| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Show root in interval |
| Difficulty | Standard +0.3 This is a structured multi-part question that guides students through standard techniques: verifying a root by substitution, showing a root exists via sign change, polynomial division to find a factor, and applying the Newton-Raphson formula. Each part is routine with clear signposting, requiring only methodical application of A-level techniques rather than problem-solving insight. Slightly easier than average due to the scaffolding. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.09b Sign change methods: understand failure cases1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(-1) = (-1)^4 + (-1)^3 - 2(-1)^2 - 4(-1) - 2 = 1 - 1 - 2 + 4 - 2 = 0\) | E1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(1) = 1 + 1 - 2 - 4 - 2 = -6\) or 'negative' | B1 | 1.1 |
| \(f(2) = 16 + 8 - 8 - 8 - 2 = 6\) or 'positive'; change of sign \(\Rightarrow\) root between 1 and 2 | E1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| long division or equating coefficients | M1 | 1.1 |
| \(\Rightarrow g(x) = x^3 - 2x - 2\) so \(a = -2,\ b = -2\) | A1, A1 | 2.2a, 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Clear explanation: e.g. \(f(x) = (x+1)g(x)\); for the root of \(f(x) = 0\) between 1 and 2, RHS is also zero hence \(g(x) = 0\) | E1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)} = x_n - \frac{x_n^3 - 2x_n - 2}{3x_n^2 - 2}\) | M1 | 1.1 |
| \(= \frac{3x_n^3 - 2x_n - x_n^3 + 2x_n + 2}{3x_n^2 - 2} = \frac{2x_n^3 + 2}{3x_n^2 - 2}\) | E1 | 2.4 |
| Root \(1.769\) (4sf) | A1 | 2.2a |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(-1) = (-1)^4 + (-1)^3 - 2(-1)^2 - 4(-1) - 2 = 1 - 1 - 2 + 4 - 2 = 0$ | E1 | 1.1 | **[1]** |
---
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(1) = 1 + 1 - 2 - 4 - 2 = -6$ or 'negative' | B1 | 1.1 | both correct |
| $f(2) = 16 + 8 - 8 - 8 - 2 = 6$ or 'positive'; change of sign $\Rightarrow$ root between 1 and 2 | E1 | 2.4 | allow no mention of continuity of f; AG **[2]** |
---
## Question 10(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| long division or equating coefficients | M1 | 1.1 |
| $\Rightarrow g(x) = x^3 - 2x - 2$ so $a = -2,\ b = -2$ | A1, A1 | 2.2a, 1.1 | **[3]** |
---
## Question 10(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Clear explanation: e.g. $f(x) = (x+1)g(x)$; for the root of $f(x) = 0$ between 1 and 2, RHS is also zero hence $g(x) = 0$ | E1 | 2.4 | **[1]** |
---
## Question 10(e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)} = x_n - \frac{x_n^3 - 2x_n - 2}{3x_n^2 - 2}$ | M1 | 1.1 |
| $= \frac{3x_n^3 - 2x_n - x_n^3 + 2x_n + 2}{3x_n^2 - 2} = \frac{2x_n^3 + 2}{3x_n^2 - 2}$ | E1 | 2.4 | AG |
| Root $1.769$ (4sf) | A1 | 2.2a | BC **[3]** |10 The function $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = x ^ { 4 } + x ^ { 3 } - 2 x ^ { 2 } - 4 x - 2$.
\begin{enumerate}[label=(\alph*)]
\item Show that $x = - 1$ is a root of $\mathrm { f } ( x ) = 0$.
\item Show that another root of $\mathrm { f } ( x ) = 0$ lies between $x = 1$ and $x = 2$.
\item Show that $\mathrm { f } ( x ) = ( x + 1 ) \mathrm { g } ( x )$, where $\mathrm { g } ( x ) = x ^ { 3 } + a x + b$ and $a$ and $b$ are integers to be determined.
\item Without further calculation, explain why $\mathrm { g } ( x ) = 0$ has a root between $x = 1$ and $x = 2$.
\item Use the Newton-Raphson formula to show that an iteration formula for finding roots of $\mathrm { g } ( x ) = 0$ may be written
$$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 2 } { 3 x _ { n } ^ { 2 } - 2 }$$
Determine the root of $\mathrm { g } ( x ) = 0$ which lies between $x = 1$ and $x = 2$ correct to 4 significant figures.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 Q10 [10]}}