Edexcel C4 (Core Mathematics 4) 2008 January

1. \begin{figure}[h] \end{figure} The curve shown in Figure 1 has equation \(y = \mathrm { e } ^ { x } \sqrt { } ( \sin x ) , 0 \leqslant x \leqslant \pi\). The finite region \(R\) bounded by the curve and the \(x\)-axis is shown shaded in Figure 1.

1. Complete the table below with the values of \(y\) corresponding to \(x = \frac { \pi } { 4 }\) and \(\frac { \pi } { 2 }\), giving your answers to 5 decimal places.

   \(x\)	0	\(\frac { \pi } { 4 }\)	\(\frac { \pi } { 2 }\)	\(\frac { 3 \pi } { 4 }\)	\(\pi\)
   \(y\)	0			8.87207	0

2. Use the trapezium rule, with all the values in the completed table, to obtain an estimate for the area of the region \(R\). Give your answer to 4 decimal places.

2. (a) Use the binomial theorem to expand $$( 8 - 3 x ) ^ { \frac { 1 } { 3 } } , \quad | x | < \frac { 8 } { 3 }$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), giving each term as a simplified fraction.
(b) Use your expansion, with a suitable value of \(x\), to obtain an approximation to \(\sqrt [ 3 ] { } ( 7.7 )\). Give your answer to 7 decimal places.

3. \begin{figure}[h] \end{figure} The curve shown in Figure 2 has equation \(y = \frac { 1 } { ( 2 x + 1 ) }\). The finite region bounded by the curve, the \(x\)-axis and the lines \(x = a\) and \(x = b\) is shown shaded in Figure 2. This region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis to generate a solid of revolution. Find the volume of the solid generated. Express your answer as a single simplified fraction, in terms of \(a\) and \(b\).

4. (i) Find \(\int \ln \left( \frac { x } { 2 } \right) \mathrm { d } x\).
(ii) Find the exact value of \(\int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 2 } } \sin ^ { 2 } x \mathrm {~d} x\).

5. A curve is described by the equation $$x ^ { 3 } - 4 y ^ { 2 } = 12 x y$$

1. Find the coordinates of the two points on the curve where \(x = - 8\).
2. Find the gradient of the curve at each of these points.

6. The points \(A\) and \(B\) have position vectors \(2 \mathbf { i } + 6 \mathbf { j } - \mathbf { k }\) and \(3 \mathbf { i } + 4 \mathbf { j } + \mathbf { k }\) respectively. The line \(l _ { 1 }\) passes through the points \(A\) and \(B\).

1. Find the vector \(\overrightarrow { A B }\).
2. Find a vector equation for the line \(l _ { 1 }\). A second line \(l _ { 2 }\) passes through the origin and is parallel to the vector \(\mathbf { i } + \mathbf { k }\). The line \(l _ { 1 }\) meets the line \(l _ { 2 }\) at the point \(C\).
3. Find the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\).
4. Find the position vector of the point \(C\).

7. \begin{figure}[h] \end{figure} The curve \(C\) has parametric equations $$x = \ln ( t + 2 ) , \quad y = \frac { 1 } { ( t + 1 ) } , \quad t > - 1$$ The finite region \(R\) between the curve \(C\) and the \(x\)-axis, bounded by the lines with equations \(x = \ln 2\) and \(x = \ln 4\), is shown shaded in Figure 3.

1. Show that the area of \(R\) is given by the integral $$\int _ { 0 } ^ { 2 } \frac { 1 } { ( t + 1 ) ( t + 2 ) } \mathrm { d } t$$
2. Hence find an exact value for this area.
3. Find a cartesian equation of the curve \(C\), in the form \(y = \mathrm { f } ( x )\).
4. State the domain of values for \(x\) for this curve.
   \(\_\_\_\_\)}

8. Liquid is pouring into a large vertical circular cylinder at a constant rate of \(1600 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is \(4000 \mathrm {~cm} ^ { 2 }\).

1. Show that at time \(t\) seconds, the height \(h \mathrm {~cm}\) of liquid in the cylinder satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - k \sqrt { } h \text {, where } k \text { is a positive constant. }$$ When \(h = 25\), water is leaking out of the hole at \(400 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Show that \(k = 0.02\)
3. Separate the variables of the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - 0.02 \sqrt { } h$$ to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by $$\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { } h } \mathrm {~d} h$$ Using the substitution \(h = ( 20 - x ) ^ { 2 }\), or otherwise,
4. find the exact value of \(\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h\).
5. Hence find the time taken to fill the cylinder from empty to a height of 100 cm , giving your answer in minutes and seconds to the nearest second.