| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Show root in interval |
| Difficulty | Moderate -0.3 This is a straightforward application of standard A-level techniques: part (a) requires simple function evaluation to show sign change, part (b) is routine differentiation using chain rule, and part (c) is direct application of the Newton-Raphson formula with one iteration. All steps are procedural with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.09a Sign change methods: locate roots1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{f(3.6)=\}\ 3.6 + \tan\!\left(\frac{1}{2}(3.6)\right) = -0.686\ldots < 0\) and \(\{f(3.7)=\}\ 3.7 + \tan\!\left(\frac{1}{2}(3.7)\right) = 0.211\ldots > 0\) | M1 | Attempts both \(f(3.6)\) and \(f(3.7)\), obtains at least one correct to 1 s.f., considers signs. Use of degrees is M0. |
| Change of sign and function is continuous in the interval \(\Rightarrow\) conclusion e.g. "there is a root in \([3.6, 3.7]\)" | A1* | Requires: both values correct to 1 s.f., reference to sign change, reference to continuity, and a minimal conclusion. Do not condone "change of sign therefore continuous". |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(\tan\!\left(\frac{1}{2}x\right) \rightarrow \ldots\sec^2\!\left(\frac{1}{2}x\right)\) | M1 | Differentiating \(\tan\!\left(\frac{1}{2}x\right)\) to obtain \(k\sec^2\!\left(\frac{1}{2}x\right)\) where \(k\) is a positive constant. |
| \(\{f'(x)=\}\ 1 + \frac{1}{2}\sec^2\!\left(\frac{1}{2}x\right)\) | A1 | May be unsimplified. Note \(\frac{3}{2}+\frac{1}{2}\tan^2\!\left(\frac{1}{2}x\right)\) is also correct. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(3.7 - \dfrac{3.7 + \tan\!\left(\frac{1}{2}(3.7)\right)}{1 + \frac{1}{2}\sec^2\!\left(\frac{1}{2}(3.7)\right)} = \ldots\) | M1 | Attempts \(3.7 - \frac{f(3.7)}{f'(3.7)}\) and obtains a value, using a "changed" function for \(f'(x)\). N.B. \(f(3.7)=0.211\ldots\) and \(f'(3.7)=7.58\ldots\) |
| \(\alpha = \text{awrt } 3.672\) | A1 | For awrt 3.672. Ignore any subsequent iterations. |
# Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{f(3.6)=\}\ 3.6 + \tan\!\left(\frac{1}{2}(3.6)\right) = -0.686\ldots < 0$ **and** $\{f(3.7)=\}\ 3.7 + \tan\!\left(\frac{1}{2}(3.7)\right) = 0.211\ldots > 0$ | M1 | Attempts both $f(3.6)$ and $f(3.7)$, obtains at least one correct to 1 s.f., considers signs. Use of degrees is M0. |
| Change of sign and function is continuous in the interval $\Rightarrow$ conclusion e.g. "there is a root in $[3.6, 3.7]$" | A1* | Requires: both values correct to 1 s.f., reference to sign change, reference to continuity, and a minimal conclusion. Do not condone "change of sign therefore continuous". |
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# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\tan\!\left(\frac{1}{2}x\right) \rightarrow \ldots\sec^2\!\left(\frac{1}{2}x\right)$ | M1 | Differentiating $\tan\!\left(\frac{1}{2}x\right)$ to obtain $k\sec^2\!\left(\frac{1}{2}x\right)$ where $k$ is a positive constant. |
| $\{f'(x)=\}\ 1 + \frac{1}{2}\sec^2\!\left(\frac{1}{2}x\right)$ | A1 | May be unsimplified. Note $\frac{3}{2}+\frac{1}{2}\tan^2\!\left(\frac{1}{2}x\right)$ is also correct. |
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# Question 3(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $3.7 - \dfrac{3.7 + \tan\!\left(\frac{1}{2}(3.7)\right)}{1 + \frac{1}{2}\sec^2\!\left(\frac{1}{2}(3.7)\right)} = \ldots$ | M1 | Attempts $3.7 - \frac{f(3.7)}{f'(3.7)}$ and obtains a value, using a "changed" function for $f'(x)$. N.B. $f(3.7)=0.211\ldots$ and $f'(3.7)=7.58\ldots$ |
| $\alpha = \text{awrt } 3.672$ | A1 | For awrt 3.672. Ignore any subsequent iterations. |3.
$$\mathrm { f } ( x ) = x + \tan \left( \frac { 1 } { 2 } x \right) \quad \pi < x < \frac { 3 \pi } { 2 }$$
Given that the equation $\mathrm { f } ( x ) = 0$ has a single root $\alpha$
\begin{enumerate}[label=(\alph*)]
\item show that $\alpha$ lies in the interval [3.6, 3.7]
\item Find $\mathrm { f } ^ { \prime } ( x )$
\item Using 3.7 as a first approximation for $\alpha$, apply the Newton-Raphson method once to obtain a second approximation for $\alpha$. Give your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q3 [6]}}