| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Multiple roots and starting value selection |
| Difficulty | Standard +0.3 This is a straightforward Newton-Raphson question requiring standard differentiation of g(x) to establish the equation, then applying the iterative formula with a sensible starting value read from a graph. Part (c) tests basic understanding of when Newton-Raphson fails (likely due to zero derivative or divergence), which is routine bookwork. The multi-part structure and graph interpretation add some complexity, but all techniques are standard A-level applications with no novel insight required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = x\cos x\) oe | B1 | |
| \(x\cos x = 1\) | M1 | For *their* \(\frac{dy}{dx} = 1\) |
| \(\frac{1}{x} = \cos x\) so \(\frac{1}{x} - \cos x = 0\) | A1 | Convincing completion to given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Iteration \(x_{n+1} = x_n - \dfrac{\left(\frac{1}{x_n} - \cos x_n\right)}{\left(-\frac{1}{x_n^2} + \sin x_n\right)}\) | A1 | oe, e.g. \(x_{n+1} = x_n - \dfrac{\left(x_n - x_n^2\cos x_n\right)}{\left(-1 + x_n^2\sin x_n\right)}\) (subscripts are needed) |
| Differentiation | M1 | Differentiation (look for power of \(x\) and term in \(\sin x\) or \(\cos x\), at least one term correct) |
| Suitable starting value | M1 | Starting values from 3.6 to 6.1 inclusive work; other values giving required root also accepted |
| \(4.9172\) | A1 | BC – candidates need not show intermediate iterations; if value outside expected range converges to 4.9172 then M1A1 awarded. awrt 4.9172 |
| Answer | Marks | Guidance |
|---|---|---|
| Iteration \(x_{n+1} = x_n - \dfrac{(x_n\cos x_n - 1)}{(\cos x_n - x_n\sin x_n)}\) | M1, A1 | oe e.g. \(x_{n+1} = \dfrac{(1 - x_n^2\sin x_n)}{(\cos x_n - x_n\sin x_n)}\) |
| Suitable starting value | M1 | |
| \(4.9172\) | A1 | awrt 4.9172 |
| Answer | Marks | Guidance |
|---|---|---|
| The gradient is close to zero so the next iteration is a long way from the root; or as it is a turning point, the starting value is invalid as you cannot divide by 0; or the iteration converges to a different root | B1 | Explanation referring to gradient of curve or convergence to a different root. Not just that it is close to another root. isw after correct answer |
## Question 11:
### Part (a):
$\frac{dy}{dx} = x\cos x$ oe | B1 |
$x\cos x = 1$ | M1 | For *their* $\frac{dy}{dx} = 1$
$\frac{1}{x} = \cos x$ so $\frac{1}{x} - \cos x = 0$ | A1 | Convincing completion to given answer
**Total: [3]**
### Part (b):
For $f(x) = \frac{1}{x} - \cos x$, $f'(x) = -\frac{1}{x^2} + \sin x$
Iteration $x_{n+1} = x_n - \dfrac{\left(\frac{1}{x_n} - \cos x_n\right)}{\left(-\frac{1}{x_n^2} + \sin x_n\right)}$ | A1 | oe, e.g. $x_{n+1} = x_n - \dfrac{\left(x_n - x_n^2\cos x_n\right)}{\left(-1 + x_n^2\sin x_n\right)}$ **(subscripts are needed)**
Differentiation | M1 | Differentiation (look for power of $x$ and term in $\sin x$ or $\cos x$, at least one term correct)
Suitable starting value | M1 | Starting values from 3.6 to 6.1 inclusive work; other values giving required root also accepted
$4.9172$ | A1 | BC – candidates need not show intermediate iterations; if value outside expected range converges to 4.9172 then **M1A1** awarded. awrt 4.9172
**Alternative method:**
For $f(x) = x\cos x - 1$, $f'(x) = \cos x - x\sin x$
Iteration $x_{n+1} = x_n - \dfrac{(x_n\cos x_n - 1)}{(\cos x_n - x_n\sin x_n)}$ | M1, A1 | oe e.g. $x_{n+1} = \dfrac{(1 - x_n^2\sin x_n)}{(\cos x_n - x_n\sin x_n)}$
Suitable starting value | M1 |
$4.9172$ | A1 | awrt 4.9172
**Total: [4]**
### Part (c):
The gradient is close to zero so the next iteration is a long way from the root; **or** as it is a turning point, the starting value is invalid as you cannot divide by 0; **or** the iteration converges to a different root | B1 | Explanation referring to gradient of curve or convergence to a different root. Not just that it is close to another root. isw after correct answer
**Total: [1]**11 Fig. 11.1 shows the curve with equation $\mathrm { y } = \mathrm { g } ( \mathrm { x } )$ where $\mathrm { g } ( x ) = x \sin x + \cos x$ and the curve of the gradient function $\mathrm { y } = \mathrm { g } ^ { \prime } ( \mathrm { x } )$ for $- 2 \pi \leqslant x \leqslant 2 \pi$.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 11.1}
\includegraphics[alt={},max width=\textwidth]{60e1e785-c34b-48ef-a63f-13a25fee186e-08_1136_1196_459_246}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinates of the points on the curve $y = g ( x )$ where the gradient is 1 satisfy the equation $\frac { 1 } { x } - \cos x = 0$.
Fig. 11.2 shows part of the curve with equation $y = \frac { 1 } { x } - \cos x$.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 11.2}
\includegraphics[alt={},max width=\textwidth]{60e1e785-c34b-48ef-a63f-13a25fee186e-09_678_1363_424_239}
\end{center}
\end{figure}
\item Use the Newton-Raphson method with a suitable starting value to find the smallest positive $x$-coordinate of a point on the curve $y = x \sin x + \cos x$ where the gradient is 1 .
You should write down at least the following.
\begin{itemize}
\item The iteration you use
\item The starting value
\item The solution correct to $\mathbf { 4 }$ decimal places
\item Explain why $x _ { 1 } = 3$ is not a suitable starting value for the Newton-Raphson method in part (b).
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2024 Q11 [8]}}