| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Intersection of curves via iteration |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring routine differentiation (chain rule for exponential, standard derivative for ln), simple algebraic manipulation to show an equation, and mechanical application of a given iterative formula. All techniques are standard A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 8xe^{4x^2-1}\) or e.g. \(\frac{8xe^{4x^2}}{e}\) | B1 | Correct derivative in any form; "\(f'(x) =\)" not required; apply isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(g'(x) = \frac{8}{x}\) or e.g. \(8x^{-1}\) | B1 | Correct derivative in any form; "\(g'(x) =\)" not required; apply isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(8xe^{4x^2-1} = \frac{8}{x} \Rightarrow e^{4x^2-1} = \frac{1}{x^2} \Rightarrow 4x^2 - 1 = \ln\frac{1}{x^2}\) | M1 | Eliminates \(e\) by setting \(f'(x) = g'(x)\); \(g'(x) = \frac{B}{x}\) with \(A \times B > 0\); proceeds via \(e^{4x^2-1} = \frac{\cdots}{x^2}\) |
| \(4x^2 - 1 = \ln\frac{1}{x^2} \Rightarrow 4x^2 - 1 = -2\ln x \Rightarrow 4x^2 + 2\ln x - 1 = 0\) * | A1* | Obtains printed answer with sufficient working; "e" must be eliminated before given answer; must follow correct derivatives; condone \(4x^2 + 2\ln |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_1 = 0.6 \Rightarrow x_2 = \sqrt{\frac{1 - 2\ln 0.6}{4}}\) | M1 | |
| \(x_2 = 0.7109\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha = 0.6706\) | B1, A1 (in ePEN) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts iterative formula with \(x_1 = 0.6\), e.g. \(x_2 = \sqrt{\dfrac{1-2\ln 0.6}{4}}\) | M1 | May be implied by awrt 0.71 provided no incorrect working seen |
| \(x_2 =\) awrt \(0.7109\) | A1 | Sight of \(x_2 =\) awrt 0.7109 scores M1A1 |
| \(\alpha = 0.6706\) (4dp) | B1 (A1 on ePEN) | Must be this exact value, not awrt 0.6706 |
# Question 6(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 8xe^{4x^2-1}$ or e.g. $\frac{8xe^{4x^2}}{e}$ | B1 | Correct derivative in any form; "$f'(x) =$" not required; apply isw |
# Question 6(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g'(x) = \frac{8}{x}$ or e.g. $8x^{-1}$ | B1 | Correct derivative in any form; "$g'(x) =$" not required; apply isw |
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $8xe^{4x^2-1} = \frac{8}{x} \Rightarrow e^{4x^2-1} = \frac{1}{x^2} \Rightarrow 4x^2 - 1 = \ln\frac{1}{x^2}$ | M1 | Eliminates $e$ by setting $f'(x) = g'(x)$; $g'(x) = \frac{B}{x}$ with $A \times B > 0$; proceeds via $e^{4x^2-1} = \frac{\cdots}{x^2}$ |
| $4x^2 - 1 = \ln\frac{1}{x^2} \Rightarrow 4x^2 - 1 = -2\ln x \Rightarrow 4x^2 + 2\ln x - 1 = 0$ * | A1* | Obtains printed answer with sufficient working; "e" must be eliminated before given answer; must follow correct derivatives; condone $4x^2 + 2\ln|x| - 1 = 0$ |
# Question 6(c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1 = 0.6 \Rightarrow x_2 = \sqrt{\frac{1 - 2\ln 0.6}{4}}$ | M1 | |
| $x_2 = 0.7109$ | A1 | |
# Question 6(c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha = 0.6706$ | B1, A1 (in ePEN) | |
## Question 6(c)(i)/(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts iterative formula with $x_1 = 0.6$, e.g. $x_2 = \sqrt{\dfrac{1-2\ln 0.6}{4}}$ | M1 | May be implied by awrt 0.71 provided no incorrect working seen |
| $x_2 =$ awrt $0.7109$ | A1 | Sight of $x_2 =$ awrt 0.7109 scores M1A1 |
| $\alpha = 0.6706$ (4dp) | B1 (A1 on ePEN) | Must be this exact value, **not** awrt 0.6706 |
**Reference table:** $x_1=0.6,\ x_2=0.7109239143,\ x_3=0.6485329086,\ x_4=0.6830236199,\ x_5=0.6637868021,\ x_6=0.6744606223,\ \alpha=0.6706416243$
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2ce10759-9ce6-47a1-b55d-d22082f88f55-12_518_670_248_740}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curves with equations $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$ where
$$\begin{array} { l l }
\mathrm { f } ( x ) = \mathrm { e } ^ { 4 x ^ { 2 } - 1 } & x > 0 \\
\mathrm {~g} ( x ) = 8 \ln x & x > 0
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { f } ^ { \prime } ( x )$
\item $\mathrm { g } ^ { \prime } ( x )$
Given that $\mathrm { f } ^ { \prime } ( x ) = \mathrm { g } ^ { \prime } ( x )$ at $x = \alpha$
\end{enumerate}\item show that $\alpha$ satisfies the equation
$$4 x ^ { 2 } + 2 \ln x - 1 = 0$$
The iterative formula
$$x _ { n + 1 } = \sqrt { \frac { 1 - 2 \ln x _ { n } } { 4 } }$$
is used with $x _ { 1 } = 0.6$ to find an approximate value for $\alpha$
\item Calculate, giving each answer to 4 decimal places,
\begin{enumerate}[label=(\roman*)]
\item the value of $x _ { 2 }$
\item the value of $\alpha$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2024 Q6 [7]}}