| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve trigonometric equation via iteration |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on fixed point iteration and calculus. Part (a) requires simple substitution to verify a root interval. Part (b) is routine calculator work applying a given iteration formula. Part (c) involves standard differentiation and solving a simple trigonometric equation. All techniques are standard P3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07n Stationary points: find maxima, minima using derivatives1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Attempts the value of \(y\) at 0.8 AND 0.9 with at least one correct to 1 sf rounded or truncated | M1 | |
| States change of sign, continuous and hence root | A1 | Note: If candidate chooses 0.8 and 0.9 the minimal conclusion does not need to mention the interval. So e.g. \(y |
| (b)(i) \(\left(x_2\right) = \frac{1}{3}\arccos(1.5 \times 0.8 - 2) = 0.8327\) | M1 A1 | |
| (ii) \(x_5 = 0.8110\) | A1 | (3) |
| (c) \(\frac{dy}{dx} = -6\sin 3x - 3\) | M1 A1 | |
| Attempts \(\frac{dy}{dx} = 0 \Rightarrow \sin 3x = -\frac{1}{2} \Rightarrow x = ...\) via correct order of operations | dM1 | |
| Achieves either \(\frac{7\pi}{18}\) or \(\frac{19\pi}{18}\) | A1 | |
| Correct attempt to find the third positive solution e.g. \(x = \frac{\frac{\pi}{6} + 3\pi}{3}\) | ddM1 | |
| \(\beta = \frac{7\pi}{18}\) and \(\lambda = \frac{19\pi}{18}\) | A1 | (6) |
**(a)** Attempts the value of $y$ at 0.8 AND 0.9 with at least one correct to 1 sf rounded or truncated | M1 |
States change of sign, continuous and hence root | A1 | Note: If candidate chooses 0.8 and 0.9 the minimal conclusion does not need to mention the interval. So e.g. $y|_{0.8} = 0.1 > 0, y|_{0.9} = -0.5 < 0$ and function is continuous, so ✓ would be acceptable | (2)
**(b)(i)** $\left(x_2\right) = \frac{1}{3}\arccos(1.5 \times 0.8 - 2) = 0.8327$ | M1 A1 |
**(ii)** $x_5 = 0.8110$ | A1 | (3)
**(c)** $\frac{dy}{dx} = -6\sin 3x - 3$ | M1 A1 |
Attempts $\frac{dy}{dx} = 0 \Rightarrow \sin 3x = -\frac{1}{2} \Rightarrow x = ...$ via correct order of operations | dM1 |
Achieves either $\frac{7\pi}{18}$ or $\frac{19\pi}{18}$ | A1 |
Correct attempt to find the third positive solution e.g. $x = \frac{\frac{\pi}{6} + 3\pi}{3}$ | ddM1 |
$\beta = \frac{7\pi}{18}$ and $\lambda = \frac{19\pi}{18}$ | A1 | (6)
**Total: 11 marks**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1c700103-ecab-4a08-b411-3f445ed88885-22_707_1047_264_463}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with equation
$$y = 2 \cos 3 x - 3 x + 4 \quad x > 0$$
where $x$ is measured in radians.
The curve crosses the $x$-axis at the point $P$, as shown in Figure 3.\\
Given that the $x$ coordinate of $P$ is $\alpha$,
\begin{enumerate}[label=(\alph*)]
\item show that $\alpha$ lies between 0.8 and 0.9
The iteration formula
$$x _ { n + 1 } = \frac { 1 } { 3 } \arccos \left( 1.5 x _ { n } - 2 \right)$$
can be used to find an approximate value for $\alpha$.
\item Using this iteration formula with $x _ { 1 } = 0.8$ find, to 4 decimal places, the value of
\begin{enumerate}[label=(\roman*)]
\item $X _ { 2 }$
\item $X _ { 5 }$
The point $Q$ and the point $R$ are local minimum points on the curve, as shown in Figure 3.\\
Given that the $x$ coordinates of $Q$ and $R$ are $\beta$ and $\lambda$ respectively, and that they are the two smallest values of $x$ at which local minima occur,
\end{enumerate}\item find, using calculus, the exact value of $\beta$ and the exact value of $\lambda$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2020 Q7 [11]}}