Problems

2012 Paper 2 Q1

D: 1600.0 B: 1500.0

generalised binomial theorem power series expansion coefficient extraction generating functions partial fractions combinatorics geometric series

Write down the general term in the expansion in powers of $x$ of $(1-x^6)^{-2}\,$.

Find the coefficient of $x^{24}$ in the expansion in powers of $x$ of \[ (1-x^6)^{-2} (1-x^3)^{-1}\,.\] Obtain also, and simplify, formulae for the coefficient of $x^n$ in the different cases that arise.
Show that the coefficient of $x^{24}$ in the expansion in powers of $x$ of \[ (1-x^6)^{-2} (1-x^3)^{-1} (1-x)^{-1}\,\] is $55$, and find the coefficients of $x^{25}$ and $x^{66}$.

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2012 Paper 2 Q2

D: 1600.0 B: 1530.0

polynomials composition of functions degree of polynomial functional equation simultaneous equations proof by contradiction quadratic

If $\p(x)$ and $\q(x)$ are polynomials of degree $m$ and $n$, respectively, what is the degree of $\p(\q(x))$?

The polynomial $\p(x)$ satisfies \[ \p(\p(\p(x)))- 3 \p(x)= -2x\, \] for all $x$. Explain carefully why $\p(x)$ must be of degree 1, and find all polynomials that satisfy this equation.
Find all polynomials that satisfy \[ 2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4 \] for all $x$.

Solution: If $\p(x)$ and $\q(x)$ are polynomials of degree $m$ and $n$, $\p(\q(x))$ has degree $mn$.

Suppose $\p(\p(\p(x)))- 3 \p(x)= -2x$, and suppose $p(x)$ has degree $n = \geq 2$, then $\p(\p(\p(x)))$ has degree $n^3$ and so the left hand side has degree higher than $1$ and the right hand side is degree $1$. Therefore $\p(x)$ is degree $1$ or $0$. If $p(x) = c$ then $c^3-3c = -2x$ but the LHS doesn't depend on $x$ which is also a contradiction. Therefore $\p(x)$ is degree $1$. Suppose $\p(x) = ax+b$ then: \begin{align*} && -2x &= \p(\p(\p(x))) - 3\p(x) \\ &&&= \p(\p(ax+b)) - 3(ax+b) \\ &&&= \p(a(ax+b)+b) - 3ax -3b \\ &&&= a(a^2x+ab+b) + b - 3ax - 3b \\ &&&= (a^3-3a)x + b(a^2+a-2) \\ \Rightarrow &&& \begin{cases} a^3-3a&=-2 \\ b(a^2+a-2) &= 0\end{cases} \\ \Rightarrow &&& \begin{cases} a^3-3a+2 = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow &&& \begin{cases} (a-1)(a^2+a-2) = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow && (a,b) &= (1, b), (-2,b) \end{align*}
Suppose $2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4$ and let $\deg \p(x) = n$, then LHS has degree $\max(n^2,2n,n)$ and the right hand side has degree $4$. Therefore $\p(x)$ must have degree $2$. Let $\p(x) = ax^2 + bx + c$, then, considering the coefficient of $x^4$ in $2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)$ we will have $2a^3+3a^2=1 \Rightarrow 2a^3+3a^2-1 = (a+1)^2(2a-1) \Rightarrow a = -1, a=\frac12$. Consider the coefficient of $x^3$ in $2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)$ we have $4a^2b+6ab = 0 \Rightarrow 2ab(2a+3) = 0$ Since $a = -1, \frac12$ this means $b = 0$. Consider the constant coefficient in $2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)$ (using $b = 0$). $2ac^2+c+3c^2-4c = 0 \Rightarrow c(2ac+3c-3) = 0$. Therefore $c = 0$ or $a = -1, c = 3, a = \frac12, c = \frac34$, so our possible polynomials are: $\p(x) = -x^2, \frac12x^2, -x^2+3, \frac12x^2+\frac34$

