Edexcel C4 (Core Mathematics 4)

Use the binomial theorem to expand $$\sqrt{(4-9x)}, \quad |x| < \frac{4}{9},$$ in ascending powers of \(x\), up to and including the term in \(x^3\), simplifying each term. [5]

A curve has equation $$x^2 + 2xy - 3y^2 + 16 = 0.$$ Find the coordinates of the points on the curve where \(\frac{dy}{dx} = 0\). [7]

1. Express \(\frac{5x + 3}{(2x - 3)(x + 2)}\) in partial fractions. [3]
2. Hence find the exact value of \(\int_0^1 \frac{5x + 3}{(2x - 3)(x + 2)} dx\), giving your answer as a single logarithm. [5]

Use the substitution \(x = \sin \theta\) to find the exact value of $$\int_0^1 \frac{1}{(1-x^2)^{3/2}} dx.$$ [7]

\includegraphics{figure_1} Figure 1 shows the graph of the curve with equation $$y = xe^x, \quad x \geq 0.$$ The finite region \(R\) bounded by the lines \(x = 1\), the \(x\)-axis and the curve is shown shaded in Figure 1.

1. Use integration to find the exact value of the area for \(R\). [5]
2. Complete the table with the values of \(y\) corresponding to \(x = 0.4\) and \(0.8\).

   \(x\)	0	0.2	0.4	0.6	0.8
   \(y = xe^x\)	0	0.29836		1.99207

   [1]
3. Use the trapezium rule with all the values in the table to find an approximate value for this area, giving your answer to 4 significant figures. [4]

A curve has parametric equations $$x = 2\cot t, \quad y = 2\sin^2 t, \quad 0 < t \leq \frac{\pi}{2}.$$

1. Find an expression for \(\frac{dy}{dx}\) in terms of the parameter \(t\). [4]
2. Find an equation of the tangent to the curve at the point where \(t = \frac{\pi}{4}\). [4]
3. Find a cartesian equation of the curve in the form \(y = f(x)\). State the domain on which the curve is defined. [4]

The line \(l_1\) has vector equation $$\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}$$ and the line \(l_2\) has vector equation $$\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix},$$ where \(\lambda\) and \(\mu\) are parameters. The lines \(l_1\) and \(l_2\) intersect at the point \(B\) and the acute angle between \(l_1\) and \(l_2\) is \(\theta\).

1. Find the coordinates of \(B\). [4]
2. Find the value of \(\cos \theta\), giving your answer as a simplified fraction. [4]

The point \(A\), which lies on \(l_1\), has position vector \(\mathbf{a} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}\). The point \(C\), which lies on \(l_2\), has position vector \(\mathbf{c} = 5\mathbf{i} - \mathbf{j} - 2\mathbf{k}\). The point \(D\) is such that \(ABCD\) is a parallelogram.

1. Show that \(|\overrightarrow{AB}| = |\overrightarrow{BC}|\). [3]
2. Find the position vector of the point \(D\). [2]

Liquid is pouring into a container at a constant rate of \(20\text{ cm}^3\text{s}^{-1}\) and is leaking out at a rate proportional to the volume of the liquid already in the container.

1. Explain why, at time \(t\) seconds, the volume, \(V\text{ cm}^3\), of liquid in the container satisfies the differential equation $$\frac{dV}{dt} = 20 - kV,$$ where \(k\) is a positive constant. [2]

The container is initially empty.

1. By solving the differential equation, show that $$V = A + Be^{-kt},$$ giving the values of \(A\) and \(B\) in terms of \(k\). [6]

Given also that \(\frac{dV}{dt} = 10\) when \(t = 5\),

1. find the volume of liquid in the container at 10 s after the start. [5]