| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a standard fixed-point iteration question requiring routine differentiation to derive the equation, straightforward calculator work to iterate the formula, and a change of sign method to verify the root. All techniques are well-practiced at A-level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 2x + \dfrac{4x-4}{2x^2 - 4x + 5}\) | M1 | Differentiates \(\ln(2x^2 - 4x + 5)\) to obtain \(\dfrac{g(x)}{2x^2-4x+5}\) where \(g(x)\) could be 1 |
| \(f'(x) = 2x + \dfrac{4x-4}{2x^2-4x+5}\) | A1 | Correct expression |
| \(2x + \dfrac{4x-4}{2x^2-4x+5} = 0 \Rightarrow 2x(2x^2-4x+5) + 4x - 4 = 0\) | dM1 | Sets \(f'(x) = ax + \dfrac{g(x)}{2x^2-4x+5} = 0\) and uses correct algebra to obtain cubic, condoning slips |
| \(2x^3 - 4x^2 + 7x - 2 = 0\) | A1* | Achieves this with no errors (dM1 must have been awarded) |
| (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_2 = \frac{1}{7}\left(2 + 4(0.3)^2 - 2(0.3)^3\right)\) | M1 | Attempts iterative formula with \(x_1 = 0.3\). If no method shown, award for \(x_2 =\) awrt \(0.33\) |
| \(x_2 = 0.3294\) | A1 | awrt \(0.3294\); note \(\frac{1153}{3500}\) is correct |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_4 = 0.3398\) | A1 | awrt \(0.3398\) |
| (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h(x) = 2x^3 - 4x^2 + 7x - 2\); \(h(0.3415) = 0.00366\ldots\), \(h(0.3405) = -0.00130\ldots\) | M1 | Substitutes \(x=0.3415\) and \(x=0.3405\) into suitable function; gets one value correct to 1 s.f. Must show evidence of the function used |
| States: change of sign, \(f'(x)\) is continuous, \(\alpha = 0.341\) to 3 d.p. | A1 | Both values correct (rounded/truncated to 1 s.f.); states change of sign and function is continuous; minimal conclusion e.g. \(\alpha = 0.341\) |
| (2 marks) |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 2x + \dfrac{4x-4}{2x^2 - 4x + 5}$ | M1 | Differentiates $\ln(2x^2 - 4x + 5)$ to obtain $\dfrac{g(x)}{2x^2-4x+5}$ where $g(x)$ could be 1 |
| $f'(x) = 2x + \dfrac{4x-4}{2x^2-4x+5}$ | A1 | Correct expression |
| $2x + \dfrac{4x-4}{2x^2-4x+5} = 0 \Rightarrow 2x(2x^2-4x+5) + 4x - 4 = 0$ | dM1 | Sets $f'(x) = ax + \dfrac{g(x)}{2x^2-4x+5} = 0$ and uses correct algebra to obtain cubic, condoning slips |
| $2x^3 - 4x^2 + 7x - 2 = 0$ | A1* | Achieves this with no errors (dM1 must have been awarded) |
| **(4 marks)** | | |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2 = \frac{1}{7}\left(2 + 4(0.3)^2 - 2(0.3)^3\right)$ | M1 | Attempts iterative formula with $x_1 = 0.3$. If no method shown, award for $x_2 =$ awrt $0.33$ |
| $x_2 = 0.3294$ | A1 | awrt $0.3294$; note $\frac{1153}{3500}$ is correct |
| **(2 marks)** | | |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_4 = 0.3398$ | A1 | awrt $0.3398$ |
| **(1 mark)** | | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h(x) = 2x^3 - 4x^2 + 7x - 2$; $h(0.3415) = 0.00366\ldots$, $h(0.3405) = -0.00130\ldots$ | M1 | Substitutes $x=0.3415$ and $x=0.3405$ into suitable function; gets one value correct to 1 s.f. Must show evidence of the function used |
| States: change of sign, $f'(x)$ is continuous, $\alpha = 0.341$ to 3 d.p. | A1 | Both values correct (rounded/truncated to 1 s.f.); states change of sign and function is continuous; minimal conclusion e.g. $\alpha = 0.341$ |
| **(2 marks)** | | |\begin{enumerate}
\item The curve with equation $y = \mathrm { f } ( x )$ where
\end{enumerate}
$$f ( x ) = x ^ { 2 } + \ln \left( 2 x ^ { 2 } - 4 x + 5 \right)$$
has a single turning point at $x = \alpha$\\
(a) Show that $\alpha$ is a solution of the equation
$$2 x ^ { 3 } - 4 x ^ { 2 } + 7 x - 2 = 0$$
The iterative formula
$$x _ { n + 1 } = \frac { 1 } { 7 } \left( 2 + 4 x _ { n } ^ { 2 } - 2 x _ { n } ^ { 3 } \right)$$
is used to find an approximate value for $\alpha$.\\
Starting with $x _ { 1 } = 0.3$\\
(b) calculate, giving each answer to 4 decimal places,\\
(i) the value of $x _ { 2 }$\\
(ii) the value of $x _ { 4 }$
Using a suitable interval and a suitable function that should be stated,\\
(c) show that $\alpha$ is 0.341 to 3 decimal places.
\hfill \mbox{\textit{Edexcel Paper 1 2021 Q4 [9]}}