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2012 Paper 2 Q3

D: 1600.0 B: 1516.0

integration substitution trigonometric substitution rational function definite integral algebraic manipulation t-substitution

Show that, for any function f (for which the integrals exist), \[ \int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x = \frac12 \int_1^\infty \left(1+\frac 1 {t^2}\right) \f(t)\, \d t \,. \] Hence evaluate \[ \int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \, \, \d x \,, \] and, using the substitution $x=\tan\theta$, \[ \int_0^{\frac12\pi} \frac{1}{(1+\sin\theta)^3}\,\d \theta \,. \]

Solution: \begin{align*} && t &= x + \sqrt{1+x^2} \\ &&\frac1t &= \frac{1}{x+\sqrt{1+x^2}} \\ &&&= \frac{\sqrt{1+x^2}-x}{1+x^2-1} \\ &&&= \sqrt{1+x^2}-x \\ \Rightarrow && x &=\frac12 \left ( t - \frac1t\right) \\ \Rightarrow && \d x &=\frac12 \left (1 + \frac1{t^2} \right)\d t \\ \\ \Rightarrow && \int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x &= \int_{t=1}^{t = \infty}f(t) \frac12\left (1 + \frac1{t^2} \right)\d t \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right)f(t) \d t \end{align*} \begin{align*} && I &= \int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \d x \\ &&&= \int_0^\infty \frac1 {(x+\sqrt{x^2+1})^2} \d x \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right) \frac{1}{t^2} \d t \\ &&&= \frac12 \left [-\frac1t-\frac13\frac1{t^3} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac43 = \frac23 \end{align*} \begin{align*} && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&x &= \tan \theta\\ && \d x &= \sec^2 \theta = (1+x^2) \d \theta\\ && \tan\theta &= \frac{s}{\sqrt{1-s^2}}\\ \Rightarrow && \tan^2 \theta &= \frac{s^2}{1-s^2} \\ \Rightarrow && \sin \theta &= \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \\ && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&&= \int_0^{\frac12 \pi} \frac{1}{\left (1+ \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \right )^3} \d \theta \\ &&&= \int_{x=0}^{x=\infty} \frac{1}{\left(1 + \frac{x}{\sqrt{1+x^2}} \right)^3} \frac{1}{1+x^2} \d x \\ &&&= \int_0^{\infty} \frac{\sqrt{1+x^2}}{(\sqrt{1+x^2}+x)^3} \d x \\ &&J_a &= \int_0^{\infty} \frac{\sqrt{1+x^2}+x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \frac23 \\ &&J_b &= \int_0^{\infty} \frac{\sqrt{1+x^2}-x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \int_0^{\infty} \frac{1}{(\sqrt{1+x^2}+x)^4} \d x\\ &&&= \frac12\int_1^{\infty} \left (1 +\frac1{t^2} \right)\frac{1}{t^4} \d t \\ &&&= \frac12 \left [-\frac13 t^{-3}-\frac15t^{-5} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac8{15} = \frac4{15} \\ \Rightarrow && J &= \frac12(J_a+J_b) = \frac7{15} \end{align*}

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2012 Paper 2 Q4

D: 1600.0 B: 1500.0

Taylor series logarithmic series inequality ln(1+x) Euler's number e convergence series comparison limits

In this question, you may assume that the infinite series \[ \ln(1+x) = x-\frac{x^2}2 + \frac{x^3}{3} -\frac {x^4}4 +\cdots + (-1)^{n+1} \frac {x^n}{n} + \cdots \] is valid for $\vert x \vert <1$.

