Edexcel AEA (Advanced Extension Award) 2004 June

Solve the equation \(\cos x + \sqrt{(1 - \frac{1}{2} \sin 2x)} = 0\), in the interval \(0° \leq x < 360°\). [9]

1. For the binomial expansion of \(\frac{1}{(1-x)^2}\), \(|x| < 1\), in ascending powers of \(x\),
   1. find the first four terms,
   2. write down the coefficient of \(x^n\). [2]
2. Hence, show that, for \(|x| < 1\), \(\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}\). [2]
3. Prove that, for \(|x| < 1\), \(\sum_{n=1}^{\infty} (an+1)x^n = \frac{(a+1)x-x^2}{(1-x)^2}\), where \(a\) is a constant. [4]
4. Hence evaluate \(\sum_{n=1}^{\infty} \frac{5n+1}{2^{3n}}\). [2]

$$f(x) = x^3 - (k+4)x + 2k,$$ where \(k\) is a constant.

1. Show that, for all values of \(k\), the curve with equation \(y = f(x)\) passes through the point \((2, 0)\). [1]
2. Find the values of \(k\) for which the equation \(f(x) = 0\) has exactly two distinct roots. [5]

Given that \(k > 0\), that the \(x\)-axis is a tangent to the curve with equation \(y = f(x)\), and that the line \(y = p\) intersects the curve in three distinct points,

1. find the set of values that \(p\) can take. [5]

\includegraphics{figure_1} The circle, with centre \(C\) and radius \(r\), touches the \(y\)-axis at \((0, 4)\) and also touches the line with equation \(4y - 3x = 0\), as shown in Fig. 1.

1. 1. Find the value of \(r\).
   2. Show that \(\arctan \left(\frac{4}{3}\right) + 2 \arctan \left(\frac{1}{2}\right) = \frac{1}{2} \pi\). [8]

The line with equation \(4x + 3y = q\), \(q > 12\), is a tangent to the circle.

1. Find the value of \(q\). [4]

1. Given that \(y = \ln [t + \sqrt{(1 + t^2)}]\), show that \(\frac{dy}{dt} = \frac{1}{\sqrt{(1+t^2)}}\). [3]

The curve \(C\) has parametric equations $$x = \frac{1}{\sqrt{(1+t^2)}}, \quad y = \ln [t + \sqrt{(1 + t^2)}], \quad t \in \mathbb{R}.$$ A student was asked to prove that, for \(t > 0\), the gradient of the tangent to \(C\) is negative. The attempted proof was as follows: $$y = \ln \left(t + \frac{1}{x}\right)$$ $$= \ln \left(\frac{tx + 1}{x}\right)$$ $$= \ln (tx + 1) - \ln x$$ $$\therefore \frac{dy}{dx} = \frac{t}{tx + 1} - \frac{1}{x}$$ $$= \frac{\frac{t}{x}}{t + \frac{1}{x}} - \frac{1}{x}$$ $$= \frac{t\sqrt{(1+t^2)}}{t + \sqrt{(1+t^2)}} - \sqrt{(1 + t^2)}$$ $$= -\frac{(1+t^2)}{t + \sqrt{(1+t^2)}}$$ As \((1 + t^2) > 0\), and \(t + \sqrt{(1 + t^2)} > 0\) for \(t > 0\), \(\frac{dy}{dx} < 0\) for \(t > 0\).

1. 1. Identify the error in this attempt.
   2. Give a correct version of the proof. [6]
2. Prove that \(\ln [-t + \sqrt{(1 + t^2)}] = -\ln [t + \sqrt{(1 + t^2)}]\). [3]
3. Deduce that \(C\) is symmetric about the \(x\)-axis and sketch the graph of \(C\). [3]

$$f(x) = x - [x], \quad x \geq 0$$ where \([x]\) is the largest integer \(\leq x\). For example, \(f(3.7) = 3.7 - 3 = 0.7\); \(f(3) = 3 - 3 = 0\).

1. Sketch the graph of \(y = f(x)\) for \(0 \leq x < 4\). [3]
2. Find the value of \(p\) for which \(\int_2^p f(x) dx = 0.18\). [3]

Given that $$g(x) = \frac{1}{1+kx}, \quad x \geq 0, \quad k > 0,$$ and that \(x_0 = \frac{1}{2}\) is a root of the equation \(f(x) = g(x)\),

1. find the value of \(k\). [2]
2. Add a sketch of the graph of \(y = g(x)\) to your answer to part \((a)\). [1]

The root of \(f(x) = g(x)\) in the interval \(n < x < n + 1\) is \(x_n\), where \(n\) is an integer.

1. Prove that $$2 x_n^2 - (2n - 1)x_n - (n + 1) = 0.$$ [4]
2. Find the smallest value of \(n\) for which \(x_n - n < 0.05\). [4]

Triangle \(ABC\), with \(BC = a\), \(AC = b\) and \(AB = c\) is inscribed in a circle. Given that \(AB\) is a diameter of the circle and that \(a^2\), \(b^2\) and \(c^2\) are three consecutive terms of an arithmetic progression (arithmetic series),

1. express \(b\) and \(c\) in terms of \(a\), [4]
2. verify that \(\cot A\), \(\cot B\) and \(\cot C\) are consecutive terms of an arithmetic progression. [3]

In an acute-angled triangle \(PQR\) the sides \(QR\), \(PR\) and \(PQ\) have lengths \(p\), \(q\) and \(r\) respectively.

1. Prove that $$\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R}.$$ [3]

Given now that triangle \(PQR\) is such that \(p^2\), \(q^2\) and \(r^2\) are three consecutive terms of an arithmetic progression,

1. use the cosine rule to prove that $$\frac{2\cos Q}{q} = \frac{\cos P}{p} + \frac{\cos R}{r}.$$ [6]
2. Using the results given in parts \((c)\) and \((d)\), prove that \(\cot P\), \(\cot Q\) and \(\cot R\) are consecutive terms in an arithmetic progression. [3]