| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with complex derivative required |
| Difficulty | Moderate -0.3 This is a straightforward application of the Newton-Raphson method with a slightly more complex derivative involving a negative power term. Part (a) is routine differentiation using the power rule, and part (b) requires one iteration of a standard formula with given starting value. While it's Further Maths content, it's a textbook exercise requiring only mechanical application of learned techniques with no problem-solving insight needed. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = x^2 + \frac{5}{2}x^{-1} - 3x - 1\) | — | Rewrite with negative index |
| \(f'(x) = 2x - \frac{5}{2}x^{-2} - 3\) | M1 | At least two of the four terms differentiated correctly |
| \(\left\{f'(x) = 2x - \frac{5}{2x^2} - 3\right\}\) | A1 | Correct differentiation (allow any correct unsimplified form) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(0.8) = 0.8^2 + \frac{5}{2(0.8)} - 3(0.8) - 1 = \frac{73}{200}\) | B1 | A correct numerical expression for \(f(0.8)\) |
| \(f'(0.8) = -5.30625\left(= \frac{-849}{160}\right)\) | M1 | Attempt to insert \(x = 0.8\) into their \(f'(x)\). Does not require evaluation. If \(f'(0.8)\) is incorrect for their derivative, no working score M0 |
| \(\alpha_2 = 0.8 - \left(\frac{\text{"0.365"}}{\text{"-5.30625"}}\right)\) | M1 | Correct application of Newton-Raphson using their values. Does not require evaluation |
| \(= 0.869\) (3dp) | A1 cao | 0.869 |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = x^2 + \frac{5}{2}x^{-1} - 3x - 1$ | — | Rewrite with negative index |
| $f'(x) = 2x - \frac{5}{2}x^{-2} - 3$ | M1 | At least two of the four terms differentiated correctly |
| $\left\{f'(x) = 2x - \frac{5}{2x^2} - 3\right\}$ | A1 | Correct differentiation (allow any correct unsimplified form) |
**(2 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0.8) = 0.8^2 + \frac{5}{2(0.8)} - 3(0.8) - 1 = \frac{73}{200}$ | B1 | A correct numerical expression for $f(0.8)$ |
| $f'(0.8) = -5.30625\left(= \frac{-849}{160}\right)$ | M1 | Attempt to insert $x = 0.8$ into their $f'(x)$. Does not require evaluation. If $f'(0.8)$ is incorrect for their derivative, no working score M0 |
| $\alpha_2 = 0.8 - \left(\frac{\text{"0.365"}}{\text{"-5.30625"}}\right)$ | M1 | Correct application of Newton-Raphson using their values. Does not require evaluation |
| $= 0.869$ (3dp) | A1 cao | 0.869 |
**(4 marks)**
---4.
$$f ( x ) = x ^ { 2 } + \frac { 5 } { 2 x } - 3 x - 1 , \quad x \neq 0$$
\begin{enumerate}[label=(\alph*)]
\item Use differentiation to find $\mathrm { f } ^ { \prime } ( x )$.
The root $\alpha$ of the equation $\mathrm { f } ( x ) = 0$ lies in the interval [0.7, 0.9].
\item Taking 0.8 as a first approximation to $\alpha$, apply the Newton-Raphson process once to $\mathrm { f } ( x )$ to obtain a second approximation to $\alpha$. Give your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2011 Q4 [6]}}