Problems

The curve \(C\) has the equation \(x^3+y^3 = 3xy\).

1. Show that there is no point of inflection on \(C\). You may assume that the origin is not a point of inflection.
2. The part of \(C\) which lies in the first quadrant is a closed loop touching the axes at the origin. By converting to polar coordinates, or otherwise, evaluate the area of this loop.

The time taken for me to set an acceptable examination question it \(T\) hours. The distribution of \(T\) is a truncated normal distribution with probability density \(\f\) where \[ \mathrm{f}(t)=\begin{cases} \dfrac{1}{k\sigma\sqrt{2\pi}}\exp\left(-\dfrac{1}{2}\left(\dfrac{t-\sigma}{\sigma}\right)^{2}\right) & \mbox{ for }t\geqslant0\\ 0 & \mbox{ for }t<0. \end{cases} \] Sketch the graph of \(\f(t)\). Show that \(k\) is approximately \(0.841\) and obtain the mean of \(T\) as a multiple of \(\sigma\). Over a period of years, I find that the mean setting time is 3 hours.

1. Find the approximate probability that none of the 16 questions on next year's paper will take more than 4 hours to set.
2. Given that a particular question is unsatisfactory after 2 hours work, find the probability that it will still be unacceptable after a further 2 hours work.

Sketch the following subsets of the complex plane using Argand diagrams. Give reasons for your answers.

1. \(\{z:\mathrm{Re}((1+\mathrm{i})z)\geqslant0\}.\)
2. \(\{z: |z^{2}| \leqslant2,\mathrm{Re}(z^{2})\geqslant0\}.\)
3. \(\{z=z_{1}+z_{2}:\left|z_{1}\right|=2,\left|z_{2}\right|=1\}.\)

In the figure, the large circle with centre \(O\) has radius \(4\) and the small circle with centre \(P\) has radius \(1\). The small circle rolls around the inside of the larger one. When \(P\) was on the line \(OA\) (before the small circle began to roll), the point \(B\) was in contact with the point \(A\) on the large circle.

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Sketch the graphs of \(y=\sec x\) and \(y=\ln(2\sec x)\) for \(0\leqslant x\leqslant\frac{1}{2}\pi\). Show graphically that the equation \[ kx=\ln(2\sec x) \] has no solution with \(0\leqslant x<\frac{1}{2}\pi\) if \(k\) is a small positive number but two solutions if \(k\) is large. Explain why there is a number \(k_{0}\) such that \[ k_{0}x=\ln(2\sec x) \] has exactly one solution with \(0\leqslant x<\frac{1}{2}\pi\). Let \(x_{0}\) be this solution, so that \(0\leqslant x_{0}<\frac{1}{2}\pi\) and \(k_{0}x_{0}=\ln(2\sec x_0)\). Show that \[ x_{0}=\cot x_{0}\ln(2\sec x_{0}). \] Use any appropriate method to find \(x_{0}\) correct to two decimal places. Hence find an approximate value for \(k_{0}\).

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Solution:

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The red line is \(y = \ln (2 \sec x)\), blue is \(y = \sec x\). We can see that if the gradient is too small it never touches the red line. If it is large it will cross the red line twice in that interval. For some value it will be perfectly tangent. Since the line is tangent we must have \begin{align*} && y &= \ln (2 \sec x) \\ \Rightarrow && \frac{\d y}{ \d x} &= \frac{1}{2 \sec x} \cdot 2\sec x \tan x \\ &&&= \tan x \\ \Rightarrow && k_0 &=\tan x_0 \\ \Rightarrow && k_0 x_0 &= \ln(2 \sec x_0 ) \\ \Rightarrow && x_0 &= \cot x_0 \ln (2 \sec x_0) \end{align*} If \(f(x) =x- \cot x \ln (2 \sec x)\), then \(f'(x) =1 - 1+\ln(2\sec x) \cosec^2x = \ln(2 \sec x)\cosec^2x \) so we should look at \begin{align*} x_{n+1} &= x_n - \frac{f(x_n)}{f'(x_n)} \\ &= x_n - \frac{x_n- \cot x_n \ln (2 \sec x_n)}{\ln(2 \sec x_n)\cosec^2x_n } \\ &= x_n \left (1 - \frac{\sin^2 x_n}{\ln (2 \sec x_n)}\right) +\sin x_n \cos x_n \end{align*} \begin{array}{c|c} n & x_n \\ \hline 1 & \frac{\pi}{4} \\ 2 & 0.907701\ldots \\ 3 & 0.91439340\ldots \\ 4 & 0.914403867\ldots \\ 5 & 0.91440386\ldots \\ 6 & 0.91440386\ldots \\ \end{array} The sign change test shows that \(x_0 \approx 0.91\) and \(k_0 = \tan(x_0) \approx 1.30\)

