Edexcel C2 (Core Mathematics 2) 2013 June

1. Using calculus, find the coordinates of the stationary point on the curve with equation

$$y = 2 x + 3 + \frac { 8 } { x ^ { 2 } } , \quad x > 0$$

2. $$y = \frac { x } { \sqrt { ( 1 + x ) } }$$

1. Complete the table below with the value of \(y\) corresponding to \(x = 1.3\), giving your answer to 4 decimal places.

   \(x\)	1	1.1	1.2	1.3	1.4	1.5
   \(y\)	0.7071	0.7591	0.8090		0.9037	0.9487

2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an approximate value for $$\int _ { 1 } ^ { 1.5 } \frac { x } { \sqrt { } ( 1 + x ) } \mathrm { d } x$$ giving your answer to 3 decimal places.
   You must show clearly each stage of your working.

3. Find the first 4 terms, in ascending powers of \(x\), of the binomial expansion of $$\left( 2 - \frac { 1 } { 2 } x \right) ^ { 8 }$$ giving each term in its simplest form.

4. \(\mathrm { f } ( x ) = a x ^ { 3 } - 11 x ^ { 2 } + b x + 4\), where \(a\) and \(b\) are constants. When \(\mathrm { f } ( x )\) is divided by ( \(x - 3\) ) the remainder is 55
When \(\mathrm { f } ( x )\) is divided by \(( x + 1 )\) the remainder is - 9

1. Find the value of \(a\) and the value of \(b\). Given that \(( 3 x + 2 )\) is a factor of \(\mathrm { f } ( x )\),
2. factorise \(\mathrm { f } ( x )\) completely.

5. The first three terms of a geometric series are \(4 p , ( 3 p + 15 )\) and ( \(5 p + 20\) ) respectively, where \(p\) is a positive constant.

1. Show that \(11 p ^ { 2 } - 10 p - 225 = 0\)
2. Hence show that \(p = 5\)
3. Find the common ratio of this series.
4. Find the sum of the first ten terms of the series, giving your answer to the nearest integer.

6. Given that \(\log _ { 3 } x = a\), find in terms of \(a\),

1. \(\log _ { 3 } ( 9 x )\)
2. \(\log _ { 3 } \left( \frac { x ^ { 5 } } { 81 } \right)\)
   giving each answer in its simplest form.
3. Solve, for \(x\), $$\log _ { 3 } ( 9 x ) + \log _ { 3 } \left( \frac { x ^ { 5 } } { 81 } \right) = 3$$ giving your answer to 4 significant figures. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4f4eac7b-8908-480f-bb39-049944203fff-10_775_1605_221_159} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} The line with equation \(y = 10\) cuts the curve with equation \(y = x ^ { 2 } + 2 x + 2\) at the points \(A\) and \(B\) as shown in Figure 1. The figure is not drawn to scale.

1. Find by calculation the \(x\)-coordinate of \(A\) and the \(x\)-coordinate of \(B\). The shaded region \(R\) is bounded by the line with equation \(y = 10\) and the curve as shown in Figure 1.
2. Use calculus to find the exact area of \(R\).

8. \begin{figure}[h] \end{figure} Figure 2 shows the design for a triangular garden \(A B C\) where \(A B = 7 \mathrm {~m} , A C = 13 \mathrm {~m}\) and \(B C = 10 \mathrm {~m}\). Given that angle \(B A C = \theta\) radians,

1. show that, to 3 decimal places, \(\theta = 0.865\) The point \(D\) lies on \(A C\) such that \(B D\) is an arc of the circle centre \(A\), radius 7 m .
   The shaded region \(S\) is bounded by the arc \(B D\) and the lines \(B C\) and \(D C\). The shaded region \(S\) will be sown with grass seed, to make a lawned area. Given that 50 g of grass seed are needed for each square metre of lawn,
2. find the amount of grass seed needed, giving your answer to the nearest 10 g .

1. (i) Solve, for \(0 \leqslant \theta < 180 ^ { \circ }\)

$$\sin \left( 2 \theta - 30 ^ { \circ } \right) + 1 = 0.4$$ giving your answers to 1 decimal place.
(ii) Find all the values of \(x\), in the interval \(0 \leqslant x < 360 ^ { \circ }\), for which $$9 \cos ^ { 2 } x - 11 \cos x + 3 \sin ^ { 2 } x = 0$$ giving your answers to 1 decimal place. You must show clearly how you obtained your answers.