Let $n$ be an integer greater than 1. Show that, for any positive integer $k$, \[ \frac1{(k+1)n^{k+1}} < \frac1{kn^{k}}\,. \] Hence show that $\displaystyle \ln\! \left(1+\frac1n\right) <\frac1n\,$. Deduce that \[ \left(1+\frac1n\right)^{\!n}<\e\,. \]
Show, using an expansion in powers of $\dfrac1y\,$, that $ \displaystyle \ln \! \left(\frac{2y+1}{2y-1}\right) > \frac 1y %= \sum _{r=0}^\infty \frac 1{(2r+1)(2y)^{2r}}\,. $ for $y>\frac12$. Deduce that, for any positive integer $n$, \[ \e < \left(1+\frac1n\right)^{\! n+\frac12}\,. \]
Use parts (i) and (ii) to show that as $n\to\infty$ \[ \left(1+\frac1n\right)^{\!n} \to \e\,. \]

Solution:

Since $k \geq 1$ we have $n^{k+1} > n^k$ and $(k+1) > k$, therefore $(k+1)n^{k+1} >kn^k \Rightarrow \frac{1}{(k+1)n^{k+1}} < \frac{1}{kn^k}$ \begin{align*} && \ln \left ( 1 + \frac1n \right) &= \frac1n -\frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \cdots \\ &&&= \frac1n - \underbrace{\left (\frac{1}{2n^2}-\frac{1}{3n^3} \right)}_{>0}- \underbrace{\left (\frac{1}{4n^4}-\frac{1}{5n^5} \right)}_{>0} - \cdot \\ &&&< \frac1n \\ \\ \Rightarrow && n \ln \left ( 1 + \frac1n \right) &< 1 \\ \Rightarrow && \ln \left ( \left ( 1 + \frac1n \right)^n \right) &< 1 \\ \Rightarrow && \left ( 1 + \frac1n \right)^n &< e \end{align*}
$\,$ \begin{align*} &&\ln \left(\frac{2y+1}{2y-1}\right) &= \ln \left (1 + \frac{1}{2y} \right)-\ln \left (1 - \frac{1}{2y} \right) \\ &&&= \frac{1}{2y} - \frac{1}{2(2y)^2} + \frac{1}{3(2y)^3} - \cdots - \left (-\frac{1}{2y} - \frac{1}{2(2y)^2} - \frac{1}{3(2y)^3} - \cdots \right) \\ &&&= \frac{1}{y} + \frac{2}{3(2y)^3} + \frac{2}{5(2y)^5} \\ &&&= \sum_{r=1}^{\infty} \frac{2}{(2r-1)(2y)^{2r-1}} \\ &&&> \frac1y \\ \\ \Rightarrow && \ln \left (1 + \frac{1}{y-\frac12} \right) &> \frac{1}{y} \\\Rightarrow && \ln \left (1 + \frac{1}{n} \right) &> \frac{1}{n+\frac12} \\ \Rightarrow &&(n+\tfrac12) \ln \left (1 + \frac{1}{n} \right) &> 1\\ \Rightarrow && \ln \left ( \left (1 + \frac{1}{n} \right)^{n+\tfrac12} \right) &> 1\\ \Rightarrow && \left (1 + \frac{1}{n} \right)^{n+\tfrac12} & > e \end{align*}

\item Since $\left (1 + \frac1n \right)^n$ is both bounded above, and increasing, it must tend to some limit $L$. \begin{align*} && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n+\frac12} \\ \Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \lim_{n \to \infty} \sqrt{1 + \frac1n} \\ \Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \\ \end{align*} And therefore equality must hold.

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2012 Paper 2 Q5

D: 1600.0 B: 1484.0

curve sketching asymptotes stationary points rational function differentiation symmetry partial fractions

Sketch the curve $y=\f(x)$, where \[ \f(x) = \frac 1 {(x-a)^2 -1} \hspace{2cm}(x\ne a\pm1), \] and $a$ is a constant.
The function $\g(x)$ is defined by \[ \g(x) = \frac 1 {\big( (x-a)^2-1 \big) \big( (x-b)^2 -1\big)} \hspace{1cm}(x\ne a\pm1, \ x\ne b\pm1), \] where $a$ and $b$ are constants, and $b>a$. Sketch the curves $y=\g(x)$ in the two cases $b>a+2$ and $b=a+2$, finding the values of $x$ at the stationary points.

