| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson error analysis |
| Difficulty | Standard +0.3 This is a straightforward application of Newton-Raphson with standard verification steps. Part (i) is routine algebraic manipulation, part (ii) involves differentiation and substitution (though verifying F'(α)=0 requires some care), and part (iii) is mechanical iteration. While it's Further Maths content, the question follows a predictable template with no novel problem-solving required, making it slightly easier than average overall. |
| Spec | 1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_{n+1}=x_n - \frac{x_n^3-5x_n+3}{3x_n^2-5}=\frac{2x_n^3-3}{3x_n^2-5}\) | M1 | For attempt at N-R formula |
| A1 | For correct N-R expression | |
| A1 3 | For correct answer (necessary details needed) AG; allow omission of suffixes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F'(x)=\frac{6x^2(3x^2-5)-6x(2x^3-3)}{(3x^2-5)^2}=\frac{6x(x^3-5x+3)}{(3x^2-5)^2}\) | M1 | For using quotient OR product rule to find \(F'(x)\) |
| M1 | For factorising numerator to show \(k(x^3-5x+3)\) | |
| \(F'(\alpha)=\frac{6\alpha(\alpha^3-5\alpha+3)}{(3\alpha^2-5)^2}=0\) since \(\alpha^3-5\alpha+3=0\) | A1 3 | For correct explanation of AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_1=2 \Rightarrow 1.85714,\; 1.83479,\; 1.83424,\; 1.83424\) | B1 | First iterate correct to at least 4 d.p. OR \(\frac{13}{7}\) |
| \((\alpha=)\; 1.8342\) | B1 | For 2 equal iterates to at least 4 d.p. |
| B1 3 | For correct \(\alpha\) to 4 d.p.; allow answer rounding to 1.8342; SR If not N-R, B0 B0 B0 |
## Question 5(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_{n+1}=x_n - \frac{x_n^3-5x_n+3}{3x_n^2-5}=\frac{2x_n^3-3}{3x_n^2-5}$ | M1 | For attempt at N-R formula |
| | A1 | For correct N-R expression |
| | A1 **3** | For correct answer (necessary details needed) **AG**; allow omission of suffixes |
## Question 5(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F'(x)=\frac{6x^2(3x^2-5)-6x(2x^3-3)}{(3x^2-5)^2}=\frac{6x(x^3-5x+3)}{(3x^2-5)^2}$ | M1 | For using quotient OR product rule to find $F'(x)$ |
| | M1 | For factorising numerator to show $k(x^3-5x+3)$ |
| $F'(\alpha)=\frac{6\alpha(\alpha^3-5\alpha+3)}{(3\alpha^2-5)^2}=0$ since $\alpha^3-5\alpha+3=0$ | A1 **3** | For correct explanation of **AG** |
## Question 5(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_1=2 \Rightarrow 1.85714,\; 1.83479,\; 1.83424,\; 1.83424$ | B1 | First iterate correct to at least 4 d.p. OR $\frac{13}{7}$ |
| $(\alpha=)\; 1.8342$ | B1 | For 2 equal iterates to at least 4 d.p. |
| | B1 **3** | For correct $\alpha$ to 4 d.p.; allow answer rounding to 1.8342; **SR** If not N-R, B0 B0 B0 |
---5 The equation
$$x ^ { 3 } - 5 x + 3 = 0$$
may be solved by the Newton-Raphson method. Successive approximations to a root are denoted by $x _ { 1 } , x _ { 2 } , \ldots , x _ { n } , \ldots$.\\
(i) Show that the Newton-Raphson formula can be written in the form $x _ { n + 1 } = \mathrm { F } \left( x _ { n } \right)$, where
$$\mathrm { F } ( x ) = \frac { 2 x ^ { 3 } - 3 } { 3 x ^ { 2 } - 5 }$$
(ii) Find $\mathrm { F } ^ { \prime } ( x )$ and hence verify that $\mathrm { F } ^ { \prime } ( \alpha ) = 0$, where $\alpha$ is any one of the roots of equation (A).\\
(iii) Use the Newton-Raphson method to find the root of equation (A) which is close to 2 . Write down sufficient approximations to find the root correct to 4 decimal places.
\hfill \mbox{\textit{OCR FP2 2011 Q5 [9]}}