Problems

In this question, the numbers \(a\), \(b\) and \(c\) may be complex.

1. Let \(p\), \(q\) and \(r\) be real numbers. Given that there are numbers \(a\) and \(b\) such that \[ a + b = p, \quad a^2 + b^2 = q \quad \text{and} \quad a^3 + b^3 = r, \qquad (*) \] show that \(3pq - p^3 = 2r\).
2. Conversely, you are given that the real numbers \(p\), \(q\) and \(r\) satisfy \(3pq - p^3 = 2r\). By considering the equation \(2x^2 - 2px + (p^2 - q) = 0\), show that there exist numbers \(a\) and \(b\) such that the three equations \((*)\) hold.
3. Let \(s\), \(t\), \(u\) and \(v\) be real numbers. Given that there are distinct numbers \(a\), \(b\) and \(c\) such that \[ a + b + c = s, \quad a^2 + b^2 + c^2 = t, \quad a^3 + b^3 + c^3 = u \quad \text{and} \quad abc = v, \] show, using part~(i), that \(c\) is a root of the equation \[ 6x^3 - 6sx^2 + 3(s^2 - t)x + 3st - s^3 - 2u = 0 \] and write down the other two roots. Deduce that \(s^3 - 3st + 2u = 6v\).
4. Find numbers \(a\), \(b\) and \(c\) such that \[ a + b + c = 3, \quad a^2 + b^2 + c^2 = 1, \quad a^3 + b^3 + c^3 = -3 \quad \text{and} \quad abc = 2, \qquad (**) \] and verify that your solution satisfies the four equations \((**)\).

The number of texts that George receives on his mobile phone can be modelled by a Poisson random variable with mean \(\lambda\) texts per hour. Given that the probability George waits between 1 and 2 hours in the morning before he receives his first text is \(p\), show that \[ p\e^{2\lambda}-\e^{\lambda}+1=0. \] Given that \(4p<1\), show that there are two positive values of \(\lambda\) that satisfy this equation. The number of texts that Mildred receives on each of her two mobile phones can be modelled by independent Poisson random variables with different means \(\lambda_{1}\) and \(\lambda_{2}\) texts per hour. Given that, for each phone, the probability that Mildred waits between 1 and 2 hours in the morning before she receives her first text is also \(p\), find an expression for \(\lambda_{1}+\lambda_{2}\) in terms of \(p\). Find the probability, in terms of \(p\), that she waits between 1 and 2 hours in the morning to receive her first text.

1. Sketch on the same axes the functions \({\rm cosec}\, x\) and \(2x/ \pi\), for \(0 < x < \pi\,\). Deduce that the equation \(x\sin x = \pi/2 \) has exactly two roots in the interval \(0 < x < \pi\,\). Show that \[ \displaystyle \int_{\pi/2}^{\pi} \left \vert x\sin x - \frac{\pi} { 2} \right \vert \; \mathrm{d}x = 2\sin\alpha +\frac{3\pi^2} 4 - \alpha \pi -\pi -2\alpha \cos\alpha -1 \] where \(\alpha\) is the larger of the roots referred to above.
2. Show that the region bounded by the positive \(x\)-axis, the \(y\)-axis and the curve \[y = \Bigl| \vert \e^x - 1 \vert - 1 \Bigr|\] has area \(\ln 4-1\).

Let $$ \f(x) = P \, {\sin x} + Q\, {\sin 2x} + R\, {\sin 3x} \;. $$ Show that if \(Q^2 < 4R(P-R)\), then the only values of \(x\) for which \(\f(x) = 0\) are given by \(x=m\pi\), where \(m\) is an integer. \newline [You may assume that \(\sin 3x = \sin x(4\cos^2 x -1)\).] Now let $$ \g(x) = {\sin 2nx} + {\sin 4nx} - {\sin 6nx}, $$ where \(n\) is a positive integer and \(0 < x < \frac{1}{2}\pi \). Find an expression for the largest root of the equation \(\g(x)=0\), distinguishing between the cases where \(n\) is even and \(n\) is odd.

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 \(q\).

A damped system with feedback is modelled by the equation \[ \mathrm{f}'(t)+\mathrm{f}(t)-k\mathrm{f}(t-1)=0,\mbox{ }\tag{\(\dagger\)} \] where \(k\) is a given non-zero constant. Show that (non-zero) solutions for \(\mathrm{f}\) of the form \(\mathrm{f}(t)=A\mathrm{e}^{pt},\) where \(A\) and \(p\) are constants, are possible provided \(p\) satisfies \[ p+1=k\mathrm{e}^{-p}.\mbox{ }\tag{*} \] Show also, by means of a sketch, or otherwise, that equation \((*)\) can have \(0,1\) or \(2\) real roots, depending on the value of \(k\), and find the set of values of \(k\) for which such solutions of \((\dagger)\) exist. For what set of values of \(k\) do such solutions tend to zero as \(t\rightarrow+\infty\)?