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2012 Paper 2 Q6

D: 1600.0 B: 1528.8

trigonometry cyclic quadrilateral cosine rule area Brahmagupta's formula Heron's formula algebraic manipulation geometric proof

A cyclic quadrilateral $ABCD$ has sides $AB$, $BC$, $CD$ and $DA$ of lengths $a$, $b$, $c$ and $d$, respectively. The area of the quadrilateral is $Q$, and angle $DAB$ is $\theta$. Find an expression for $\cos\theta$ in terms of $a$, $b$, $c$ and $d$, and an expression for $\sin\theta$ in terms of $a$, $b$, $c$, $d$ and $Q$. Hence show that \[ 16Q^2 = 4(ad+bc)^2 - (a^2+d^2-b^2-c^2)^2 \,, \] and deduce that \[ Q^2 = (s-a)(s-b)(s-c)(s-d)\,, \] where $s= \frac12(a+b+c+d)$. Deduce a formula for the area of a triangle with sides of length $a$, $b$ and $c$.

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2012 Paper 2 Q7

D: 1600.0 B: 1516.0

vectors circle dot product position vectors geometric proof scalar parameter centroid unit circle

Three distinct points, $X_1$, $X_2$ and $X_3$, with position vectors ${\bf x}_1$, ${\bf x}_2$ and ${\bf x}_3$ respectively, lie on a circle of radius 1 with its centre at the origin $O$. The point $G$ has position vector $\frac13({\bf x}_1+{\bf x}_2+{\bf x}_3)$. The line through $X_1$ and $G$ meets the circle again at the point $Y_1$ and the points $Y_2$ and $Y_3$ are defined correspondingly. Given that $\overrightarrow{GY_1} =-\lambda_1\overrightarrow{GX_1}$, where $\lambda_1$ is a positive scalar, show that \[ \overrightarrow{OY_1}= \tfrac13 \big( (1-2\lambda_1){\bf x}_1 +(1+\lambda_1)({\bf x}_2+{\bf x}_3)\big) \] and hence that \[ \lambda_1 = \frac {3-\alpha-\beta-\gamma} {3+\alpha -2\beta-2\gamma} \,,\] where $\alpha = {\bf x}_2 \,.\, {\bf x}_3$, $\beta = {\bf x}_3\,.\, {\bf x}_1$ and $\gamma = {\bf x}_1\,.\, {\bf x}_2$. Deduce that $\dfrac {GX_1}{GY_1} + \dfrac {GX_2}{GY_2} + \dfrac {GX_3}{GY_3} =3 \,$.

Solution:

