Edexcel AEA (Advanced Extension Award) 2002 June

Solve the following equation, for \(0 \leq x \leq \pi\), giving your answers in terms of \(\pi\). $$\sin 5x - \cos 5x = \cos x - \sin x.$$ [8]

In the binomial expansion of $$(1 - 4x)^p, \quad |x| < \frac{1}{4},$$ the coefficient of \(x^2\) is equal to the coefficient of \(x^4\) and the coefficient of \(x^3\) is positive. Find the value of \(p\). [9]

The curve \(C\) has parametric equations $$x = 15t - t^3, \quad y = 3 - 2t^2.$$ Find the values of \(t\) at the points where the normal to \(C\) at \((14, 1)\) cuts \(C\) again. [11]

Find the coordinates of the stationary points of the curve with equation $$x^3 + y^3 - 3xy = 48$$ and determine their nature. [14]

\includegraphics{figure_1} Figure 1 shows a sketch of part of the curve with equation $$y = \sin (\cos x).$$ The curve cuts the \(x\)-axis at the points \(A\) and \(C\) and the \(y\)-axis at the point \(B\).

1. Find the coordinates of the points \(A\), \(B\) and \(C\). [3]
2. Prove that \(B\) is a stationary point. [2]

Given that the region \(OCB\) is convex,

1. show that, for \(0 \leq x \leq \frac{\pi}{2}\), $$\sin (\cos x) \leq \cos x$$ and $$(1 - \frac{2}{\pi} x) \sin 1 \leq \sin (\cos x)$$ and state in each case the value or values of \(x\) for which equality is achieved. [6]
2. Hence show that $$\frac{\pi}{4} \sin 1 < \int_0^{\frac{\pi}{2}} \sin(\cos x) \, dx < 1.$$ [4]

\includegraphics{figure_2} Figure 2 shows a sketch of part of two curves \(C_1\) and \(C_2\) for \(y \geq 0\). The equation of \(C_1\) is \(y = m_1 - x^{n_1}\) and the equation of \(C_2\) is \(y = m_2 - x^{n_2}\), where \(m_1\), \(m_2\), \(n_1\) and \(n_2\) are positive integers with \(m_2 > m_1\). Both \(C_1\) and \(C_2\) are symmetric about the line \(x = 0\) and they both pass through the points \((3, 0)\) and \((-3, 0)\). Given that \(n_1 + n_2 = 12\), find

1. the possible values of \(n_1\) and \(n_2\), [4]
2. the exact value of the smallest possible area between \(C_1\) and \(C_2\), simplifying your answer, [8]
3. the largest value of \(x\) for which the gradients of the two curves can be the same. Leave your answer in surd form. [5]

A student was attempting to prove that \(x = \frac{1}{2}\) is the only real root of $$x^3 + \frac{1}{4}x - \frac{1}{2} = 0.$$ The attempted solution was as follows. $$x^3 + \frac{1}{4}x = \frac{1}{2}$$ $$\therefore \quad x(x^2 + \frac{1}{4}) = \frac{1}{2}$$ $$\therefore \quad x = \frac{1}{2}$$ or $$x^2 + \frac{1}{4} = \frac{1}{2}$$ i.e. $$x^2 = -\frac{1}{4} \quad \text{no solution}$$ $$\therefore \quad \text{only real root is } x = \frac{1}{2}$$

1. Explain clearly the error in the above attempt. [2]
2. Give a correct proof that \(x = \frac{1}{2}\) is the only real root of \(x^3 + \frac{1}{4}x - \frac{1}{2} = 0\). [3]

The equation $$x^3 + \beta x - \alpha = 0 \quad \text{(I)}$$ where \(\alpha\), \(\beta\) are real, \(\alpha \neq 0\), has a real root at \(x = \alpha\).

1. Find and simplify an expression for \(\beta\) in terms of \(\alpha\) and prove that \(\alpha\) is the only real root provided \(|\alpha| < 2\). [6]

An examiner chooses a positive number \(\alpha\) so that \(\alpha\) is the only real root of equation (I) but the incorrect method used by the student produces 3 distinct real "roots".

1. Find the range of possible values for \(\alpha\). [7]