| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a slightly above-average C3 question combining logarithms, iteration, and differentiation. Part (a) requires straightforward algebraic manipulation of logarithms (routine). Part (b) is standard iterative root-finding with accuracy justification (textbook exercise). Part (c) involves differentiating a composite function and solving, which is standard C3 fare but requires careful algebra. The question tests multiple techniques but follows predictable patterns without requiring novel insight. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.07q Product and quotient rules: differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(2x^2 + 3\ln(2-x) = 0 \Rightarrow 3\ln(2-x) = -2x^2\) | M1 | |
| \(\ln(2-x) = -\frac{2}{3}x^2\) | M1 | |
| \(2 - x = e^{-\frac{2}{3}x^2}\) | M1 | |
| \(x = 2 - e^{-\frac{2}{3}x^2}\) | A1 | \([k = -\frac{2}{3}]\) |
| (b) \(x_1 = 1.90988\), \(x_2 = 1.91212\), \(x_3 = 1.91262\), \(x_4 = 1.91273\) | M1 A1 | |
| \(\therefore \alpha = 1.913\) (3dp) | A1 | |
| \(f(1.91125) = 0.0070\), \(f(1.91135) = -0.020\) | M1 | |
| sign change, \(f(x)\) continuous ∴ root | A1 | |
| (c) \(f'(x) = 4x + \frac{3}{2-x} \times (-1) = 4x - \frac{3}{2-x}\) | M1 A1 | |
| \(\therefore 4x - \frac{3}{2-x} = 0\), \(4x = \frac{3}{2-x}\), \(4x(2-x) = 3\) | M1 | |
| \(4x^2 - 8x + 3 = 0\), \((2x-3)(2x-1) = 0\) | M1 | |
| \(x = \frac{1}{2}, \frac{3}{2}\) | A1 | (13) |
(a) $2x^2 + 3\ln(2-x) = 0 \Rightarrow 3\ln(2-x) = -2x^2$ | M1 |
$\ln(2-x) = -\frac{2}{3}x^2$ | M1 |
$2 - x = e^{-\frac{2}{3}x^2}$ | M1 |
$x = 2 - e^{-\frac{2}{3}x^2}$ | A1 | $[k = -\frac{2}{3}]$ |
(b) $x_1 = 1.90988$, $x_2 = 1.91212$, $x_3 = 1.91262$, $x_4 = 1.91273$ | M1 A1 |
$\therefore \alpha = 1.913$ (3dp) | A1 |
$f(1.91125) = 0.0070$, $f(1.91135) = -0.020$ | M1 |
sign change, $f(x)$ continuous ∴ root | A1 |
(c) $f'(x) = 4x + \frac{3}{2-x} \times (-1) = 4x - \frac{3}{2-x}$ | M1 A1 |
$\therefore 4x - \frac{3}{2-x} = 0$, $4x = \frac{3}{2-x}$, $4x(2-x) = 3$ | M1 |
$4x^2 - 8x + 3 = 0$, $(2x-3)(2x-1) = 0$ | M1 |
$x = \frac{1}{2}, \frac{3}{2}$ | A1 | **(13)**$$f(x) = 2x^2 + 3 \ln (2 - x), \quad x \in \mathbb{R}, \quad x < 2.$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $f(x) = 0$ can be written in the form
$$x = 2 - e^{kx^2},$$
where $k$ is a constant to be found. [3]
\end{enumerate}
The root, $\alpha$, of the equation $f(x) = 0$ is $1.9$ correct to $1$ decimal place.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Use the iteration formula
$$x_{n+1} = 2 - e^{kx_n^2},$$
with $x_0 = 1.9$ and your value of $k$, to find $\alpha$ to $3$ decimal places and justify the accuracy of your answer. [5]
\item Solve the equation $f'(x) = 0$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q6 [13]}}