Sketch the curve \(C_{1}\) whose parametric equations are \(x=t^{2},\) \(y=t^{3}.\) The circle \(C_{2}\) passes through the origin \(O\). The points \(R\) and \(S\) with real non-zero parameters \(r\) and \(s\) respectively are other intersections of \(C_{1}\) and \(C_{2}.\) Show that \(r\) and \(s\) are roots of an equation of the form \[ t^{4}+t^{2}+at+b=0, \] where \(a\) and \(b\) are real constants. By obtaining a quadratic equation, with coefficients expressed in terms of \(r\) and \(s\), whose roots would be the parameters of any further intersections of \(C_{1}\) and \(C_{2},\) or otherwise, show that \(O\), \(R\) and \(S\) are the only real intersections of \(C_{1}\) and \(C_{2}.\)

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Solution:

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Suppose the circle has centre \((c,d)\), then \begin{align*} && c^2+d^2 &= (t^2-c)^2+(t^3-d)^2 \\ \Rightarrow && 0 &= t^4-2ct^2+t^6-2t^3d \\ \Rightarrow && 0 &= t^4+t^2-2td-2c \end{align*} So by setting \(a = -2d\) and \(b = -2c\) we have the desired equation. By matching the coefficients of \(t^4, t^3, t^2\) we must have: \begin{align*} && 0 &= (t^2-(r+s)t+rs)(t^2+t(r+s)-rs+(r+s)^2+1) \\ \Rightarrow && 0 &= t^2+(r+s)t-rs+(r+s)^2+1 \\ && \Delta &= (r+s)^2 -4(1-rs+(r+s)^2) \\ &&&= -4+4rs-3(r+s)^2 \\ &&&=-4-2(r+s)^2-(r-s)^2 < 0 \end{align*} Therefore there are no further (real) solutions. Hence \(O, R, S\) are the only solutions.

A set of curves \(S_{1}\) is defined by the equation \[ y=\frac{x}{x-a}, \] where \(a\) is a constant which is different for different members of \(S_{1}.\) Sketch on the same axes the curves for which \(a=-2,-1,1\) and \(2\). A second of curves \(S_{2}\) is such that at each intersection between a member of \(S_{2}\) and a member of \(S_{1}\) the tangents of the intersecting curves are perpendicular. On the same axes as the already sketched members of \(S_{1},\) sketch the member of \(S_{2}\) that passes through the point \((1,-1)\). Obtain the first order differential equation for \(y\) satisfied at all points on all members of \(S_{1}\) (i.e. an equation connecting \(x,y\) and \(\mathrm{d}y/\mathrm{d}x\) which does not involve \(a\)). State the relationship between the values of \(\mathrm{d}y/\mathrm{d}x\) on two intersecting curves, one from \(S_{1}\) and one from \(S_{2},\) at their intersection. Hence show that the differential equation for the curves of \(S_{2}\) is \[ x=y(y-1)\dfrac{\mathrm{d}y}{\mathrm{d}x}. \] Find an equation for the member of \(S_{2}\) that you have sketched.

Sketch the curve \(C\) whose polar equation is \[ r=4a\cos2\theta\qquad\mbox{ for }-\tfrac{1}{4}\pi<\theta<\tfrac{1}{4}\pi. \] The ellipse \(E\) has parametric equations \[ x=2a\cos\phi,\qquad y=a\sin\phi. \] Show, without evaluating the integrals, that the perimeters of \(C\) and \(E\) are equal. Show also that the areas of the regions enclosed by \(C\) and \(E\) are equal.

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Solution:

