A uniform fine chain of length \(l\) is suspended with its lower end just touching a horizontal table. The chain is allowed to fall freely. If the mass of the chain is \(M\), find the force on the table when a length \(x\) has reached it. [You may assume that the part of the chain on the table does not interfere with the subsequent motion.]
Let \(f(x)\) be a polynomial in \(x\). Explain why, if \(z\) is an approximation to a root of \(f(x)\), then \(z-f(z)/f'(z)\) is often a closer approximation. By considering polynomials of the form \(x^r+a\), and suitable real values of \(z_0\), show that the iteration \[z_n = z_{n-1}-f(z_{n-1})/f'(z_{n-1}) \quad (n = 1, 2, \ldots)\] may exhibit any of the following three behaviours.
Show that, for \(r \geq 10\), \[(r-\frac{1}{2})(r+\frac{1}{2}) < r^2 < (r-\frac{39}{80})(r+\frac{41}{80}).\] Deduce that \[\frac{80}{761} \leq \sum_{r=10}^{\infty} \frac{1}{r^2} \leq \frac{2}{19}.\]
Lady Bracknell is holding a dinner party. She has arranged the six diners around a circular table, with Algernon next to Cecily. It is the custom at Bracknell Hall for those dining to change places several times during the meal, in order to vary the conversation. Let \(G\) be the set of those rearrangements of the six diners after which Algernon and Cecily are still sitting next to each other. (Two rearrangements are to be considered the same if one can be obtained from the other merely by rotating the diners around the table.) Show that \(G\) forms a group, under the operation of performing one rearrangement after another. How many elements does \(G\) have? Now suppose that, in the initial arrangement, Cecily is seated on Algernon's right. Let \(H\) be the set of those elements of \(G\) after which Cecily is still on Algernon's right. Show that whenever \(g\) is an element of \(G\) and \(h\) an element of \(H\), \(ghg^{-1}\) is an element of \(H\).
Evaluate the determinant \[A = \begin{vmatrix} 1 & z & z^2 & 0 \\ 0 & 1 & z & z^2 \\ z^2 & 0 & 1 & z \\ z & z^2 & 0 & 1 \end{vmatrix}.\] Plot in the Argand diagram the points satisfying \(A = 0\).
State Pythagoras's Theorem. Two circles \(\alpha\), \(\beta\) with centres \(A\) and \(B\) and radii \(a\) and \(b\), lie in different planes \(\pi\) and \(\varpi\) respectively which meet in a line \(l\). Show that the two circles will lie on the same sphere if and only if \(AB\) is perpendicular to \(l\) and \[AP^2-BP^2 = a^2-b^2\] for every point \(P\) on \(l\).
Prove that the straight line \[ty = x+at^2\] touches the parabola \(y^2 = 4ax\) (\(a \neq 0\)), and find the coordinates of the point of contact. The tangents from a point to the parabola meet the directrix in points \(L\) and \(M\). Show that, if \(LM\) is of a fixed length \(l\), the point must lie on \[(x+a)^2(y^2-4ax) = l^2x^2.\]
Positive numbers \(p\) and \(q\) satisfy \[\frac{1}{p}+\frac{1}{q} = 1,\] and \(y\) is defined by \(y = x^{p-1}\), for \(x > 0\). Express \(x\) in terms of \(y\) and \(q\). By considering \(\int_0^s ydx\) and \(\int_0^t xdy\) as areas, or otherwise, show that if \(s > 0\) and \(t > 0\) then \[st \leq \frac{s^p}{p}+\frac{t^q}{q}.\] When does equality hold?
A circular arc subtends an angle \(2\alpha(< \pi)\) at the centre of a circle of radius \(R\). A surface is generated by rotating the arc about the line through its end points. Prove that the area of this surface is \(4\pi R^2(\sin\alpha-\alpha\cos\alpha)\).
A function \(f(x)\) is defined, for \(x > 0\), by \[f(x) = \int_{-1}^1 \frac{dt}{\sqrt{(1-2xt+x^2)}}.\] Prove that, if \(0 \leq x \leq 1\), then \(f(x) = 2\). What is the value of \(f(x)\) if \(x > 1\)? Has \(f(x)\) a derivative at \(x = 1\)?
Solution: \begin{align*} f(x) &= \int_{-1}^1 \frac{\d t}{\sqrt{1-2xt+x^2}}\\ &= \left [-\frac{\sqrt{1-2xt+x^2}}{x} \right] _{-1}^1 \\ &= \left ( -\frac{\sqrt{1-2x+x^2}}{x}\right) - \left ( -\frac{\sqrt{1+2x+x^2}}{x}\right) \\ &= \frac{|1+x|}{x}-\frac{|1-x|}{x} \\ &= \begin{cases} \frac{1+x}{x} - \frac{1-x}{x} & \text{if } 0 < x \leq 1 \\ \frac{1+x}{x} - \frac{x-1}{x} & \text{if } x > 1 \\ \end{cases} \\ &= \begin{cases} 2 & \text{if } 0 < x \leq 1 \\ \frac{2}{x} & \text{if } x > 1 \\ \end{cases} \end{align*} \(f(x)\) does not have a derivative at \(x = 1\) since: \begin{align*} \lim_{x \to 1^-} \frac{f(x)-f(1)}{x-1} &= \frac{2-2}{x-1} \\ &= 0 \\ \lim_{x \to 1^+} \frac{f(x)-f(1)}{x-1} &= \frac{2/x-2}{x-1} \\ &= \frac{2-2x}{x-1} \\ &= -2 \neq 0 \end{align*}