| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Find stationary point coordinate |
| Difficulty | Standard +0.3 This is a structured multi-part question involving differentiation using quotient/chain rules, algebraic manipulation to show a given result, and straightforward application of an iteration formula. All steps are routine A-level techniques with clear guidance. The differentiation requires care but follows standard methods, and the iteration is purely computational. Slightly easier than average due to the scaffolding provided. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln x \rightarrow \frac{1}{x}\) | B1 | Seen or implied |
| Method to differentiate \(\frac{4x^2+x}{2\sqrt{x}}\) | M1 | Look for \(\frac{4x^2+x}{2\sqrt{x}} \to \ldots x^{3/2} + \ldots x^{1/2}\) differentiated, or quotient rule applied |
| e.g. \(2 \times \frac{3}{2}x^{1/2} + \frac{1}{2} \times \frac{1}{2}x^{-1/2}\) | A1 | Correct differentiation (unsimplified) |
| \(\frac{dy}{dx} = 3\sqrt{x} + \frac{1}{4\sqrt{x}} - \frac{4}{x} = \frac{12x^2+x-16\sqrt{x}}{4x\sqrt{x}}\) | A1* | No errors including missing brackets; sufficient working shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(12x^2 + x - 16\sqrt{x} = 0 \Rightarrow 12x^{3/2} + x^{1/2} - 16 = 0\) | M1 | Sets numerator \(= 0\); divides by \(\sqrt{x}\) or equivalent |
| e.g. \(12x^{3/2} = 16 - \sqrt{x}\) | dM1 | Makes \(x^{3/2}\) the subject |
| \(x^{3/2} = \frac{4}{3} - \frac{\sqrt{x}}{12} \Rightarrow x = \left(\frac{4}{3} - \frac{\sqrt{x}}{12}\right)^{2/3}\) | A1* | Correct rigorous argument to given solution; no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_2 = \sqrt[3]{\left(\frac{4}{3} - \frac{\sqrt{2}}{12}\right)^2}\) | M1 | Uses iterative formula with \(x_1 = 2\) |
| \(x_2 = \text{awrt } 1.13894\) | A1 | |
| \(x = 1.15650\) | A1 |
# Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln x \rightarrow \frac{1}{x}$ | B1 | Seen or implied |
| Method to differentiate $\frac{4x^2+x}{2\sqrt{x}}$ | M1 | Look for $\frac{4x^2+x}{2\sqrt{x}} \to \ldots x^{3/2} + \ldots x^{1/2}$ differentiated, or quotient rule applied |
| e.g. $2 \times \frac{3}{2}x^{1/2} + \frac{1}{2} \times \frac{1}{2}x^{-1/2}$ | A1 | Correct differentiation (unsimplified) |
| $\frac{dy}{dx} = 3\sqrt{x} + \frac{1}{4\sqrt{x}} - \frac{4}{x} = \frac{12x^2+x-16\sqrt{x}}{4x\sqrt{x}}$ | A1* | No errors including missing brackets; sufficient working shown |
---
# Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $12x^2 + x - 16\sqrt{x} = 0 \Rightarrow 12x^{3/2} + x^{1/2} - 16 = 0$ | M1 | Sets numerator $= 0$; divides by $\sqrt{x}$ or equivalent |
| e.g. $12x^{3/2} = 16 - \sqrt{x}$ | dM1 | Makes $x^{3/2}$ the subject |
| $x^{3/2} = \frac{4}{3} - \frac{\sqrt{x}}{12} \Rightarrow x = \left(\frac{4}{3} - \frac{\sqrt{x}}{12}\right)^{2/3}$ | A1* | Correct rigorous argument to given solution; no errors |
---
# Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2 = \sqrt[3]{\left(\frac{4}{3} - \frac{\sqrt{2}}{12}\right)^2}$ | M1 | Uses iterative formula with $x_1 = 2$ |
| $x_2 = \text{awrt } 1.13894$ | A1 | |
| $x = 1.15650$ | A1 | |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e28350e9-5090-4079-97da-e669ef9a5a7a-16_621_799_246_630}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C$ with equation
$$y = \frac { 4 x ^ { 2 } + x } { 2 \sqrt { x } } - 4 \ln x \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 12 x ^ { 2 } + x - 16 \sqrt { x } } { 4 x \sqrt { x } }$$
The point $P$, shown in Figure 1, is the minimum turning point on $C$.
\item Show that the $x$ coordinate of $P$ is a solution of
$$x = \left( \frac { 4 } { 3 } - \frac { \sqrt { x } } { 12 } \right) ^ { \frac { 2 } { 3 } }$$
\item Use the iteration formula
$$x _ { n + 1 } = \left( \frac { 4 } { 3 } - \frac { \sqrt { x _ { n } } } { 12 } \right) ^ { \frac { 2 } { 3 } } \quad \text { with } x _ { 1 } = 2$$
to find (i) the value of $x _ { 2 }$ to 5 decimal places,\\
(ii) the $x$ coordinate of $P$ to 5 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q7 [10]}}