\begin{align*} && \mathbf{y}_1 &= \overrightarrow{OG}+\overrightarrow{GY_1} \\ &&&= \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3) -\lambda_1 \left (\mathbf{x}_1 - \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ && 1 &= \mathbf{y}_1 \cdot \mathbf{y}_1 \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \cdot \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac19\left ( (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\mathbf{x}_1 \cdot \mathbf{x}_2+\mathbf{x}_1 \cdot \mathbf{x}_3) + 2(1+\lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3 \right) \\ \Rightarrow && 9 &= (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\beta+\gamma) + 2(1+\lambda_1)^2 \alpha \\ &&&= 3+6\lambda_1^2+2(\beta+\gamma)-2(\beta+\gamma)\lambda_1 - 4\lambda_1^2(\beta+\gamma) + 2\alpha+4\lambda_1\alpha + 2\lambda_1^2 \alpha \\ && 0 &= (-6+2(\alpha+\beta+\gamma))+2(2\alpha-(\beta+\gamma))\lambda_1 + (6+2(\alpha-2(\beta+\gamma)))\lambda_1^2 \\ \Rightarrow && 0 &= ((\alpha+\beta+\gamma)-3)+(2\alpha-(\beta+\gamma))\lambda_1 + (3+\alpha-2(\beta+\gamma))\lambda_1^2 \\ &&&= (\lambda_1+1)((3+\alpha-2(\beta+\gamma))\lambda_1+ ((\alpha+\beta+\gamma)-3)) \\ \Rightarrow && \lambda_1 &= \frac{3-(\alpha+\beta+\gamma)}{3+\alpha-2(\beta+\gamma)} \end{align*} as required. Since $\dfrac {GX_1}{GY_1} = \frac1{\lambda_1}$ we must have, \begin{align*} && \frac {GX_1}{GY_1} + \frac {GX_2}{GY_2} + \frac {GX_3}{GY_3} &= \frac1{\lambda_1}+\frac1{\lambda_2}+\frac1{\lambda_3} \\ &&&= \frac{(3+\alpha-2\beta-2\gamma)+(3+\beta-2\gamma-2\alpha)+3+\gamma-2\alpha-2\beta)}{3-\alpha-\beta-\gamma} \\ &&&= \frac{9-3(\alpha+\beta+\gamma)}{3-(\alpha+\beta+\gamma)} \\ &&&= 3 \end{align*}

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2012 Paper 2 Q8

D: 1600.0 B: 1485.7

sequences recurrence relation inequality strictly increasing sequence algebraic manipulation second order recurrence quadratic recurrence

The positive numbers $\alpha$, $\beta$ and $q$ satisfy $\beta-\alpha >q$. Show that \[ \frac{\alpha^2+\beta^2 -q^2}{\alpha\beta}-2> 0\,. \] The sequence $u_0$, $u_1$, $\ldots$ is defined by $u_0=\alpha$, $u_1=\beta$ and \[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u_{n+1} = \frac {u_{n}^2 -q^2}{u_{n-1}} \ \ \ \ \ \ \ \ \ \ \ (n\ge1), \] where $\alpha$, $\beta$ and $q$ are given positive numbers (and $\alpha$ and $\beta$ are such that no term in the sequence is zero). Prove that $u_n(u_n+u_{n+2}) = u_{n+1}(u_{n-1}+u_{n+1})\,$. Prove also that \[ u_{n+1} -pu_n + u_{n-1}=0 \] for some number $p$ which you should express in terms of $\alpha$, $\beta$ and $q$. Hence, or otherwise, show that if $\beta> \alpha+q$, the sequence is strictly increasing (that is, $u_{n+1}-u_n > 0$ for all $n$). Comment on the case $\beta =\alpha +q$.

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2012 Paper 2 Q9

D: 1600.0 B: 1500.0

projectiles inequality quadratic trigonometric trajectory equation range net clearance optimisation

A tennis ball is projected from a height of $2h$ above horizontal ground with speed $u$ and at an angle of $\alpha$ below the horizontal. It travels in a plane perpendicular to a vertical net of height $h$ which is a horizontal distance of $a$ from the point of projection. Given that the ball passes over the net, show that \[ \frac 1{u^2}< \frac {2(h-a\tan\alpha)}{ga^2\sec^2\alpha}\,. \] The ball lands before it has travelled a horizontal distance of $b$ from the point of projection. Show that \[ \sqrt{u^2\sin^2\alpha +4gh \ } < \frac{bg}{u\cos\alpha} + u \sin\alpha\,. \] Hence show that \[ \tan\alpha < \frac{h(b^2-2a^2)}{ab(b-a)}\,. \]

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2012 Paper 2 Q10

D: 1600.0 B: 1500.0

friction rigid body equilibrium circular cylinder rod moments normal reaction trigonometric inequality geometry

