1-\frac12 x)(1+x^2) = 1\) and we have already found \(3\) roots. \begin{align*} && A &= \int_0^1 \frac{1}{1 + x^2} = \frac{\pi}{4} \\ && A &> \frac12 \cdot 1 \cdot (1 + \tfrac12) = \frac34 \\ \Rightarrow && \pi &> 3 \\ \\ && V &=\pi \int_{\frac12}^1 x^2 \d y \\ &&&= \pi \int_{\frac12}^1 \left ( \frac{1}{y}-1 \right) \d y \\ &&&= \pi \left [\ln y \right]_{1/2}^1-\frac12 \\ &&&= \pi \ln 2 - \frac{\pi}{2} \\ && V &> \frac13 \pi 1^2 \frac{1}{2} \\ &&&= \frac{\pi}{6} \\ \Rightarrow && \ln 2 &> \frac{2}{3} \end{align*}

Let \[ \f(x) = x^n + a_1 x^{n-1} + \cdots + a_n\;, \] where \(a_1\), \(a_2\), \(\ldots\), \(a_n\) are given numbers. It is given that \(\f(x)\) can be written in the form \[ \f(x) = (x+k_1)(x+k_2)\cdots(x+k_n)\;. \] By considering \(\f(0)\), or otherwise, show that \(k_1k_2 \ldots k_n =a_n\). Show also that $$(k_1+1)(k_2+1)\cdots(k_n+1)= 1+a_1+a_2+\cdots+a_n$$ and give a corresponding result for \((k_1-1)(k_2-1)\cdots(k_n-1)\). Find the roots of the equation \[ x^4 +22x^3 +172x^2 +552x+576=0\;, \] given that they are all integers.

A pyramid stands on horizontal ground. Its base is an equilateral triangle with sides of length~\(a\), the other three sides of the pyramid are of length \(b\) and its volume is \(V\). Given that the formula for the volume of any pyramid is $ \textstyle \frac13 \times \mbox{area of base} \times \mbox {height} \,, $ show that \[ V= \frac1{12} {a^2(3b^2-a^2)}^{\frac12}\;. \] The pyramid is then placed so that a non-equilateral face lies on the ground. Show that the new height, \(h\), of the pyramid is given by \[ h^2 = \frac{a^2(3b^2-a^2)}{4b^2-a^2}\;. \] Find, in terms of \(a\) and \(b\,\), the angle between the equilateral triangle and the horizontal.

Let \[ I= \int_0^a \frac {\cos x}{\sin x + \cos x} \; \d x \, \quad \mbox{ and } \quad J= \int_0^a \frac {\sin x}{\sin x + \cos x} \; \d x \;, \] where \(0\le a < \frac{3}{4}\pi\,\). By considering \(I+J\) and \(I-J\), show that $ 2I= a + \ln (\sin a +\cos a)\;. $ Find also:

1. \(\displaystyle \int_0^{\frac{1}{2}\pi} \frac {\cos x}{p\sin x + q\cos x} \; \d x \,\), where \(p\) and \(q\) are positive numbers; %
2. [(ii)] %\(\displaystyle \int_0^{\frac{1}{2}\pi/2} \frac {\cos x}{\sin (x+k)} \; \d x \,\), where \(0 < k < \pi/2\,\);
3. \(\displaystyle \int_0^{\frac{1}{2}\pi} \frac {\cos x+4}{3\sin x + 4\cos x+ 25} \; \d x \,\).

I borrow \(C\) pounds at interest rate \(100\alpha \,\%\) per year. The interest is added at the end of each year. Immediately after the interest is added, I make a repayment. The amount I repay at the end of the \(k\)th year is \(R_k\) pounds and the amount I owe at the beginning of \(k\)th year is \(C_k\) pounds (with \(C_1=C\)). Express \(C_{n+1}\) in terms of \(R_k\) (\(k= 1\), \(2\), \(\ldots\), \(n\)), \(\alpha\) and \(C\) and show that, if I pay off the loan in \(N\) years with repayments given by \(R_k= (1+\alpha)^kr\,\), where \(r\) is constant, then \(r=C/N\,\). If instead I pay off the loan in \(N\) years with \(N\) equal repayments of \(R\) pounds, show that \[ \frac R C = \frac{\alpha (1+\alpha)^{N} }{(1+\alpha)^N-1} \;, \] and that \(R/C\approx 27/103\) in the case \(\alpha =1/50\), \(N=4\,\).

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1. If the normal reactions on the lower and higher wheels of the lorry are equal, show that the sum of the frictional forces between the wheels and the ground is zero.
2. If \(P\) is such that the lorry does not tip over (but the normal reactions on the lower and higher wheels of the lorry need not be equal), show that \[ W\tan(\alpha - \beta) \le P \le W\tan(\alpha+\beta)\;, \] where \(\tan\beta = d/h\,\).

A bicycle pump consists of a cylinder and a piston. The piston is pushed in with steady speed~\(u\). A particle of air moves to and fro between the piston and the end of the cylinder, colliding perfectly elastically with the piston and the end of the cylinder, and always moving parallel with the axis of the cylinder. Initially, the particle is moving towards the piston at speed \(v\). Show that the speed, \(v_n\), of the particle just after the \(n\)th collision with the piston is given by \(v_n=v+2nu\). Let \(d_n\) be the distance between the piston and the end of the cylinder at the \(n\)th collision, and let \(t_n\) be the time between the \(n\)th and \((n+1)\)th collisions. Express \(d_n - d_{n+1}\) in terms of \(u\) and \(t_n\), and show that \[ d_{n+1} = \frac{v+(2n-1)u}{v+(2n+1)u} \, d_n \;. \] Express \(d_n\) in terms of \(d_1\), \(u\), \(v\) and \(n\). In the case \(v=u\), show that \(ut_n = \displaystyle \frac {d_1} {n(n+1)}\). %%%%%Verify that \(\sum\limits_1^\infty t_n = d/u\).