Edexcel AEA (Advanced Extension Award) 2015 June

1. Sketch the graph of the curve with equation $$y = \ln(2x + 5), \quad x > -\frac{5}{2}$$ On your sketch you should clearly state the equations of any asymptotes and mark the coordinates of points where the curve meets the coordinate axes. [3]
2. Solve the equation \(\ln(2x + 5) = \ln 9\) [3]

1. Show that \((x + 1)\) is a factor of \(2x^3 + 3x^2 - 1\) [1]
2. Solve the equation $$\sqrt{x^2 + 2x + 5} = x + \sqrt{2x + 3}$$ [8]

Solve for \(0 < x < 360°\) $$\cot 2x - \tan 78° = \frac{(\sec x)(\sec 78°)}{2}$$ where \(x\) is not an integer multiple of \(90°\) [9]

1. Find the binomial series expansion for \((4 + y)^{\frac{1}{2}}\) in ascending powers of \(y\) up to and including the term in \(y^3\). Simplify the coefficient of each term. [3]
2. Hence show that the binomial series expansion for \((4 + 5x + x^2)^{\frac{1}{2}}\) in ascending powers of \(x\) up to and including the term in \(x^3\) is $$2 + \frac{5x}{4} - \frac{9x^2}{64} + \frac{45x^3}{512}$$ [3]
3. Show that the binomial series expansion of \((4 + 5x + x^2)^{\frac{1}{2}}\) will converge for \(-\frac{1}{2} < x \leq \frac{1}{2}\) [6]
4. Use the result in part (b) to estimate $$\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{4 + 5x + x^2} \, dx$$ Give your answer as a single fraction. [3]

% Figure shows a curve with maximum at point A, passing through origin O, with horizontal asymptote \includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = f(x)\) where $$f(x) = \frac{x^2 + 16}{3x} \quad x \neq 0$$ The curve has a maximum at the point \(A\) with coordinates \((a, b)\).

1. Find the value of \(a\) and the value of \(b\). [4] The function g is defined as $$g : x \mapsto \frac{x^2 + 16}{3x} \quad a \leq x < 0$$ where \(a\) is the value found in part (a).
2. Write down the range of g. [1]
3. On the same axes sketch \(y = g(x)\) and \(y = g^{-1}(x)\). [3]
4. Find an expression for \(g^{-1}(x)\) and state the domain of \(g^{-1}\) [5]
5. Solve the equation \(g(x) = g^{-1}(x)\). [3]

The lines \(L_1\) and \(L_2\) have vector equations $$L_1 : \mathbf{r} = \begin{pmatrix} 1 \\ 10 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -5 \\ 4 \end{pmatrix}$$ $$L_2 : \mathbf{r} = \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$$

1. Show that \(L_1\) and \(L_2\) are perpendicular. [2]
2. Show that \(L_1\) and \(L_2\) are skew lines. [3] The point \(A\) with position vector \(-\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\) lies on \(L_2\) and the point \(X\) lies on \(L_1\) such that \(\overrightarrow{AX}\) is perpendicular to \(L_1\)
3. Find the position vector of \(X\). [5]
4. Find \(|\overrightarrow{AX}|\) [2] The point \(B\) (distinct from \(A\)) also lies on \(L_2\) and \(|\overrightarrow{BX}| = |\overrightarrow{AX}|\)
5. Find the position vector of \(B\). [5]
6. Find the cosine of angle \(AXB\). [2]

1. Use the substitution \(x = \sec\theta\) to show that $$\int_{\sqrt{2}}^{2} \frac{1}{(x^2 - 1)^{\frac{3}{2}}} \, dx = \frac{\sqrt{6} - 2}{\sqrt{3}}$$ [5]
2. Use integration by parts to show that $$\int \cos\theta \cot^2\theta \, d\theta = \frac{1}{2}[\ln|\cos\theta + \cot\theta| - \cos\theta \cot\theta] + c$$ [6] % Figure shows a curve y = 1/(x^2-1)^(1/2) for x > 1, with shaded region R between x = sqrt(2) and x = 2 \includegraphics{figure_2} Figure 2 shows a sketch of part of the curve with equation \(y = \frac{1}{(x^2 - 1)^{\frac{1}{2}}}\) for \(x > 1\) The region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the lines \(x = \sqrt{2}\) and \(x = 2\) The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis.
3. Show that the volume of the solid formed is $$\pi \left[\frac{3}{8}\ln\left(\frac{1 + \sqrt{2}}{\sqrt{3}}\right) + \frac{7}{36} - \frac{\sqrt{2}}{8}\right]$$ [8]