A triangle \(ABC\) has area \(\Delta\), and \(P\) is an interior point. The line through \(P\) parallel to \(BC\) cuts \(AB\) in \(W\) and \(AC\) in \(T\); the line through \(P\) parallel to \(CA\) cuts \(AB\) in \(V\) and \(BC\) in \(S\); and the line through \(P\) parallel to \(AB\) cuts \(AC\) in \(U\) and \(BC\) in \(R\). The triangles \(PBC\), \(PCA\), \(PAB\) have areas \(\alpha\), \(\beta\), \(\gamma\) respectively and the triangles \(AVU\), \(BWR\), \(CST\) have areas \(\alpha'\), \(\beta'\), \(\gamma'\) respectively. By first showing that $$\frac{\beta'}{\gamma} = \frac{\alpha}{\Delta},$$ prove that the area of the hexagon \(RSTUVW\) is at least \(\frac{9}{8}\Delta\). [You may assume that if positive numbers \(p\), \(q\), \(r\) add up to 1, then \(pq +qr+rp\) cannot exceed \(\frac{1}{3}\).]
Let \(I_n = \int_0^{\pi/4} \tan^n\theta d\theta\). Obtain an expression for \(I_n\) in terms of \(I_{n-2}\), and hence evaluate \(I_4\) and \(I_5\). Show that for all \(n \geq 1\), \(0 \leq I_{4n} \leq \frac{1}{4}n - \frac{3}{8}\), and \(0 \leq I_{4n-2} \leq 1- \frac{1}{4}n\).
By considering the integral \(\int_1^x \frac{dt}{t}\) or otherwise, prove that \(0 < \log x < x\) for all \(x > 1\). Hence show that for fixed \(k > 0\), \(\frac{\log x}{x^k}\) tends towards 0 as \(x\) tends towards infinity. (You may find it helpful to use the substitution \(y = x^n\) in the first inequality.) Deduce that \(x^k \log x\) tends towards 0 as \(x\) tends towards 0 through positive values. Use this theory to investigate the behaviour of the function \(y = x^x\) (\(x > 0\)) when \(x\) is near to 0. Sketch the graph of \(y = x^x\) for values of \(x > 0\).
Sketch on the same diagram the curves given in polar co-ordinates \((r, \theta)\) by the equations \(r = \frac{1}{2}a(1 + \cos \theta)\) and \(r = a\theta\) (\(a > 0\), \(0 \leq \theta \leq 2\pi\)). Find the area of the region consisting of all those points \((r,\theta)\) such that \(\frac{1}{3}\pi \leq \theta \leq 2\pi\) and \(\frac{1}{2}a(1 + \cos \theta) \leq r \leq a\theta\).
The section of the curve \(y = \cosh x\) between \(x = 0\) and \(x = a\) is rotated about the \(x\)-axis. Prove that the numerical value of the curved surface area thus obtained is twice that of the volume enclosed. The curve is now rotated about the \(y\)-axis. Calculate the ratio of the numerical values of volume to curved surface area, and show that in this case it depends on \(a\).
In a relay race the baton cannot be passed successfully between two runners unless they are in the same place, travelling at the same speed. The race is run on a straight track and the team manager watches from a point \(O\). Runner \(A\), carrying the baton, passes the manager at time \(t = 0\). \(A\) is running at the speed \(\lambda v_0\), but as he passes the point \(O\) he begins to slow down. His deceleration at any subsequent point is numerically equal to his distance \(s\) from \(O\). As \(A\) passes him, the manager observes \(B\) (to whom the baton is to be passed) at a distance \(s_0\) ahead of \(A\) travelling at a steady speed \(v_0\). Assuming that \(B\) will maintain this speed, prove that the baton can be passed successfully provided \(\lambda \geq 1\) and $$s_0 = v_0\{(\lambda^2-1)^{\frac{1}{2}} - \cos^{-1}(\frac{1}{\lambda})\}.$$ Hence show that if \(\lambda > 1\), a value of \(s_0 > 0\) can be chosen in such a way that a successful hand-over can be made.