Edexcel P4 (Pure Mathematics 4) 2024 June

1. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Find $$\int _ { 0 } ^ { \frac { \pi } { 6 } } x \cos 3 x d x$$ giving your answer in simplest form.

1. With respect to a fixed origin, \(O\), the point \(A\) has position vector

$$\overrightarrow { O A } = \left( \begin{array} { r }

4
3 \end{array} \right)$$

1. find the coordinates of the point \(B\). The point \(C\) has position vector $$\overrightarrow { O C } = \left( \begin{array} { r } a

5
- 1 \end{array} \right)$$ where \(a\) is a constant.
Given that \(\overrightarrow { O C }\) is perpendicular to \(\overrightarrow { B C }\)
(b) find the possible values of \(a\).

1. The curve \(C\) is defined by the equation

$$8 x ^ { 3 } - 3 y ^ { 2 } + 2 x y = 9$$ Find an equation of the normal to \(C\) at the point ( 2,5 ), giving your answer in the form \(a x + b y + c = 0\), where \(a\), \(b\) and \(c\) are integers. 4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of a segment \(P Q R P\) of a circle with centre \(O\) and radius 5 cm .
Given that \begin{itemize} \item angle \(P O R\) is \(\theta\) radians \item \(\theta\) is increasing, from 0 to \(\pi\), at a constant rate of 0.1 radians per second \item the area of the segment \(P Q R P\) is \(A \mathrm {~cm} ^ { 2 }\)

1. show that \end{itemize} $$\frac { \mathrm { d } A } { \mathrm {~d} \theta } = K ( 1 - \cos \theta )$$ where \(K\) is a constant to be found.
2. Find, in \(\mathrm { cm } ^ { 2 } \mathrm {~s} ^ { - 1 }\), the rate of increase of the area of the segment when \(\theta = \frac { \pi } { 3 }\) 5. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e583bf92-d6a9-4f1a-b3c8-372afa6e0a0e-10_803_1086_248_493} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows a sketch of the curve defined by the parametric equations $$x = t ^ { 2 } + 2 t \quad y = \frac { 2 } { t ( 3 - t ) } \quad a \leqslant t \leqslant b$$ where \(a\) and \(b\) are constants.
   The ends of the curve lie on the line with equation \(y = 1\)
3. Find the value of \(a\) and the value of \(b\) The region \(R\), shown shaded in Figure 2, is bounded by the curve and the line with equation \(y = 1\)
4. Show that the area of region \(R\) is given by $$M - k \int _ { a } ^ { b } \frac { t + 1 } { t ( 3 - t ) } \mathrm { d } t$$ where \(M\) and \(k\) are constants to be found.
5. 1. Write \(\frac { t + 1 } { t ( 3 - t ) }\) in partial fractions.
   2. Use algebraic integration to find the exact area of \(R\), giving your answer in simplest form.

7
2
- 5 \end{array} \right)$$ Given that $$\overrightarrow { A B } = \left( \begin{array} { r } - 2
4
3 \end{array} \right)$$

