Edexcel C4 (Core Mathematics 4) 2015 June

1. (a) Find the binomial expansion of

$$( 4 + 5 x ) ^ { \frac { 1 } { 2 } } , \quad | x | < \frac { 4 } { 5 }$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 2 }\).
Give each coefficient in its simplest form.
(b) Find the exact value of \(( 4 + 5 x ) ^ { \frac { 1 } { 2 } }\) when \(x = \frac { 1 } { 10 }\) Give your answer in the form \(k \sqrt { 2 }\), where \(k\) is a constant to be determined.
(c) Substitute \(x = \frac { 1 } { 10 }\) into your binomial expansion from part (a) and hence find an approximate value for \(\sqrt { 2 }\) Give your answer in the form \(\frac { p } { q }\) where \(p\) and \(q\) are integers.

2. The curve \(C\) has equation $$x ^ { 2 } - 3 x y - 4 y ^ { 2 } + 64 = 0$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Find the coordinates of the points on \(C\) where \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\)
   (Solutions based entirely on graphical or numerical methods are not acceptable.)

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0\)
The curve meets the \(x\)-axis at the origin \(O\) and cuts the \(x\)-axis at the point \(A\).

1. Find, in terms of \(\ln 2\), the \(x\) coordinate of the point \(A\).
2. Find $$\int x \mathrm { e } ^ { \frac { 1 } { 2 } x } \mathrm {~d} x$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the \(x\)-axis and the curve with equation $$y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0$$
3. Find, by integration, the exact value for the area of \(R\). Give your answer in terms of \(\ln 2\)

1. With respect to a fixed origin \(O\), the lines \(l _ { 1 }\) and \(l _ { 2 }\) are given by the equations

$$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r }

5
- 3
p \end{array} \right) + \lambda \left( \begin{array} { r } 0
1
- 3 \end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r }

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

1. Use the substitution \(x = 1 + 2 \sin \theta\) to show that $$\int _ { 0 } ^ { 3 } \sqrt { ( 3 - x ) ( x + 1 ) } d x = k \int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } \theta d \theta$$ where \(k\) is a constant to be determined.
2. Hence find, by integration, the exact area of \(R\).

8
5
- 2 \end{array} \right) + \mu \left( \begin{array} { r } 3
4
- 5 \end{array} \right)$$ where \(\lambda\) and \(\mu\) are scalar parameters and \(p\) is a constant. The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(A\).

1. Find the coordinates of \(A\).
2. Find the value of the constant \(p\).
3. Find the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\), giving your answer in degrees to 2 decimal places. The point \(B\) lies on \(l _ { 2 }\) where \(\mu = 1\)
4. Find the shortest distance from the point \(B\) to the line \(l _ { 1 }\), giving your answer to 3 significant figures.
   1. A curve \(C\) has parametric equations
   $$x = 4 t + 3 , \quad y = 4 t + 8 + \frac { 5 } { 2 t } , \quad t \neq 0$$
5. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point on \(C\) where \(t = 2\), giving your answer as a fraction in its simplest form.
6. Show that the cartesian equation of the curve \(C\) can be written in the form $$y = \frac { x ^ { 2 } + a x + b } { x - 3 } , \quad x \neq 3$$ where \(a\) and \(b\) are integers to be determined.
   6. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{89d4a7a5-3f4f-4d16-b14e-a27243cedd78-11_666_993_244_392} \captionsetup{labelformat=empty} \caption{Diagram not to scale}
   \end{figure} Figure 2 Figure 2 shows a sketch of the curve with equation \(y = \sqrt { ( 3 - x ) ( x + 1 ) } , 0 \leqslant x \leqslant 3\)
   The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis, and the \(y\)-axis.
7. Use the substitution \(x = 1 + 2 \sin \theta\) to show that $$\int _ { 0 } ^ { 3 } \sqrt { ( 3 - x ) ( x + 1 ) } d x = k \int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } \theta d \theta$$ where \(k\) is a constant to be determined.
8. Hence find, by integration, the exact area of \(R\). 7. (a) Express \(\frac { 2 } { P ( P - 2 ) }\) in partial fractions. A team of biologists is studying a population of a particular species of animal. The population is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ( P - 2 ) \cos 2 t , t \geqslant 0$$ where \(P\) is the population in thousands, and \(t\) is the time measured in years since the start of the study. Given that \(P = 3\) when \(t = 0\),
9. solve this differential equation to show that $$P = \frac { 6 } { 3 - \mathrm { e } ^ { \frac { 1 } { 2 } \sin 2 t } }$$
10. find the time taken for the population to reach 4000 for the first time. Give your answer in years to 3 significant figures.
    8. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{89d4a7a5-3f4f-4d16-b14e-a27243cedd78-15_696_1418_287_262} \captionsetup{labelformat=empty} \caption{Figure 3}
    \end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = 3 ^ { x }$$ The point \(P\) lies on \(C\) and has coordinates \(( 2,9 )\).
    The line \(l\) is a tangent to \(C\) at \(P\). The line \(l\) cuts the \(x\)-axis at the point \(Q\).
11. Find the exact value of the \(x\) coordinate of \(Q\). The finite region \(R\), shown shaded in Figure 3, is bounded by the curve \(C\), the \(x\)-axis, the \(y\)-axis and the line \(l\). This region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
12. Use integration to find the exact value of the volume of the solid generated. Give your answer in the form \(\frac { p } { q }\) where \(p\) and \(q\) are exact constants.
    [0pt] [You may assume the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) for the volume of a cone.]