1531 problems found
Oxtown and Camville are connected by three roads, which are at risk of being blocked by flooding. On two of the three roads there are two sections which may be blocked. On the third road there is only one section which may be blocked. The probability that each section is blocked is \(p\). Each section is blocked independently of the other four sections. Show that the probability that Oxtown is cut off from Camville is \(p^3 \l 2-p \r^2\). I want to travel from Oxtown to Camville. I choose one of the three roads at random and find that my road is not blocked. Find the probability that I would not have reached Camville if I had chosen either of the other two roads. You should factorise your answer as fully as possible. Comment briefly on the value of this probability in the limit \(p\to1\).
A very generous shop-owner is hiding small diamonds in chocolate bars. Each diamond is hidden independently of any other diamond, and on average there is one diamond per kilogram of chocolate.
Solution: Not that the number of diamonds per kilogram is \(1\) so we are assuming it is \(Po(M)\) where \(M\) is the mass in kg. In particular \(\E[X] = M\) and \(\mathbb{P}(X = 0) = e^{-M}\)
Solution:
The sequence of real numbers \(u_1\), \(u_2\), \(u_3\), \(\ldots\) is defined by \begin{equation*} u_1=2 \,, \qquad\text{and} \qquad u_{n+1} = k - \frac{36}{u_n} \quad \text{for } n\ge1, \tag{\(*\)} \end{equation*} where \(k\) is a constant.
Using the series \[ \e^x = 1 + x +\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+\cdots\,, \] show that \(\e>\frac83\). Show that \(n!>2^n\) for \(n\ge4\) and hence show that \(\e<\frac {67}{24}\). Show that the curve with equation \[ y= 3\e^{2x} +14 \ln (\tfrac43-x)\,, \qquad {x<\tfrac43} \] has a minimum turning point between \(x=\frac12\) and \(x=1\) and give a sketch to show the shape of the curve.
Solution: \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&> 1 + 1+ \frac12 + \frac16 \\ &&&= \frac{12+3+1}{6} = \frac83 \end{align*} \(4! = 24 > 16 = 2^4\), notice that \(n! = \underbrace{n \cdot (n-1) \cdots 5}_{>2^{n-4}} \cdot \underbrace{4!}_{>2^4} >2^n\). \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&< \frac83 + \frac{1}{2^4} + \frac{1}{2^5} + \cdots \\ &&&= \frac83 + \frac{1}{2^4} \frac{1}{1-\tfrac12} \\ &&&= \frac83 + \frac1{8} \\ &&&= \frac{67}{24} \end{align*} \begin{align*} && y &= 3e^{2x} +14 \ln(\tfrac43-x) \\ && y' &= 6e^{2x} - \frac{14}{\tfrac43-x} \\ && y'(\tfrac12) &= 6e - \frac{14}{\tfrac43-\tfrac12} \\ &&&= 6e -\tfrac{84}{5} = 6(e-\tfrac{14}5) < 0 \\ && y'(1) &= 6e^2 - \frac{14}{\tfrac43-1} \\ &&&= 6e^2 - 42 = 6(e^2-7) \\ &&&> 6(\tfrac{64}{9} - 7) > 0 \end{align*} Therefore \(y'\) changes from negative (decreasing) to positive (increasing) in our range, and therefore there is a minima in this range.
Solution:
By making the substitution \(x=\pi-t\,\), show that \[ \! \int_0^\pi x\f(\sin x) \d x = \tfrac12 \pi \! \int_0^\pi \f(\sin x) \d x\,, \] where \(\f(\sin x)\) is a given function of \(\sin x\). Evaluate the following integrals:
The notation \({\lfloor } x \rfloor\) denotes the greatest integer less than or equal to the real number \(x\). Thus, for example, \(\lfloor \pi\rfloor =3\,\), \(\lfloor 18\rfloor =18\,\) and \(\lfloor-4.2\rfloor = -5\,\).
