| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve exponential equation via iteration |
| Difficulty | Standard +0.3 This is a multi-part question that guides students through standard techniques: sketching ln x, finding a tangent equation, explaining an intersection graphically, performing straightforward iteration calculations, and verifying accuracy via change of sign. While it has many parts (5 total), each individual step is routine C3 material with no novel problem-solving required. The iteration formula is given directly, and part (e) explicitly tells students to use change of sign method. This is slightly easier than average due to the scaffolded structure and standard techniques throughout. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07m Tangents and normals: gradient and equations1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| 7. continued | Leave blank |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Correct shape of \(y = \ln x\) | B1 | shape |
| \(x\)-intercept at \((1, 0)\) labelled | B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\frac{dy}{dx} = \frac{1}{x}\), tangent at \((e,1)\) is \(y = \frac{1}{e}x + C\) | M1 | |
| Line passes through \((e,1)\): \(1 = \frac{1}{e}\cdot e + C\), so \(C = 0\) | M1 | |
| Line passes through the origin | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Lines \(y=mx\) with gradient \(< \frac{1}{e}\) lie below the tangent line | B1 | |
| \(y = \frac{x}{e}\) cuts \(y = \ln x\) between \(x=1\) and \(x=e\) | B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(x_0 = 1.86\) | ||
| \(x_1 = e^{\frac{x_0}{3}} = 1.859\) | M1 | |
| \(x_2 = 1.858\) | A1 | |
| \(x_3 = 1.858\) | ||
| \(x_4 = 1.858\) | ||
| \(x_5 = 1.857\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| When \(x = 1.8575\): \(\ln x - \frac{1}{3}x = 0.000\,064\,8\ldots > 0\) | M1 | |
| When \(x = 1.8565\): \(\ln x - \frac{1}{3}x = -0.000\,140\ldots < 0\) | A1 | |
| Change of sign implies root between these values | A1 | (3 marks) |
# Question 7:
## Part (a)
| Answer/Working | Marks | Notes |
|---|---|---|
| Correct shape of $y = \ln x$ | B1 | shape |
| $x$-intercept at $(1, 0)$ labelled | B1 | (2 marks) |
## Part (b)
| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{x}$, tangent at $(e,1)$ is $y = \frac{1}{e}x + C$ | M1 | |
| Line passes through $(e,1)$: $1 = \frac{1}{e}\cdot e + C$, so $C = 0$ | M1 | |
| Line passes through the origin | A1 | (3 marks) |
## Part (c)
| Answer/Working | Marks | Notes |
|---|---|---|
| Lines $y=mx$ with gradient $< \frac{1}{e}$ lie below the tangent line | B1 | |
| $y = \frac{x}{e}$ cuts $y = \ln x$ between $x=1$ and $x=e$ | B1 | (2 marks) |
## Part (d)
| Answer/Working | Marks | Notes |
|---|---|---|
| $x_0 = 1.86$ | | |
| $x_1 = e^{\frac{x_0}{3}} = 1.859$ | M1 | |
| $x_2 = 1.858$ | A1 | |
| $x_3 = 1.858$ | | |
| $x_4 = 1.858$ | | |
| $x_5 = 1.857$ | A1 | (3 marks) |
## Part (e)
| Answer/Working | Marks | Notes |
|---|---|---|
| When $x = 1.8575$: $\ln x - \frac{1}{3}x = 0.000\,064\,8\ldots > 0$ | M1 | |
| When $x = 1.8565$: $\ln x - \frac{1}{3}x = -0.000\,140\ldots < 0$ | A1 | |
| Change of sign implies root between these values | A1 | (3 marks) |
**Total: 13 marks**
---7. (a) Sketch the curve with equation $y = \ln x$.\\
(b) Show that the tangent to the curve with equation $y = \ln x$ at the point ( $\mathrm { e } , 1$ ) passes through the origin.\\
(c) Use your sketch to explain why the line $y = m x$ cuts the curve $y = \ln x$ between $x = 1$ and $x = \mathrm { e }$ if $0 < m < \frac { 1 } { \mathrm { e } }$.
Taking $x _ { 0 } = 1.86$ and using the iteration $x _ { n + 1 } = \mathrm { e } ^ { \frac { 1 } { 3 } x _ { n } }$,\\
(d) calculate $x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 }$ and $x _ { 5 }$, giving your answer to $x _ { 5 }$ to 3 decimal places.
The root of $\ln x - \frac { 1 } { 3 } x = 0$ is $\alpha$.\\
(e) By considering the change of sign of $\ln x - \frac { 1 } { 3 } x$ over a suitable interval, show that your answer for $x _ { 5 }$ is an accurate estimate of $\alpha$, correct to 3 decimal places.
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7. continued & Leave blank \\
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\hfill \mbox{\textit{Edexcel C3 Q7 [13]}}