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.
If \(f(x,y)\) is a function of the two independent variables \(x\) and \(y\), define the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\). \newline The identical relation \(g(u^2-x^2, u^2-y^2, u^2-z^2)=0\) defines \(u\) as a function of \(x, y\) and \(z\). Shew that \[ \frac{1}{x}\frac{\partial u}{\partial x} + \frac{1}{y}\frac{\partial u}{\partial y} + \frac{1}{z}\frac{\partial u}{\partial z} = \frac{1}{u}. \]