Edexcel C4 (Core Mathematics 4) 2012 January

1. The curve \(C\) has the equation \(2 x + 3 y ^ { 2 } + 3 x ^ { 2 } y = 4 x ^ { 2 }\).

The point \(P\) on the curve has coordinates \(( - 1,1 )\).

1. Find the gradient of the curve at \(P\).
2. Hence find the equation of the normal to \(C\) at \(P\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

2. (a) Use integration by parts to find \(\int x \sin 3 x \mathrm {~d} x\).
(b) Using your answer to part (a), find \(\int x ^ { 2 } \cos 3 x \mathrm {~d} x\).

3. (a) Expand $$\frac { 1 } { ( 2 - 5 x ) ^ { 2 } } , \quad | x | < \frac { 2 } { 5 }$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 2 }\), giving each term as a simplified fraction. Given that the binomial expansion of \(\frac { 2 + k x } { ( 2 - 5 x ) ^ { 2 } } , | x | < \frac { 2 } { 5 }\), is $$\frac { 1 } { 2 } + \frac { 7 } { 4 } x + A x ^ { 2 } + \ldots$$ (b) find the value of the constant \(k\),
(c) find the value of the constant \(A\).

4. \begin{figure}[h] \end{figure} Figure 1 shows the curve with equation $$y = \sqrt { } \left( \frac { 2 x } { 3 x ^ { 2 } + 4 } \right) , x \geqslant 0$$ The finite region \(S\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the line \(x = 2\) The region \(S\) is rotated \(360 ^ { \circ }\) about the \(x\)-axis.
Use integration to find the exact value of the volume of the solid generated, giving your answer in the form \(k \ln a\), where \(k\) and \(a\) are constants.

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve \(C\) with parametric equations $$x = 4 \sin \left( t + \frac { \pi } { 6 } \right) , \quad y = 3 \cos 2 t , \quad 0 \leqslant t < 2 \pi$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find the coordinates of all the points on \(C\) where \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\)

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve with equation \(y = \frac { 2 \sin 2 x } { ( 1 + \cos x ) } , 0 \leqslant x \leqslant \frac { \pi } { 2 }\).
The finite region \(R\), shown shaded in Figure 3, is bounded by the curve and the \(x\)-axis. The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 2 \sin 2 x } { ( 1 + \cos x ) }\).

1. Complete the table above giving the missing value of \(y\) to 5 decimal places.
2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(R\), giving your answer to 4 decimal places.
3. Using the substitution \(u = 1 + \cos x\), or otherwise, show that $$\int \frac { 2 \sin 2 x } { ( 1 + \cos x ) } d x = 4 \ln ( 1 + \cos x ) - 4 \cos x + k$$ where \(k\) is a constant.
4. Hence calculate the error of the estimate in part (b), giving your answer to 2 significant figures.

7. Relative to a fixed origin \(O\), the point \(A\) has position vector ( \(2 \mathbf { i } - \mathbf { j } + 5 \mathbf { k }\) ), the point \(B\) has position vector \(( 5 \mathbf { i } + 2 \mathbf { j } + 10 \mathbf { k } )\), and the point \(D\) has position vector \(( - \mathbf { i } + \mathbf { j } + 4 \mathbf { k } )\). The line \(l\) passes through the points \(A\) and \(B\).

1. Find the vector \(\overrightarrow { A B }\).
2. Find a vector equation for the line \(l\).
3. Show that the size of the angle \(B A D\) is \(109 ^ { \circ }\), to the nearest degree. The points \(A , B\) and \(D\), together with a point \(C\), are the vertices of the parallelogram \(A B C D\), where \(\overrightarrow { A B } = \overrightarrow { D C }\).
4. Find the position vector of \(C\).
5. Find the area of the parallelogram \(A B C D\), giving your answer to 3 significant figures.
6. Find the shortest distance from the point \(D\) to the line \(l\), giving your answer to 3 significant figures.

1. (a) Express \(\frac { 1 } { P ( 5 - P ) }\) in partial fractions.

A team of conservationists is studying the population of meerkats on a nature reserve. The population is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 15 } P ( 5 - P ) , \quad t \geqslant 0$$ where \(P\), in thousands, is the population of meerkats and \(t\) is the time measured in years since the study began. Given that when \(t = 0 , P = 1\),
(b) solve the differential equation, giving your answer in the form, $$P = \frac { a } { b + c \mathrm { e } ^ { - \frac { 1 } { 3 } t } }$$ where \(a\), \(b\) and \(c\) are integers.
(c) Hence show that the population cannot exceed 5000