| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Find stationary point coordinate |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard A-level techniques: finding a stationary point by solving f'(x)=0 (likely requiring a calculator), applying the intermediate value theorem, and performing one iteration of Newton-Raphson. All parts are routine applications of well-practiced methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.09a Sign change methods: locate roots1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 4\cos\!\left(\frac{1}{2}x\right) - 3\) | M1, A1 | M1: differentiates to obtain \(k\cos\!\left(\frac{1}{2}x\right) \pm \alpha\). A1: correct derivative; allow unsimplified e.g. \(\frac{1}{2}\times 8\cos\!\left(\frac{1}{2}x\right) - 3x^0\) |
| Sets \(f'(x) = 4\cos\!\left(\frac{1}{2}x\right) - 3 = 0 \Rightarrow x =\) | dM1 | For complete strategy proceeding to a value for \(x\). Look for \(a\cos\!\left(\frac{1}{2}x\right)+b=0,\ a,b\neq 0\) with correct method for a valid solution. |
| \(x = 14.0\) (cao) | A1 | Must be this value only. Correct answer with no working scores no marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Explains that \(f(4) > 0\), \(f(5) < 0\) and the function is continuous | B1 | Must be a full reason including change of sign and continuity. "Because \(x\) is continuous" or "because the interval is continuous" is not acceptable. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(x_1 = 5 - \dfrac{8\sin 2.5 - 15 + 9}{\text{"}\,4\cos 2.5 - 3\text{"}}\) (where \(f(5)=-1.212\ldots\), \(f'(5)=-6.204\ldots\)) | M1 | Must be correct N-R formula with their \(f'(x)\). Allow if attempted in degrees. |
| \(x_1 = \text{awrt } 4.80\) | A1 | Not awrt 4.8; isw if awrt 4.80 seen. Ignore subsequent iterations. \(5 - \frac{f(5)}{f'(5)} \neq \text{awrt } 4.80\) with no evidence scores M0. |
## Question 6:
### Part 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 4\cos\!\left(\frac{1}{2}x\right) - 3$ | M1, A1 | M1: differentiates to obtain $k\cos\!\left(\frac{1}{2}x\right) \pm \alpha$. A1: correct derivative; allow unsimplified e.g. $\frac{1}{2}\times 8\cos\!\left(\frac{1}{2}x\right) - 3x^0$ |
| Sets $f'(x) = 4\cos\!\left(\frac{1}{2}x\right) - 3 = 0 \Rightarrow x =$ | dM1 | For complete strategy proceeding to a value for $x$. Look for $a\cos\!\left(\frac{1}{2}x\right)+b=0,\ a,b\neq 0$ with correct method for a valid solution. |
| $x = 14.0$ (cao) | A1 | Must be this value only. Correct answer with no working scores no marks. |
### Part 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Explains that $f(4) > 0$, $f(5) < 0$ **and** the function is continuous | B1 | Must be a full reason including change of sign and continuity. "Because $x$ is continuous" or "because the interval is continuous" is not acceptable. |
### Part 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $x_1 = 5 - \dfrac{8\sin 2.5 - 15 + 9}{\text{"}\,4\cos 2.5 - 3\text{"}}$ (where $f(5)=-1.212\ldots$, $f'(5)=-6.204\ldots$) | M1 | Must be correct N-R formula with their $f'(x)$. Allow if attempted in degrees. |
| $x_1 = \text{awrt } 4.80$ | A1 | Not awrt 4.8; isw if awrt 4.80 seen. Ignore subsequent iterations. $5 - \frac{f(5)}{f'(5)} \neq \text{awrt } 4.80$ with no evidence scores M0. |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{824d73c5-525c-4876-ad66-33c8f1664277-12_634_741_251_662}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$ where
$$f ( x ) = 8 \sin \left( \frac { 1 } { 2 } x \right) - 3 x + 9 \quad x > 0$$
and $x$ is measured in radians.\\
The point $P$, shown in Figure 2, is a local maximum point on the curve.\\
Using calculus and the sketch in Figure 2,
\begin{enumerate}[label=(\alph*)]
\item find the $x$ coordinate of $P$, giving your answer to 3 significant figures.
The curve crosses the $x$-axis at $x = \alpha$, as shown in Figure 2 .\\
Given that, to 3 decimal places, $f ( 4 ) = 4.274$ and $f ( 5 ) = - 1.212$
\item explain why $\alpha$ must lie in the interval $[ 4,5 ]$
\item Taking $x _ { 0 } = 5$ as a first approximation to $\alpha$, apply the Newton-Raphson method once to $\mathrm { f } ( x )$ to obtain a second approximation to $\alpha$.
Show your method and give your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2022 Q6 [7]}}