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\begin{align*} && \text{Perimeter}(C) &= \int_{-\pi/4}^{\pi/4} \sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &&&= \int_{-\pi/4}^{\pi/4} \sqrt{16a^2 \cos^2 2 \theta + 64a^2 \sin^2 2 \theta } \d \theta \\ &&&= \int_{-\pi/4}^{\pi/4} 4a\sqrt{1 + 3 \sin^2 2 \theta } \d \theta \\ \\ \\ && \text{Perimeter}(D) &= \int_0^{2 \pi} \sqrt{\left ( \frac{\d x}{\d \phi}\right)^2+\left ( \frac{\d y}{\d \phi}\right)^2} \d \phi \\ &&&= \int_0^{2 \pi} \sqrt{ 4a^2 \sin^2 \phi+a^2 \cos^2 \phi} \d \phi \\ &&&= a^2\int_0^{2 \pi} \sqrt{ 1+3 \sin^2 \phi} \d \phi \\ \end{align*} But clearly these two integrals are equal. \begin{align*} && \text{A}(C) &= \frac12 \int_{-\pi/4}^{\pi/4} r^2 \d \theta \\ &&&= \frac12 \int_{-\pi/4}^{\pi/4} 16a^2 \cos^2 2 \theta \d \theta \\ &&&= 8a^2\int_{-\pi/4}^{\pi/4} \cos^2 2 \theta \d \theta \\ &&&= 8a^2 \frac{\pi}{4} = 2\pi a^2 \\ && \text{A}(D) &= 2\pi a^2 \end{align*}

Criticise each step of the following arguments. You should correct the arguments where necessary and possible, and say (with justification) whether you think the conclusion are true even though the argument is incorrect.

1. The function \(g\) defined by \[ \mathrm{g}(x)=\frac{2x^{3}+3}{x^{4}+4} \] satisfies \(\mathrm{g}'(x)=0\) only for \(x=0\) or \(x=\pm1.\) Hence the stationary values are given by \(x=0\), \(\mathrm{g}(x)=\frac{3}{4}\) and \(x=\pm1,\) \(\mathrm{g}(x)=1.\) Since \(\frac{3}{4}<1,\) there is a minimum at \(x=0\) and maxima at \(x=\pm1.\) Thus we must have \(\frac{3}{4}\leqslant\mathrm{g}(x)\leqslant1\) for all \(x\).
2. \({\displaystyle \int(1-x)^{-3}\,\mathrm{d}x=-3(1-x)^{-4}}\quad\) and so \(\quad{\displaystyle \int_{-1}^{3}(1-x)^{-3}\,\mathrm{d}x=0.}\)

The equation of a hyperbola (with respect to axes which are displaced and rotated with respect to the standard axes) is \[ 3y^{2}-10xy+3x^{2}+16y-16x+15=0.\tag{\(\dagger\)} \] By differentiating \((\dagger)\), or otherwise, show that the equation of the tangent through the point \((s,t)\) on the curve is \[ y=\left(\frac{5t-3s+8}{3t-5s+8}\right)x-\left(\frac{8t-8s+15}{3t-5s+8}\right). \] Show that the equations of the asymptote (the limiting tangents as \(s\rightarrow\infty\)) are \[ y=3x-4\qquad\mbox{ and }\qquad3y=x-4. \] {[}Hint: You will need to find a relationship between \(s\) and \(t\) which is valid in the limit as \(s\rightarrow\infty.\){]} Show that the angle between one asymptote and the \(x\)-axis is the same as the angle between the other asymptote and the \(y\)-axis. Deduce the slopes of the lines that bisect the angles between the asymptotes and find the equations of the axes of the hyperbola.

Give a rough sketch of the function \(\tan^{k}\theta\) for \(0\leqslant\theta\leqslant\frac{1}{4}\pi\) in the two cases \(k=1\) and \(k\gg1\) (i.e. \(k\) is much greater than 1). Show that for any positive integer \(n\) \[ \int_{0}^{\frac{1}{4}\pi}\tan^{2n+1}\theta\,\mathrm{d}\theta=(-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right), \] and deduce that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m}=\tfrac{1}{2}\ln2. \] Show similarly that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m-1}=\frac{\pi}{4}. \]

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Solution:

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Let \(\displaystyle I_n = \int_0^{\pi/4} \tan^{n} \theta \, \d \theta\), then \begin{align*} I_0 &= \int_0^{\pi/4} \tan \theta \d \theta \\ &= \left [ -\ln \cos \theta \right]_0^{\pi/4} \\ &= -\ln \frac{1}{\sqrt{2}} - 0 \\ &= \frac12 \ln 2 \\ \\ \\ I_{2n+1} &= \int_0^{\pi/4} \tan^{2n+1} \theta \, \d \theta \\&= \int_0^{\pi/4} \tan^{2n-1} \theta \tan ^2 \theta \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta (\sec^2 \theta - 1) \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta \sec^2 \theta - \tan^{2n-1} \theta \, \d \theta \\ &= \left[ \frac{1}{2n} \tan^{2n} \theta \right]_0^{\pi/4} - I_{2n-1} \\ &= \frac1{2n} - I_{2n-1} \end{align*} Therefore we can see that \(\displaystyle I_{2n+1} = (-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right)\). As we can see as \(n \to \infty\), \(I_n \to 0\) Therefore \begin{align*} && 0 &= \tfrac{1}{2}\ln2+\lim_{n \to \infty} \sum_{m=1}^{n}\frac{(-1)^{m}}{2m} \\ \Rightarrow && \tfrac{1}{2}\ln2 &= \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m} \end{align*} \begin{align*} && I_{-1} &= \int_0^{\pi/4} 1 \d \theta \\ &&&= \frac{\pi}{4} \end{align*} Therefore \(\displaystyle I_{2n} = (-1)^n \left ( \frac{\pi}{4} + \sum_{m=1}^n \frac{(-1)^m}{2m-1} \right)\) and since \(I_{2m} \to 0\) the same result follows.

