| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with trigonometric or exponential functions |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard A-level techniques: sign change verification requires simple substitution, finding maxima uses routine differentiation and solving f'(x)=0, and Newton-Raphson is a standard algorithm application. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.07i Differentiate x^n: for rational n and sums1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Part a. | (2 marks) | |
| M1 | Attempts the value of y at 0.7 and 0.8 with at least one correct to 1 significant figure rounded or truncated. Example: \(y = 4 \cos(0.7) - 2(0.7) + 1 = 0.28\) and \(y = 4 \cos(0.8) - 2(0.8) + 1 = -0.72\) | |
| A1 | Both values correct to 1 significant figure rounded or truncated with reason (sign change and continuous function) and minimal conclusion (root). Example: \(y | _{0.7} = 0.3 > 0\), \(y |
| Answer | Marks |
|---|---|
| Part b. | (5 marks) |
| M1 | Differentiates to obtain \(A \sin 2x + B\) |
| A1 | Correct derivative \(\frac{dy}{dx} = -8 \sin 2x - 2\). Allow unsimplified e.g. \(\frac{dy}{dx} = -4 \times 2 \times \sin 2x - 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| dM1 | For the complete strategy of proceeding to a value for x. \(\frac{dy}{dx} = 0\) \(\Rightarrow \sin 2x = a\), \( | a |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | For recognising that either 3.02 or 6.16 is a solution to \(\sin 2x = -\frac{1}{4}\) | |
| A1 | States that \(\beta = 3.02\) and \(\gamma = 6.16\) | labels must be correct |
| Answer | Marks |
|---|---|
| Part c. | (2 marks) |
| M1 | Attempts \(x_1 = 0.7 - \frac{f(0.7)}{f'(0.7)}\) to obtain a value following through on their f'(x) as long as it is a "changed" function. Must be correct N-R formula used-may need to check their values. Allow if attempted in degrees. For reference in degrees f(0.7) = 3.5998\(\ldots\) and f'(0.7) = -2.1954 \(\ldots\) and gives \(x_1 = 2.3392\) \(\ldots\) |
| Answer | Marks |
|---|---|
| M1 | Attempts \(x_1 = 0.8 - \frac{f(0.8)}{f'(0.8)}\) to obtain a value following through on their f'(x) as long as it is a "changed" function. Must be correct N-R formula used-may need to check their values. Allow if attempted in degrees. For reference in degrees f(0.8) = 3.3984\(\ldots\) and f'(0.8) = -2.2233 \(\ldots\) and gives \(x_1 = 2.3285\) \(\ldots\) |
| Answer | Marks |
|---|---|
| A1 | \(x_1 = \) awrt 0.73 |
(9 marks)
**Part a.** | (2 marks)
**M1** | Attempts the value of y at 0.7 and 0.8 with at least one correct to 1 significant figure rounded or truncated. Example: $y = 4 \cos(0.7) - 2(0.7) + 1 = 0.28$ and $y = 4 \cos(0.8) - 2(0.8) + 1 = -0.72$
**A1** | Both values correct to 1 significant figure rounded or truncated with reason (sign change and continuous function) and minimal conclusion (root). Example: $y|_{0.7} = 0.3 > 0$, $y|_{0.8} = -0.7 < 0$ and function is continuous so there is a root.
---
**Part b.** | (5 marks)
**M1** | Differentiates to obtain $A \sin 2x + B$
**A1** | Correct derivative $\frac{dy}{dx} = -8 \sin 2x - 2$. Allow unsimplified e.g. $\frac{dy}{dx} = -4 \times 2 \times \sin 2x - 2$
There is no need for derivative $\frac{dy}{dx}$ $\cdots$ just look for the expression
**dM1** | For the complete strategy of proceeding to a value for x. $\frac{dy}{dx} = 0$ $\Rightarrow \sin 2x = a$, $|a| < 1$ $\Rightarrow x = \cdots$
Look for correct order of operations, invsing then $\div 2$ leading to a value of x. When $\sin 2x = -\frac{1}{4}$ it is implied, for example, by awrt -0.253, awrt 3.394, awrt 6.031, awrt 9.677 or awrt 12.314
For $\sin 2x = \frac{1}{4}$ it is implied, for example, by awrt 0.253 or awrt 2.889
The calculations must be using radians. If degrees are used initially they must be converted to radians.
Example: $2x = 6.031$ $\cdots$ $\Rightarrow x = 3.02$ or $2x = 12.314$ $\cdots$ $\Rightarrow x = 6.16$
**A1** | For recognising that either 3.02 or 6.16 is a solution to $\sin 2x = -\frac{1}{4}$
**A1** | States that $\beta = 3.02$ and $\gamma = 6.16$ | labels must be correct
---
**Part c.** | (2 marks)
**M1** | Attempts $x_1 = 0.7 - \frac{f(0.7)}{f'(0.7)}$ to obtain a value following through on their f'(x) as long as it is a "changed" function. Must be correct N-R formula used-may need to check their values. Allow if attempted in degrees. For reference in degrees f(0.7) = 3.5998$\ldots$ and f'(0.7) = -2.1954 $\ldots$ and gives $x_1 = 2.3392$ $\ldots$
There must be clear evidence that $0.7 - \frac{f(0.7)}{f'(0.7)}$ is being attempted.
So e.g. $x_1 = 0.7 - \frac{f(0.7)}{f'(0.7)} = 0.7 - \frac{0.2798685716}{-9.88359784} = 0.7283164667 = 0.728$
**OR**
**M1** | Attempts $x_1 = 0.8 - \frac{f(0.8)}{f'(0.8)}$ to obtain a value following through on their f'(x) as long as it is a "changed" function. Must be correct N-R formula used-may need to check their values. Allow if attempted in degrees. For reference in degrees f(0.8) = 3.3984$\ldots$ and f'(0.8) = -2.2233 $\ldots$ and gives $x_1 = 2.3285$ $\ldots$
There must be clear evidence that $0.8 - \frac{f(0.8)}{f'(0.8)}$ is being attempted.
So e.g. $x_1 = 0.8 - \frac{f(0.8)}{f'(0.8)} = 0.8 - \frac{-0.7167980892}{-9.996588824} = 0.7282957315 = 0.728$
**A1** | $x_1 = $ awrt 0.73
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d37eaba2-0a25-4abf-b2c8-1e08673229fb-10_1287_988_278_340}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation
$$f ( x ) = 4 \cos 2 x - 2 x + 1 \quad x > 0$$
and where $x$ is measured in radians.\\
The curve crosses the $x$-axis at the point $A$, as shown in figure 1 .\\
Given that $x$-coordinate of $A$ is $\alpha$\\
a. show that $\alpha$ lies between 0.7 and 0.8
Given that $x$-coordinates of $B$ and $C$ are $\beta$ and $\gamma$ respectively and they are two smallest values of $x$ at which local maxima occur\\
b. find, using calculus, the value of $\beta$ and the value of $\gamma$, giving your answers to 3 significant figures.\\
c. taking $x _ { 0 } = 0.7$ or 0.8 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 2 significant figures.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q6 [9]}}