A bead of mass \(m\) can slide freely on a straight rod, which can rotate in a horizontal plane about a vertical axis fixed to a point \(O\) of the rod and mounted in smooth bearings. The rod is kept in motion with constant angular velocity \(\omega\) by a couple of moment \(G\) applied to the axis. At time \(t=0\) the bead is at a distance \(a\) from \(O\) and is at rest relative to the rod. Find the distance of the bead from \(O\) at time \(t\). \newline Find how \(G\) must vary with \(t\) to maintain the motion, and verify that the increase of kinetic energy in any interval of time is equal to the work done by the couple in that interval.
Solve: \begin{align*} x+y+z &= 1, \\ x^2+y^2+z^2 &= 21, \\ x^3+y^3+z^3 &= 55. \end{align*}
The equation \(x^4+ax^3+bx^2+cx+d=0\) is such that the sum of two of its roots is equal to the sum of the remaining two. Shew that \(a^3-4ab+8c=0\). \newline If, in particular, \(a=2, b=-1, c=-2, d=-3\), find all the roots.
Shew that the geometric mean of \(n\) positive numbers is not greater than their arithmetic mean. \newline By expressing \((n!)^2\) as a product of \(n\) factors, or otherwise, shew that \[ n! < \left\{\frac{1}{6}(n+1)(n+2)\right\}^{\frac{1}{2}n}. \]
Find, to the nearest minute, all angles \(x\) and \(y\) for which \begin{align*} \tan \tfrac{1}{2}x + \tan \tfrac{1}{2}y &= \frac{2}{7}, \\ \tan x + \tan y &= \frac{28}{45}. \end{align*}
Solution: Let \(t = \tan \tfrac12 x, s = \tan \tfrac12 y\), then we have \begin{align*} && t + s &= \frac27 \\ && \frac{2t}{1-t^2} + \frac{2s}{1-s^2} &= \frac{28}{45} \\ \Rightarrow && \frac{t(1-s^2)+s(1-t^2)}{(1-t^2)(1-s^2)} &= \frac{14}{45} \\ \Rightarrow && \frac27 -st(s+t) &= \frac{14}{45}(1-t^2)(1-s^2) \\ \Rightarrow && \frac27(1-st) &= \frac{14}{45}(1 - t^2-s^2+s^2t^2) \\ &&&= \frac{14}{45}(1-(s+t)^2+2st+s^2t^2) \\ &&&= \frac{14}{45}(1-\frac{4}{49}+2st + s^2t^2) \\ \Rightarrow && 90-90st &= 98-8+196st+98s^2t^2 \\ \Rightarrow && 0 &= 286st+98s^2t^2 \\ \Rightarrow && 0 &= st(143+49st) \\ \Rightarrow && st &= 0, st =-\frac{143}{49} \end{align*} Case 1: \(st = 0\), WLOG \(s = 0\), so \(t = \tfrac27 \Rightarrow \tan \tfrac12x = \tfrac27 \Rightarrow x = 2\tan^{-1} \tfrac27 + 2n\pi, y = 2m\pi\) Case 2: \(st = -\frac{143}{49}\) so \(s,t\) are roots of: \begin{align*} && 0 &= x^2-\frac27x -\frac{143}{49} \\ && 0 &= 49x^2-14x-143 \\ && 0 &= (7x+11)(7x-13) \\ \Rightarrow && s,t &= \frac{13}{7}, -\frac{11}{7} \end{align*} So \(x = 2\tan^{-1} \tfrac{13}{7}+2n\pi, y = -2\tan^{-1}\frac{11}{7}+2m\pi\)
The function \(f(x)\) and the constant \(a\) are defined by \[ f(x) = \int_0^x \frac{dt}{1+t^2}, \quad a = \lim_{x \to \infty} f(x) = \int_0^\infty \frac{dt}{1+t^2}. \] By forming the difference \(f(x)-f(y)\) and making a suitable change of variable in the integrand, shew that \[ f(x)-f(y)=f\left(\frac{x-y}{1+xy}\right). \] Prove also that \(f(x)+f(1/x)=a\), and that \(f(1)=\frac{1}{2}a, f(\sqrt{3})=\frac{2}{3}a\). The definition of \(f(x)\) given above should be used and no properties of trigonometrical functions should be assumed.
Find the numerical values of \[ y = \sin\left(x+\frac{\pi}{4}\right) + \frac{1}{4}\sin 4x \] at its stationary values in the range \(-\pi \le x \le \pi\). Distinguish between maxima, minima and points of inflexion, and give a rough sketch of the curve.
Evaluate the integrals \[ \int_0^\infty \frac{x \tan^{-1}x}{(1+x^2)^2} \, dx, \quad \int_0^\pi \frac{\cos^2\theta \, d\theta}{a^2\cos^2\theta+b^2\sin^2\theta}, \quad \int \frac{dx}{(1-x)(1+x)^{\frac{1}{2}}}. \]
Trace the curve \(y^2 = x^2(x-a)\) for \(a=1, 0, -1\). Find the area enclosed by the loop in the case \(a=-1\).
If \[ I_{p,q} = \int_0^\pi \sin^p x \cos^q x \, dx \] shew that \[ (p+q)I_{p,q} = \begin{cases} (q-1)I_{p,q-2} & (q \ge 2) \\ (p-1)I_{p-2,q} & (p \ge 2) \end{cases}, \] and evaluate \(I_{\alpha-1,5}\) where \(\alpha\) is any positive real number.