The function \(\mathrm{f}\) is defined for \(x<2\) by \[ \mathrm{f}(x)=2| x^{2}-x|+|x^{2}-1|-2|x^{2}+x|. \] Find the maximum and minimum points and the points of inflection of the graph of \(\mathrm{f}\) and sketch this graph. Is \(\mathrm{f}\) continuous everywhere? Is \(\mathrm{f}\) differentiable everywhere? Find the inverse of the function \(\mathrm{f}\), i.e. expressions for \(\mathrm{f}^{-1}(x),\) defined in the various appropriate intervals.

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Solution: \[ f(x) = 2|x(x-1)| + |(x-1)(x+1)|-2|x(x+1)| \] Therefore the absolute value terms will change behaviour at \(x = -1, 0, 1\). Then \begin{align*} f(x) &= \begin{cases} 2(x^2-x)+(x^2-1)-2(x^2+x) & x \leq -1 \\ 2(x^2-x)-(x^2-1)+2(x^2+x) & -1 < x \leq 0 \\ -2(x^2-x)-(x^2-1)-2(x^2+x) & 0 < x \leq 1 \\ 2(x^2-x)+(x^2-1)-2(x^2+x) & 1 < x\end{cases} \\ &= \begin{cases} x^2-4x-1 & x \leq -1 \\ 3x^2+1& -1 < x \leq 0 \\ -5x^2+1& 0 < x \leq 1 \\ x^2-4x-1 & 1 < x\end{cases} \\ \\ f'(x) &= \begin{cases} 2x-4 & x <-1 \\ 6x & -1 < x < 0 \\ -10x & 0 < x < 1 \\ 2x-4 & 1 < x\end{cases} \\ \end{align*} Therefore \(f'(x) = 0 \Rightarrow x = 0, 2\) and so we should check all the turning points. Therefore the minimum is \(x = 2, y = -5\), maximum is \(x = -2, y = 11\) (assuming the range is actually \(|x| < 2\). There is a point of inflection at \(x = 0, y = 1\).

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\(f\) is continuous everywhere as a sum of continuous functions. \(f\) is not differentiable at \(x = -1, 1\) Suppose \begin{align*} &&y &=x^2-4x-1 \\ &&&= (x-2)^2 -5 \\ \Rightarrow &&x &= 2\pm \sqrt{y+5} \\ \\ && y &= 3x^2+1 \\ \Rightarrow && x &= \pm \sqrt{\frac{y-1}{3}} \\ \\ && y &= -5x^2+1 \\ \Rightarrow && x &=\pm \sqrt{\frac{1-y}{5}} \\ \\ \Rightarrow && f^{-1}(y) &= \begin{cases} 2 - \sqrt{y+5} & y > 4 \\ -\sqrt{\frac{y-1}{3}} & 1 < y < 4 \\ \sqrt{\frac{1-y}{5}} & -4 < y < 1 \\ 2 + \sqrt{y+5} & y < -4 \end{cases} \end{align*}

A smooth tube whose axis is horizontal has an elliptic cross-section in the form of the curve with parametric equations \[ x=a\cos\theta\qquad y=b\sin\theta \] where the \(x\)-axis is horizontal and the \(y\)-axis is vertically upwards. A particle moves freely under gravity on the inside of the tube in the plane of this cross-section. By first finding \(\ddot{x}\) and \(\ddot{y},\) or otherwise, show that the acceleration along the inward normal at the point with parameter \(\theta\) is \[ \frac{ab\dot{\theta}^{2}}{\sqrt{a^{2}\sin^{2}\theta+b^{2}\cos^{2}\theta}}. \] The particle is projected along the surface in the vertical cross-section plane, with speed \(2\sqrt{bg},\) from the lowest point. Given that \(2a=3b,\) show that it will leave the surface at the point with parameter \(\theta\) where \[ 5\sin^{3}\theta+12\sin\theta-8=0. \]

1. Evaluate \[ \int_{1}^{3}\frac{1}{6x^{2}+19x+15}\,\mathrm{d}x\,. \]
2. Sketch the graph of the function \(\mathrm{f}\), where \(\mathrm{f}(x)=x^{1760}-x^{220}+q\), and \(q\) is a constant. Find the possible numbers of \textit{distinct }roots of the equation \(\mathrm{f}(x)=0\), and state the inequalities satisfied by \(q\).

