Solution: \begin{questionparts} \item The tangent to \(y = \ln x\) is \begin{align*} && \frac{y - \ln x_1}{x - x_1} &= \frac{1}{x_1} \\ \Rightarrow && \frac{x_1y -x_1 \ln x_1}{ x- x_1} &= 1 \\ \Rightarrow && x_1 y - x_1 \ln x_1 &= x - x_1 \end{align*} So to run through the origin, we need \(\ln x_1 = 1 \Rightarrow x_1 = e\) so the line will be \(y = \frac1{e} x\) If \(ma = \ln a \Rightarrow m = \frac{\ln a}{a} = \frac{\ln b}{b} \Rightarrow b \ln a = a \ln b \Rightarrow a^b = b^a\). \item
For \(n\ge 0\), let \[ I_n = \int_0^1 x^n(1-x)^n\d x\,. \]
Solution:
The line passing through the point \((a,0)\) with gradient \(b\) intersects the circle of unit radius centred at the origin at \(P\) and \(Q\), and \(M\) is the midpoint of the chord \(PQ\). Find the coordinates of \(M\) in terms of \(a\) and \(b\).
Solution: \begin{align*} && y &= bx-ba \\ && 1 &= x^2 + y^2 \\ \Rightarrow && 1 &= x^2 + b^2(x-a)^2 \\ \Rightarrow && 0 &= (1+b^2)x^2-2ab^2x+b^2a^2-1 \end{align*} This will have roots which sum to \(\frac{2ab^2}{1+b^2}\), therefore \(M = \left ( \frac{ab^2}{1+b^2}, \frac{ab^3}{1+b^2}-ba \right)=\left ( \frac{ab^2}{1+b^2}, \frac{-ba}{1+b^2} \right)\)
In this question, the following theorem may be used.\newline {\sl Let \(u_1\), \(u_2\), \(\ldots\) be a sequence of (real) numbers. If the sequence is bounded above (that is, \(u_n\le b\) for all \(n\), where \(b\) is some fixed number) and increasing (that is, \(u_n\ge u_{n-1}\) for all \(n\)), then the sequence tends to a limit (that is, converges).} The sequence \(u_1\), \(u_2\), \(\ldots\) is defined by \(u_1=1\) and \[ u_{n+1} = 1+\frac 1{u_n} \ \ \ \ \ \ \ \ \ \ (n\ge1)\,. \tag{\(*\)} \]
The function \(\f\) satisfies \(\f(x)>0\) for \(x\ge0\) and is strictly decreasing (which means that \(\f(b)<\f(a)\) for \(b>a\)).
The diagram shows three identical discs in equilibrium in a vertical plane. Two discs rest, not in contact with each other, on a horizontal surface and the third disc rests on the other two. The angle at the upper vertex of the triangle joining the centres of the discs is \(2\theta\).
A particle is projected at an angle of elevation \(\alpha\) (where \(\alpha>0\)) from a point \(A\) on horizontal ground. At a general point in its trajectory the angle of elevation of the particle from \(A\) is \(\theta\) and its direction of motion is at an angle \(\phi\) above the horizontal (with \(\phi\ge0\) for the first half of the trajectory and \(\phi\le0\) for the second half). Let \(B\) denote the point on the trajectory at which \(\theta = \frac12 \alpha\) and let \(C\) denote the point on the trajectory at which \(\phi = -\frac12\alpha\).