1. find the coordinates of the point \(B\). The point \(C\) has position vector $$\overrightarrow { O C } = \left( \begin{array} { r } a
   5
   - 1 \end{array} \right)$$ where \(a\) is a constant.
   Given that \(\overrightarrow { O C }\) is perpendicular to \(\overrightarrow { B C }\)
2. find the possible values of \(a\).
   1. The curve \(C\) is defined by the equation
   $$8 x ^ { 3 } - 3 y ^ { 2 } + 2 x y = 9$$ Find an equation of the normal to \(C\) at the point ( 2,5 ), giving your answer in the form \(a x + b y + c = 0\), where \(a\), \(b\) and \(c\) are integers. 4. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e583bf92-d6a9-4f1a-b3c8-372afa6e0a0e-08_558_542_258_749} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} Figure 1 shows a sketch of a segment \(P Q R P\) of a circle with centre \(O\) and radius 5 cm .
   Given that
   • angle \(P O R\) is \(\theta\) radians
   • \(\theta\) is increasing, from 0 to \(\pi\), at a constant rate of 0.1 radians per second
   • the area of the segment \(P Q R P\) is \(A \mathrm {~cm} ^ { 2 }\)
   • show that
   $$\frac { \mathrm { d } A } { \mathrm {~d} \theta } = K ( 1 - \cos \theta )$$ where \(K\) is a constant to be found.
3. Find, in \(\mathrm { cm } ^ { 2 } \mathrm {~s} ^ { - 1 }\), the rate of increase of the area of the segment when \(\theta = \frac { \pi } { 3 }\) 5. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e583bf92-d6a9-4f1a-b3c8-372afa6e0a0e-10_803_1086_248_493} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows a sketch of the curve defined by the parametric equations $$x = t ^ { 2 } + 2 t \quad y = \frac { 2 } { t ( 3 - t ) } \quad a \leqslant t \leqslant b$$ where \(a\) and \(b\) are constants.
   The ends of the curve lie on the line with equation \(y = 1\)
4. Find the value of \(a\) and the value of \(b\) The region \(R\), shown shaded in Figure 2, is bounded by the curve and the line with equation \(y = 1\)
5. Show that the area of region \(R\) is given by $$M - k \int _ { a } ^ { b } \frac { t + 1 } { t ( 3 - t ) } \mathrm { d } t$$ where \(M\) and \(k\) are constants to be found.
6. 1. Write \(\frac { t + 1 } { t ( 3 - t ) }\) in partial fractions.
   2. Use algebraic integration to find the exact area of \(R\), giving your answer in simplest form.
      1. With respect to a fixed origin \(O\), the line \(l _ { 1 }\) is given by the equation
   $$\mathbf { r } = \mathbf { i } + 2 \mathbf { j } + 5 \mathbf { k } + \lambda ( 8 \mathbf { i } - \mathbf { j } + 4 \mathbf { k } )$$ where \(\lambda\) is a scalar parameter.
   The point \(A\) lies on \(l _ { 1 }\)
   Given that \(| \overrightarrow { O A } | = 5 \sqrt { 10 }\)
7. show that at \(A\) the parameter \(\lambda\) satisfies $$81 \lambda ^ { 2 } + 52 \lambda - 220 = 0$$ Hence
8. 1. show that one possible position vector for \(A\) is \(- 15 \mathbf { i } + 4 \mathbf { j } - 3 \mathbf { k }\)
   2. find the other possible position vector for \(A\). The line \(l _ { 2 }\) is parallel to \(l _ { 1 }\) and passes through \(O\).
      Given that
      • \(\overrightarrow { O A } = - 15 \mathbf { i } + 4 \mathbf { j } - 3 \mathbf { k }\)
9. point \(B\) lies on \(l _ { 2 }\) where \(| \overrightarrow { O B } | = 4 \sqrt { 10 }\)
10. find the area of triangle \(O A B\), giving your answer to one decimal place.
1. The current, \(x\) amps, at time \(t\) seconds after a switch is closed in a particular electric circuit is modelled by the equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = k - 3 x$$ where \(k\) is a constant.
Initially there is zero current in the circuit.
11. Solve the differential equation to find an equation, in terms of \(k\), for the current in the circuit at time \(t\) seconds.
    Give your answer in the form \(x = \mathrm { f } ( t )\). Given that in the long term the current in the circuit approaches 7 amps,
12. find the value of \(k\).
13. Hence find the time in seconds it takes for the current to reach 5 amps, giving your answer to 2 significant figures.

8. $$f ( x ) = ( 8 - 3 x ) ^ { \frac { 4 } { 3 } } \quad 0 < x < \frac { 8 } { 3 }$$

1. Show that the binomial expansion of \(\mathrm { f } ( x )\) in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\) is $$A - 8 x + \frac { x ^ { 2 } } { 2 } + B x ^ { 3 } + \ldots$$ where \(A\) and \(B\) are constants to be found.
2. Use proof by contradiction to prove that the curve with equation $$y = 8 + 8 x - \frac { 15 } { 2 } x ^ { 2 }$$ does not intersect the curve with equation $$y = A - 8 x + \frac { x ^ { 2 } } { 2 } + B x ^ { 3 } \quad 0 < x < \frac { 8 } { 3 }$$ where \(A\) and \(B\) are the constants found in part (a).
   (Solutions relying on calculator technology are not acceptable.)

9. \begin{figure}[h] \end{figure} The curve \(C\), shown in Figure 3, has equation $$y = \frac { x ^ { - \frac { 1 } { 4 } } } { \sqrt { 1 + x } ( \arctan \sqrt { x } ) }$$ The region \(R\), shown shaded in Figure 3, is bounded by \(C\), the line with equation \(x = 3\), the \(x\)-axis and the line with equation \(x = \frac { 1 } { 3 }\) The region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis to form a solid.
Using the substitution \(\tan u = \sqrt { x }\)

1. show that the volume \(V\) of the solid formed is given by $$k \int _ { a } ^ { b } \frac { 1 } { u ^ { 2 } } \mathrm {~d} u$$ where \(k , a\) and \(b\) are constants to be found.
2. Hence, using algebraic integration, find the value of \(V\) in simplest form.