| Exam Board | WJEC |
|---|---|
| Module | Unit 3 (Unit 3) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show root in interval |
| Difficulty | Standard +0.3 This is a standard numerical methods question covering three routine techniques: sign change verification, fixed-point iteration, and Newton-Raphson. All three parts follow textbook procedures with no novel insight required. The iteration converges well and Newton-Raphson is straightforward differentiation and formula application. Slightly above average difficulty due to the 6 decimal place requirement and multiple methods, but still a typical C3/C4 numerical methods question. |
| Spec | 1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = 1 + 5x - x^4\) | M1 | (Use of Intermediate Value Theorem.) |
| \(f(1) = 5, f(2) = -5\) | A1 | (correct values and conclusions) |
| There is a change of sign indicating there is a root between 1 and 2. |
| Answer | Marks |
|---|---|
| \(x_{n+1} = \sqrt[4]{1 + 5x_n}\), \(x_0 = 1.5\) | B1 |
| \(x_1 = 1.707476485\) | B1 |
| \(x_2 = 1.75734609\) | |
| \(x_3 = 1.7687213\), \(x_4 = 1.7712854\) | B1 |
| \(x_5 = 1.7718619948\), \(\alpha \approx 1.77\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{1 + 5x_n - x_n^4}{5 - 4x_n^3}\) | M1 | (Attempt to use Newton-Raphson. All terms correct) |
| \(x_0 = 1.5\) | A1 | |
| \(x_1 = 1.904411765\) | M1, A1 | |
| \(x_2 = 1.788115338\) | ||
| \(x_3 = 1.772305156\) | ||
| \(x_4 = 1.772029085\) | ||
| \(x_5 = 1.7720288972\) | A1 | |
| Root \(\alpha \approx 1.772029\) | A1 | (Correct to 6 decimal places) |
### Part (a):
$f(x) = 1 + 5x - x^4$ | M1 | (Use of Intermediate Value Theorem.)
$f(1) = 5, f(2) = -5$ | A1 | (correct values and conclusions)
There is a change of sign indicating there is a root between 1 and 2. | |
### Part (b):
$x_{n+1} = \sqrt[4]{1 + 5x_n}$, $x_0 = 1.5$ | B1 |
$x_1 = 1.707476485$ | B1 |
$x_2 = 1.75734609$ | |
$x_3 = 1.7687213$, $x_4 = 1.7712854$ | B1 |
$x_5 = 1.7718619948$, $\alpha \approx 1.77$ | |
### Part (c):
$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{1 + 5x_n - x_n^4}{5 - 4x_n^3}$ | M1 | (Attempt to use Newton-Raphson. All terms correct)
$x_0 = 1.5$ | A1 |
$x_1 = 1.904411765$ | M1, A1 |
$x_2 = 1.788115338$ | |
$x_3 = 1.772305156$ | |
$x_4 = 1.772029085$ | |
$x_5 = 1.7720288972$ | A1 |
Root $\alpha \approx 1.772029$ | A1 | (Correct to 6 decimal places)
**Total: [11]**The equation
$$1 + 5x - x^4 = 0$$
has a positive root $\alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ lies between 1 and 2. [2]
\item Use the iterative sequence based on the arrangement
$$x = \sqrt[4]{1+5x}$$
with starting value 1.5 to find $\alpha$ correct to two decimal places. [3]
\item Use the Newton-Raphson method to find $\alpha$ correct to six decimal places. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 3 Q10 [15]}}