Show Solution

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Solution: Suppose \(f(t) = Ae^{pt}\) is a solution, then \begin{align*} && 0 &= Ape^{pt} + Ae^{pt} - Ake^{p(t-1)} \\ \Leftrightarrow && 0 &= p +1 - ke^{-p} \\ \Leftrightarrow && p+1 &= ke^{-p} \end{align*}

+ - ↺

If we sketch \(y = x+1\) and \(y = ke^{-x}\) we can see that when \(k \geq 0\) there will be exactly one solution. If \(k < 0\) there can be no solutions (if \(k\) is large and negative) and two solutions if \(k\) is small and negative. There will be exactly one real root if \(y = x+1\) is tangent to \(y = ke^{-x}\). The gradient of \(y = ke^{-x}\) is \(-ke^{-x}\) so to have a gradient \(1\) we must have \(x+1 = -1 \Rightarrow x = -2\), but also \(x = -\ln(-k) \Rightarrow k = -e^{-2}\). Therefore there will be so solution as long as \(k \geq -e^{-2}\). The solutions will tend to \(0\) as \(t \to + \infty\) as long as the intersection point is less than zero. For \(k \geq 0\) they intersect at \(0\) when \(k =1\), so we would want \(k < 1\). All negative values of k will work since the intersection has to happen for negative \(p\). Therefore the range is \(-e^{-2} \leq k < 1\)

By considering the graphs of \(y=kx\) and \(y=\sin x,\) show that the equation \(kx=\sin x,\) where \(k>0,\) may have \(0,1,2\) or \(3\) roots in the interval \((4n+1)\frac{\pi}{2} < x < (4n+5)\frac{\pi}{2},\) where \(n\) is a positive integer. For a certain given value of \(n\), the equation has exactly one root in this interval. Show that \(k\) lies in an interval which may be written \(\sin\delta < k < \dfrac{2}{(4n+1)\pi},\) where \(0 < \delta < \frac{1}{2}\pi\) and \[ \cos\delta=\left((4n+5)\frac{\pi}{2}-\delta\right)\sin\delta. \] Show that, if \(n\) is large, then \(\delta\approx\dfrac{2}{(4n+5)\pi}\) and obtain a second, improved, approximation.

Show Solution

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Solution:

+ - ↺

Clearly we can achieve \(0\), \(1\), and \(2\) intersections by never entering the range, entering too flat, or entering before hitting the second branch. To achieve \(3\) we can go at a flat enough slope that we hit somewhere near the top of the second branch, and since the gradient there will be \(\approx 0\), and our gradient is positive, we must intersect before that point as well, ie \(3\) intersections. Clearly we cannot intersect the second branch \(3\) times or the first branch twice, therefore there are at most \(3\) intersections. To intersect the graph only once, we need to:

• be below \(((4n+1)\tfrac{\pi}{2}, 1)\) and
• not touch the second gradient

The first condition means that \(k (4n+1)\tfrac{\pi}{2} < 1 \Rightarrow k < \frac{2}{(4n+1)\pi}\). For the second condition, consider a point on the curve \(\sin x\) whose tangent line goes through the origin, ie \(\frac{y - \sin t}{x - t} = \cos t \Rightarrow y = (\cos t)x - t \cos t+\sin t\) ie \(\sin t = t \cos t\). For this point \(t\) to be in the required interval, we need \((4n+5) \tfrac{\pi}{2} -t \in (0, \frac{\pi}{2})\), so let's call this value \(\delta\). Then our result is: The gradient needs to be steeper than \(\cos t = \cos \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) = \sin \delta\) and \(\cos \delta =\left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \sin \delta \). If \(n\) is large, then, \begin{align*} && 1 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1 &\approx (4n+5) \tfrac{\pi}{2} \delta \\ \Rightarrow && \delta &\approx \frac{2}{(4n+5)\pi} \end{align*}. To higher order: \begin{align*} && 1-\frac12 \delta^2 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1-\frac12 \delta^2 &\approx (4n+5) \tfrac{\pi}{2} \delta - \delta^2 \\ \Rightarrow && 0 &\approx 1 - (4n + 5)\tfrac{\pi}{2} \delta + \frac12 \delta^2 \\ \Rightarrow && \delta &\approx (4n+5) \tfrac{\pi}{2} - \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2} \\ &&&= \frac{2}{(4n+5) \tfrac{\pi}{2} + \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2}} \end{align*}.