Edexcel C4 (Core Mathematics 4) 2005 June

1. Use the binomial theorem to expand

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

2. A curve has equation $$x ^ { 2 } + 2 x y - 3 y ^ { 2 } + 16 = 0 .$$ Find the coordinates of the points on the curve where \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).

3. (a) Express \(\frac { 5 x + 3 } { ( 2 x - 3 ) ( x + 2 ) }\) in partial fractions.
(b) Hence find the exact value of \(\int _ { 2 } ^ { 6 } \frac { 5 x + 3 } { ( 2 x - 3 ) ( x + 2 ) } \mathrm { d } x\), giving your answer as a single logarithm.

4. Use the substitution \(x = \sin \theta\) to find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$$

5. \begin{figure}[h] \end{figure} Figure 1 shows the graph of the curve with equation $$y = x \mathrm { e } ^ { 2 x } , \quad x \geqslant 0$$ 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 for the area of \(R\).
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	1
   \(y = x \mathrm { e } ^ { 2 x }\)	0	0.29836		1.99207		7.38906

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.

1. A curve has parametric equations

$$x = 2 \cot t , \quad y = 2 \sin ^ { 2 } t , \quad 0 < t \leqslant \frac { \pi } { 2 }$$

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

1. The line \(l _ { 1 }\) has vector equation

$$\mathbf { r } = \left( \begin{array} { l } 3
1
2 \end{array} \right) + \lambda \left( \begin{array} { r } 1
- 1
4 \end{array} \right)$$ and the line \(l _ { 2 }\) has vector equation $$\mathbf { r } = \left( \begin{array} { r } 0
4
- 2 \end{array} \right) + \mu \left( \begin{array} { r } 1
- 1
0 \end{array} \right) ,$$ 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 \(B\).
2. Find the value of \(\cos \theta\), giving your answer as a simplified fraction. 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 \(A B C D\) is a parallelogram.
3. Show that \(| \overrightarrow { A B } | = | \overrightarrow { B C } |\).
4. Find the position vector of the point \(D\).

1. Liquid is pouring into a container at a constant rate of \(20 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) and is leaking out at a rate proportional to the volume of liquid already in the container.
   1. Explain why, at time \(t\) seconds, the volume, \(V \mathrm {~cm} ^ { 3 }\), of liquid in the container satisfies the differential equation
   $$\frac { \mathrm { d } V } { \mathrm {~d} t } = 20 - k V$$ where \(k\) is a positive constant. The container is initially empty.
2. By solving the differential equation, show that $$V = A + B \mathrm { e } ^ { - k t }$$ giving the values of \(A\) and \(B\) in terms of \(k\). Given also that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = 10\) when \(t = 5\),
3. find the volume of liquid in the container at 10 s after the start.