Let $$f(x) = k\cos x - \cos 2x,$$ where \(k\) is a constant, \(k > 0\). By considering the sign of \(f'(x)\), or otherwise, find the greatest and least values taken by \(f(x)\) for \(0 \leq x \leq \frac{1}{2}\pi\), distinguishing the various cases that arise according to the value of \(k\). Sketch the graph of \(y = f(x)\) for \(0 \leq x \leq \frac{1}{2}\pi\) in each case.
Two numbers \(a\) and \(b\) are given such that \(a > b > 0\). Two sequences \(a_n\) and \(b_n\) (\(n = 0, 1, 2, \ldots\)) are defined by the rules:
Complex numbers \(z = re^{i\theta}\) (\(r > 0\), \(\theta\) real) and \(w = u + iv\) (\(u\), \(v\) real) are connected by the relation $$2w = z + \frac{1}{z},$$ and \(z\) and \(w\) are represented by points in complex planes. Find the loci described by \(w\) when \(z\) describes the following curves:
Show that the increment in the radius \(R\) of the circumcircle of a triangle \(ABC\) due to small increments in the sides \(a\), \(b\), \(c\) is given by $$\delta R = \Sigma \frac{\delta a}{a} \cot B \cot C.$$ \(R\) is calculated from measurements of \(a\), \(b\), \(c\), and each measurement is liable to a small relative error \(\epsilon\) (so that, for example, \(\delta a\) can lie anywhere between \(\pm \epsilon a\)). Show that, when \(A\), \(B\), \(C\) are all acute, the calculated value of \(R\) is likewise liable to a relative error \(\epsilon\). How must this result be modified when \(A\) is obtuse?
Prove that $$\log \frac{n}{n-1} - \frac{1}{n} = \int_0^1 \frac{t}{(n-t)^n} dt \quad (n = 2, 3, \ldots).$$ Denoting the right-hand side by \(u_n\), prove that $$0 < u_n < \frac{1}{2(n-1)^n},$$ and that the series \(\sum_{n=2}^\infty u_n\) is convergent, with a sum \(U\) satisfying \(0 < U < \frac{1}{2}\). Deduce (or prove otherwise) that $$\sum_{n=1}^N \frac{1}{n} - \log N$$ tends to a limit \(\gamma\) as \(N \to \infty\), and that \(\frac{1}{2} < \gamma < 1\).
The function \(f(x)\) is defined, for \(x > 0\), by the formula $$f(x) = \int_0^{\pi/2} \frac{d\theta}{x + \cos\theta}.$$ Evaluate \(f(x)\), distinguishing the cases (i) \(0 < x < 1\), (ii) \(x = 1\), (iii) \(x > 1\), and expressing the results in cases (i) and (iii) in terms of the variable \(u\) defined by $$u^2 = \frac{|x-1|}{x+1}, \quad u > 0.$$ Prove from your results, or from the original definition, that \(f(x) \to f(1)\) when \(x \to 1\) from below and from above.
Let $$S_r = \int_0^{\pi/2} \sin^r\theta \, d\theta \quad (r \geq 0),$$ $$P_r = rS_rS_{r-1} \quad (r \geq 1),$$ where \(r\) is not necessarily an integer. Prove that
Sketch the curve \(C\) whose equation in polar coordinates is $$r^2 = a^2\cos 2\theta,$$ where \(a > 0\) and it is understood that \(r\) may take negative as well as positive values. Show that the perimeter \(s\) of \(C\) is given by $$s = 4a \int_0^{\pi/4} \frac{d\theta}{\sqrt{\cos 2\theta}}.$$ By means of the substitution \(t = \tan^4\theta\), or otherwise, express \(s\) in terms of the function \(B(p, q)\) defined by $$B(p, q) = \int_0^1 t^{p-1}(1-t)^{q-1} dt \quad (p > 0, q > 0).$$
The functions \(u = u(x, y)\) and \(v = v(x, y)\) satisfy the equations $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x},$$ identically. By means of the substitution \(x = X + Y\), \(y = X - Y\), or otherwise, prove that \(u + v\) is a function of \(x + y\), and \(u - v\) is a function of \(x - y\). What can be said about \(f(x, y)\) if it satisfies the equation $$\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 f}{\partial y^2}$$ identically? Illustrate your conclusion in the cases
Verify that the differential equation $$y'' = (x^2 - 1)y,$$ where dashes denote differentiations with respect to \(x\), is satisfied by \(y = e^{-\frac{1}{2}x^2}\). By writing \(y = ue^{-\frac{1}{2}x^2}\) and forming the differential equation for \(u\), or otherwise, obtain an expression for the general solution of the equation in \(y\). Show that the solution for which \(y = a\) and \(y' = b\) when \(x = 0\) may be written in the form $$y = ae^{-\frac{1}{2}x^2} + b \int_0^x e^{-\frac{1}{2}t^2} dt.$$