| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve exponential equation via iteration |
| Difficulty | Standard +0.3 This is a straightforward multi-part C3 question requiring standard techniques: algebraic rearrangement to find k, applying a given iterative formula (mechanical calculation), and differentiation with equation solving. All parts follow routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(2x^2 + 3\ln(2-x) = 0 \Rightarrow 3\ln(2-x) = -2x^2\) | ||
| \(\ln(2-x) = -\frac{2}{3}x^2\) | M1 | |
| \(2 - x = e^{-\frac{2}{3}x^2}\) | M1 | |
| \(x = 2 - e^{-\frac{2}{3}x^2}\) \([k = -\frac{2}{3}]\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(x_1 = 1.90988,\ x_2 = 1.91212,\ x_3 = 1.91262,\ x_4 = 1.91273,\ x_5 = 1.91275\) | M1 A1 | |
| \(\therefore \alpha = 1.913\) (3dp) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(f'(x) = 4x + \frac{3}{2-x} \times (-1) = 4x - \frac{3}{2-x}\) | M1 A1 | |
| \(\therefore 4x - \frac{3}{2-x} = 0,\quad 4x = \frac{3}{2-x},\quad 4x(2-x) = 3\) | M1 | |
| \(4x^2 - 8x + 3 = 0,\quad (2x-3)(2x-1) = 0\) | M1 | |
| \(x = \frac{1}{2},\ \frac{3}{2}\) | A1 | (11) |
# Question 6:
## Part (i)
| Answer | Mark | Notes |
|--------|------|-------|
| $2x^2 + 3\ln(2-x) = 0 \Rightarrow 3\ln(2-x) = -2x^2$ | | |
| $\ln(2-x) = -\frac{2}{3}x^2$ | M1 | |
| $2 - x = e^{-\frac{2}{3}x^2}$ | M1 | |
| $x = 2 - e^{-\frac{2}{3}x^2}$ $[k = -\frac{2}{3}]$ | A1 | |
## Part (ii)
| Answer | Mark | Notes |
|--------|------|-------|
| $x_1 = 1.90988,\ x_2 = 1.91212,\ x_3 = 1.91262,\ x_4 = 1.91273,\ x_5 = 1.91275$ | M1 A1 | |
| $\therefore \alpha = 1.913$ (3dp) | A1 | |
## Part (iii)
| Answer | Mark | Notes |
|--------|------|-------|
| $f'(x) = 4x + \frac{3}{2-x} \times (-1) = 4x - \frac{3}{2-x}$ | M1 A1 | |
| $\therefore 4x - \frac{3}{2-x} = 0,\quad 4x = \frac{3}{2-x},\quad 4x(2-x) = 3$ | M1 | |
| $4x^2 - 8x + 3 = 0,\quad (2x-3)(2x-1) = 0$ | M1 | |
| $x = \frac{1}{2},\ \frac{3}{2}$ | A1 | **(11)** |
---6. $\quad f ( x ) = 2 x ^ { 2 } + 3 \ln ( 2 - x ) , \quad x \in \mathbb { R } , \quad x < 2$.\\
(i) Show that the equation $\mathrm { f } ( x ) = 0$ can be written in the form
$$x = 2 - \mathrm { e } ^ { k x ^ { 2 } }$$
where $k$ is a constant to be found.
The root, $\alpha$, of the equation $\mathrm { f } ( x ) = 0$ is 1.9 correct to 1 decimal place.\\
(ii) Use the iterative formula
$$x _ { n + 1 } = 2 - \mathrm { e } ^ { k x _ { n } ^ { 2 } }$$
with $x _ { 0 } = 1.9$ and your value of $k$, to find $\alpha$ correct to 3 decimal places.\\
You should show the result of each iteration.\\
(iii) Solve the equation $\mathrm { f } ^ { \prime } ( x ) = 0$.\\
\hfill \mbox{\textit{OCR C3 Q6 [11]}}