Let \(A\) and \(B\) be the points \((1,1)\) and \((b,1/b)\) respectively, where \(b>1\). The tangents at \(A\) and \(B\) to the curve \(y=1/x\) intersect at \(C\). Find the coordinates of \(C\). Let \(A',B'\) and \(C'\) denote the projections of \(A,B\) and \(C\), respectively, to the \(x\)-axis. Obtain an expression for the sum of the areas of the quadrilaterals \(ACC'A'\) and \(CBB'C'\). Hence or otherwise prove that, for \(z>0\), \[ \frac{2z}{2+z}\leqslant\ln\left(1+z\right)\leqslant z. \]

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Solution:

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\begin{align*} && y &= 1/x \\ \Rightarrow && \frac{\d y}{\d x} &= -1/x^2 \end{align*} Therefore the tangent at \((1,1)\) will be \(\frac{y - 1}{x-1} = -1 \Rightarrow y = -x + 2\) and at \((b, 1/b)\) will be \(\frac{y-1/b}{x-b} = -\frac{1}{b^2} \Rightarrow y = -\frac{x}{b^2} + \frac{2}{b}\) The intersection will be at \begin{align*} && x + y & = 2 \\ && x + b^2 y &= 2b \\ \Rightarrow && (b^2-1)y &= 2(b-1) \\ \Rightarrow && y &= \frac{2}{b+1} \\ && x &= \frac{2b}{b+1} \end{align*} Therefore \(\displaystyle C = \left (\frac{2b}{b+1}, \frac{2}{b+1} \right)\). The areas of the two trapeziums will be: \begin{align*} [ACC'A'] &= \frac12 \left (1 + \frac{2}{b+1} \right) \left (\frac{2b}{b+1} - 1 \right) \\ &= \frac12 \cdot \frac{b+3}{b+1} \cdot \frac{2b - b - 1}{b+1} \\ &= \frac 12 \frac{(b+3)(b-1)}{(b+1)^2} \end{align*} \begin{align*} [CBB'C'] &= \frac12 \left (\frac{2}{b+1} + \frac{1}{b} \right) \left (b- \frac{2b}{b+1} \right) \\ &= \frac12 \cdot \frac{3b+1}{b(b+1)} \cdot \frac{b^2+b-2b}{b+1} \\ &= \frac 12 \frac{(3b+1)b(b-1)}{b(b+1)^2} \\ &= \frac12 \frac{(3b+1)(b-1)}{(b+1)^2} \end{align*} The area under the curve between \(A\) and \(B\) will be: \begin{align*} \int_1^b \frac{1}{x} \d x &= \left [\ln x \right]_1^b \\ &= \ln b \end{align*} The area of a rectangle of height \(1\) from \(A\) will clearly be above the curve and will have area \(b-1\). The area of \(ACBB'C'A'\) will be: \begin{align*} [ACBB'C'A'] &= [ACC'A']+[CBB'C'] \\ &=\frac 12 \frac{(b+3)(b-1)}{(b+1)^2}+ \frac12 \frac{(3b+1)(b-1)}{(b+1)^2} \\ &= \frac12 \frac{(b-1)(4b+4)}{(b+1)^2} \\ &= \frac{2(b-1)}{b+1} \end{align*} By comparing areas, we must have: \(\frac{2(b-1)}{b+1} \leq \ln b \leq b-1\) and since \(b > 1\) we can write it as \(1 + z\) for \(z >0\), ie: \(\displaystyle \frac{2z}{2+z} \leq \ln (1 + z) \leq z\). [By considering the area of \(ABB'A'\) which is \begin{align*} [ABB'A'] &= \frac12 \left (1 + \frac{1}{b} \right) \left ( b- 1 \right) \\ &= \frac12 \frac{(b+1)(b-1)}{b} \end{align*} we can tighten the right hand bound to \(\displaystyle \frac{(2+z)z}{2(z+1)} = \left (1 - \frac{z}{2z+2} \right)z\)