| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a straightforward fixed point iteration question requiring sign change verification, algebraic rearrangement (raising both sides to the fourth power), and applying a given iterative formula three times with a calculator. All steps are routine C3 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| \(x + (1 + 3x)^4 = 0\) | M1 | AWRT; allow +ve, −ve |
| \(f(-0.32) = 0.1\) | ||
| \(f(-0.33) = -0.01\) | ||
| Change of sign \(\therefore -0.33 < x < -0.32\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = -(1 + 3x)^{\frac{1}{4}}\) | M1 | Attempt to isolate \(x^4\) |
| \(x^4 = 1 + 3x\) | ||
| \(\frac{x^4 - 1}{3} = x\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_1 = -0.3\) | M1 | |
| \((x_2 = -0.331)\) AWRT | A1 | |
| \((x_3 = -0.329)\) AWRT | ||
| \(x_4 = -0.329\) | A1 |
### 3(a)
$x + (1 + 3x)^4 = 0$ | M1 | AWRT; allow +ve, −ve |
$f(-0.32) = 0.1$ | | |
$f(-0.33) = -0.01$ | | |
Change of sign $\therefore -0.33 < x < -0.32$ | A1 | | **Total: 2**
### 3(b)
$x = -(1 + 3x)^{\frac{1}{4}}$ | M1 | Attempt to isolate $x^4$ |
$x^4 = 1 + 3x$ | | |
$\frac{x^4 - 1}{3} = x$ | A1 | AG | **Total: 2**
### 3(c)
$x_1 = -0.3$ | M1 | |
$(x_2 = -0.331)$ AWRT | A1 | |
$(x_3 = -0.329)$ AWRT | | |
$x_4 = -0.329$ | A1 | | **Total: 3**3 The equation
$$x + ( 1 + 3 x ) ^ { \frac { 1 } { 4 } } = 0$$
has a single root, $\alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ lies between - 0.33 and - 0.32 .
\item Show that the equation $x + ( 1 + 3 x ) ^ { \frac { 1 } { 4 } } = 0$ can be rearranged into the form
$$x = \frac { 1 } { 3 } \left( x ^ { 4 } - 1 \right)$$
\item Use the iteration $x _ { n + 1 } = \frac { \left( x _ { n } ^ { 4 } - 1 \right) } { 3 }$ with $x _ { 1 } = - 0.3$ to find $x _ { 4 }$, giving your answer to three significant figures.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2008 Q3 [7]}}