| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a standard C3 question covering routine differentiation, tangent equations, sign-change method, and fixed-point iteration. All parts follow predictable patterns with no novel insight required—slightly easier than average due to the structured, step-by-step guidance through each technique. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(\frac{dy}{dx} = 2x - \frac{1}{2}(4 + \ln x)^{-\frac{1}{2}} \times \frac{1}{x} = 2x - \frac{1}{2x\sqrt{4+\ln x}}\) | M1 A1 | |
| \(x = 1\), \(y = -1\), \(\text{grad} = \frac{7}{4}\) | A1 | |
| \(y + 1 = \frac{7}{4}(x-1)\) | M1 | |
| \(7x - 4y = 11\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| SP: \(2x - \frac{1}{2x\sqrt{4+\ln x}} = 0\) | M1 | |
| \(f(0.3) = -0.40\), \(f(0.4) = 0.088\) | M1 | |
| Sign change, \(f(x)\) continuous \(\therefore\) root | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(2x - \frac{1}{2x\sqrt{4+\ln x}} = 0 \Rightarrow 2x = \frac{1}{2x\sqrt{4+\ln x}}\) | ||
| \(x^2 = \frac{1}{4\sqrt{4+\ln x}} = \frac{1}{4}(4+\ln x)^{-\frac{1}{2}}\) | M1 | |
| \(x = \sqrt{\frac{1}{4}(4+\ln x)^{-\frac{1}{2}}} = \frac{1}{2}(4+\ln x)^{-\frac{1}{4}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(x_1 = 0.381512\), \(x_2 = 0.378775\), \(x_3 = 0.378999\) | M1 A1 | |
| \(x_4 = 0.378981\), \(x_5 = 0.378982\), \(\therefore \alpha = 0.37898\) (5dp) | A1 | (13) |
# Question 8:
## Part (i):
| Working/Answer | Mark | Notes |
|---|---|---|
| $\frac{dy}{dx} = 2x - \frac{1}{2}(4 + \ln x)^{-\frac{1}{2}} \times \frac{1}{x} = 2x - \frac{1}{2x\sqrt{4+\ln x}}$ | M1 A1 | |
| $x = 1$, $y = -1$, $\text{grad} = \frac{7}{4}$ | A1 | |
| $y + 1 = \frac{7}{4}(x-1)$ | M1 | |
| $7x - 4y = 11$ | A1 | |
## Part (ii):
| Working/Answer | Mark | Notes |
|---|---|---|
| SP: $2x - \frac{1}{2x\sqrt{4+\ln x}} = 0$ | M1 | |
| $f(0.3) = -0.40$, $f(0.4) = 0.088$ | M1 | |
| Sign change, $f(x)$ continuous $\therefore$ root | A1 | |
## Part (iii):
| Working/Answer | Mark | Notes |
|---|---|---|
| $2x - \frac{1}{2x\sqrt{4+\ln x}} = 0 \Rightarrow 2x = \frac{1}{2x\sqrt{4+\ln x}}$ | | |
| $x^2 = \frac{1}{4\sqrt{4+\ln x}} = \frac{1}{4}(4+\ln x)^{-\frac{1}{2}}$ | M1 | |
| $x = \sqrt{\frac{1}{4}(4+\ln x)^{-\frac{1}{2}}} = \frac{1}{2}(4+\ln x)^{-\frac{1}{4}}$ | A1 | |
## Part (iv):
| Working/Answer | Mark | Notes |
|---|---|---|
| $x_1 = 0.381512$, $x_2 = 0.378775$, $x_3 = 0.378999$ | M1 A1 | |
| $x_4 = 0.378981$, $x_5 = 0.378982$, $\therefore \alpha = 0.37898$ (5dp) | A1 | **(13)** |
**Total: (72)**\begin{enumerate}
\item A curve has the equation $y = x ^ { 2 } - \sqrt { 4 + \ln x }$.\\
(i) Show that the tangent to the curve at the point where $x = 1$ has the equation
\end{enumerate}
$$7 x - 4 y = 11$$
The curve has a stationary point with $x$-coordinate $\alpha$.\\
(ii) Show that $0.3 < \alpha < 0.4$\\
(iii) Show that $\alpha$ is a solution of the equation
$$x = \frac { 1 } { 2 } ( 4 + \ln x ) ^ { - \frac { 1 } { 4 } }$$
(iv) Use the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \left( 4 + \ln x _ { n } \right) ^ { - \frac { 1 } { 4 } }$$
with $x _ { 0 } = 0.35$, to find $\alpha$ correct to 5 decimal places.\\
You should show the result of each iteration.
\hfill \mbox{\textit{OCR C3 Q8 [13]}}