Solution:
By considering a suitable scalar product, prove that \[ (ax+by+cz)^2 \le (a^2+b^2+c^2)(x^2+y^2+z^2) \] for any real numbers \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\). Deduce a necessary and sufficient condition on \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\) for the following equation to hold: \[ (ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2) \,. \]
Solution: Consider \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\), then we know that \begin{align*} && \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= \sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2} \cos \theta \\ \Rightarrow && (ax+by+cz)^2 &= (a^2+b^2+c^2)(x^2+y^2+z^2) \cos^2 \theta \\ &&&\leq (a^2+b^2+c^2)(x^2+y^2+z^2) \end{align*} For equality to hold, we must have that the vectors are parallel, ie \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)
An ellipse has equation $\dfrac{x^2}{a^2} +\dfrac {y^2}{b^2} = 1$. Show that the equation of the tangent at the point \((a\cos\alpha, b\sin\alpha)\) is \[ y=- \frac {b \cot \alpha} a \, x + b\, {\rm cosec\,}\alpha\,. \] The point \(A\) has coordinates \((-a,-b)\), where \(a\) and \(b\) are positive. The point \(E\) has coordinates \((-a,0)\) and the point \(P\) has coordinates \((a,kb)\), where \(0 < k < 1\). The line through \(E\) parallel to \(AP\) meets the line \(y=b\) at the point \(Q\). Show that the line \(PQ\) is tangent to the above ellipse at the point given by \(\tan(\alpha/2)=k\). Determine by means of sketches, or otherwise, whether this result holds also for \(k=0\) and \(k=1\).
Show that the line through the points with position vectors \(\bf x\) and \(\bf y\) has equation \[{\bf r} = (1-\alpha){\bf x} +\alpha {\bf y}\,, \] where \(\alpha\) is a scalar parameter. The sides \(OA\) and \(CB\) of a trapezium \(OABC\) are parallel, and \(OA>CB\). The point \(E\) on \(OA\) is such that \(OE : EA = 1:2\), and \(F\) is the midpoint of \(CB\). The point \(D\) is the intersection of \(OC\) produced and \(AB\) produced; the point \(G\) is the intersection of \(OB\) and \(EF\); and the point \(H\) is the intersection of \(DG\) produced and \(OA\). Let \(\bf a\) and \(\bf c\) be the position vectors of the points \(A\) and \(C\), respectively, with respect to the origin \(O\).
A painter of weight \(kW\) uses a ladder to reach the guttering on the outside wall of a house. The wall is vertical and the ground is horizontal. The ladder is modelled as a uniform rod of weight \(W\) and length \(6a\). The ladder is not long enough, so the painter stands the ladder on a uniform table. The table has weight \(2W\) and a square top of side \(\frac12 a\) with a leg of length \(a\) at each corner. The foot of the ladder is at the centre of the table top and the ladder is inclined at an angle \(\arctan 2\) to the horizontal. The edge of the table nearest the wall is parallel to the wall. The coefficient of friction between the foot of the ladder and the table top is \(\frac12\). The contact between the ladder and the wall is sufficiently smooth for the effects of friction to be ignored.
Three particles, \(A\), \(B\) and \(C\), of masses \(m\), \(km\) and \(3m\) respectively, are initially at rest lying in a straight line on a smooth horizontal surface. Then \(A\) is projected towards \(B\) at speed \(u\). After the collision, \(B\) collides with \(C\). The coefficient of restitution between \(A\) and \(B\) is \(\frac12\) and the coefficient of restitution between \(B\) and \(C\) is \(\frac14\).
Solution:
A projectile of unit mass is fired in a northerly direction from a point on a horizontal plain at speed \(u\) and an angle \(\theta\) above the horizontal. It lands at a point \(A\) on the plain. In flight, the projectile experiences two forces: gravity, of magnitude \(g\); and a horizontal force of constant magnitude \(f\) due to a wind blowing from North to South. Derive an expression, in terms of \(u\), \(g\), \(f\) and \(\theta\) for the distance \(OA\).
A cricket team has only three bowlers, Arthur, Betty and Cuba, each of whom bowls 30 balls in any match. Past performance reveals that, on average, Arthur takes one wicket for every 36 balls bowled, Betty takes one wicket for every 25 balls bowled, and Cuba takes one wicket for every 41 balls bowled.
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