A hollow circular cylinder of internal radius $r$ is held fixed with its axis horizontal. A uniform rod of length $2a$ (where $a < r$) rests in equilibrium inside the cylinder inclined at an angle of $\theta$ to the horizontal, where $\theta\ne0$. The vertical plane containing the rod is perpendicular to the axis of the cylinder. The coefficient of friction between the cylinder and each end of the rod is $\mu$, where $\mu > 0$. Show that, if the rod is on the point of slipping, then the normal reactions $R_1$ and $R_2$ of the lower and higher ends of the rod, respectively, on the cylinder are related by \[ \mu(R_1+R_2) = (R_1-R_2)\tan\phi \] where $\phi$ is the angle between the rod and the radius to an end of the rod. Show further that \[ \tan\theta = \frac {\mu r^2}{r^2 -a^2(1+\mu^2)}\,. \] Deduce that $\lambda <\phi $, where $\tan\lambda =\mu$.

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2012 Paper 2 Q11

D: 1600.0 B: 1484.9

momentum collisions perfectly inelastic collision sequential collisions kinetic energy loss summation proof

A small block of mass $km$ is initially at rest on a smooth horizontal surface. Particles $P_1$, $P_2$, $P_3$, $\ldots$ are fired, in order, along the surface from a fixed point towards the block. The mass of the $i$th particle is $im$ ($i = 1, 2, \ldots$)and the speed at which it is fired is $u/i\,$. Each particle that collides with the block is embedded in it. Show that, if the $n$th particle collides with the block, the speed of the block after the collision is \[ \frac{2nu}{2k +n(n+1)}\,. \] In the case $2k = N(N+1)$, where $N$ is a positive integer, determine the number of collisions that occur. Show that the total kinetic energy lost in all the collisions is \[ \tfrac12 mu^2\bigg( \sum_{n=2}^{N+1} \frac 1 n \bigg)\,. \]

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2012 Paper 2 Q12

D: 1600.0 B: 1500.7

probability conditional Bayes' theorem tree diagram binomial distribution mode inequality optimisation

A modern villa has complicated lighting controls. In order for the light in the swimming pool to be on, a particular switch in the hallway must be on and a particular switch in the kitchen must be on. There are four identical switches in the hallway and four identical switches in the kitchen. Guests cannot tell whether the switches are on or off, or what they control. Each Monday morning a guest arrives, and the switches in the hallway are either all on or all off. The probability that they are all on is $p$ and the probability that they are all off is $1-p$. The switches in the kitchen are each on or off, independently, with probability $\frac12$.

On the first Monday, a guest presses one switch in the hallway at random and one switch in the kitchen at random. Find the probability that the swimming pool light is on at the end of this process. Show that the probability that the guest has pressed the swimming pool light switch in the hallway, given that the light is on at the end of the process, is $\displaystyle \frac{1-p}{1+2p}$.
On each of seven Mondays, guests go through the above process independently of each other, and each time the swimming pool light is found to be on at the end of the process. Given that the most likely number of days on which the swimming pool light switch in the hallway was pressed is 3, show that $\frac14 < p < \frac{5}{14}$.

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2012 Paper 2 Q13

D: 1600.0 B: 1516.0

Poisson distribution probability density function continuous random variable expected value variance Gaussian integral CDF to PDF spatial Poisson process nearest neighbour distance

In this question, you may assume that $\displaystyle \int_0^\infty \!\!\! \e^{-x^2/2} \d x = \sqrt{\tfrac12 \pi}\,$. The number of supermarkets situated in any given region can be modelled by a Poisson random variable, where the mean is $k$ times the area of the given region. Find the probability that there are no supermarkets within a circle of radius $y$. The random variable $Y$ denotes the distance between a randomly chosen point in the region and the nearest supermarket. Write down $\P(Y < y)$ and hence show that the probability density function of $Y$ is $\displaystyle 2\pi y k \e^{-\pi k y^2}$ for $y\ge0$. Find $\E(Y)$ and show that $\var(Y) = \dfrac{4-\pi}{4\pi k}$.

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