| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Sketch graphs to show root existence |
| Difficulty | Standard +0.3 This is a standard C3 iterative methods question with routine curve sketching, straightforward iteration, and a transformation problem requiring careful but methodical application of rules. All parts follow predictable patterns with no novel insight required, making it slightly easier than the average A-level question. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.02w Graph transformations: simple transformations of f(x)1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch more or less correct \(y = \ln x\) | B1 | existing for positive and negative \(y\); no need to indicate \((1, 0)\); ignore any scales given on axes; condone graph touching \(y\)-axis but B0 if it crosses \(y\)-axis |
| Sketch more or less correct \(y = 8 - 2x^2\) | B1 | (roughly) symmetrical about \(y\)-axis; extending, if minimally, into quadrants for which \(y < 0\); no need to indicate \((\pm2, 0), (0, 8)\); assess each curve separately |
| Indicate intersection by some mark on diagram (just a 'blob' sufficient) or by statement in words away from diagram | B1 | needs each curve to be (more or less) correct in the first quadrant and on curves being related to each other correctly there |
| Answer | Marks | Guidance |
|---|---|---|
| Refer, in some way, to graphs crossing \(x\)-axis at \(x = 1\) and \(x = 2\) and that intersection is between these values | B1 | AG; the values 1 and 2 may be assumed from part (i) if clearly marked there; dependent on curves being (more or less) correct in first quadrant; carrying out the sign-change routine is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain correct first iterate | B1 | to at least 3 dp (except in the case of starting value 1 leading to 2) |
| Show correct iterative process | M1 | involving at least 3 iterates in all; may be implied by plausible converging values |
| Obtain at least 3 correct iterates | A1 | allowing recovery after error; iterates given to at least 3 dp; values may be rounded or truncated |
| Conclude with 1.917 | A1 | answer required to exactly 3 dp; answer only with no evidence of process is 0/4 |
| 1 → 2 → 1.91139 → 1.91731… → 1.91690… → 1.91693… | ||
| 1.5 → 1.94865… → 1.91479… → 1.91707… → 1.91692… | ||
| 2 → 1.91139… → 1.91731… → 1.91690… → 1.91693… |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain 3.92 or greater accuracy | B1√ | following their answer to part (iii) |
| Attempt \(4 \times \ln(\text{part (iii) answer})\) | M1 | |
| Obtain \(y\)-coordinate 2.60 | A1 | value required to exactly 2 dp (so A0 for 2.6 and 2.603) |
## (i)
Sketch more or less correct $y = \ln x$ | B1 | existing for positive and negative $y$; no need to indicate $(1, 0)$; ignore any scales given on axes; condone graph touching $y$-axis but B0 if it crosses $y$-axis
Sketch more or less correct $y = 8 - 2x^2$ | B1 | (roughly) symmetrical about $y$-axis; extending, if minimally, into quadrants for which $y < 0$; no need to indicate $(\pm2, 0), (0, 8)$; assess each curve separately
Indicate intersection by some mark on diagram (just a 'blob' sufficient) or by statement in words away from diagram | B1 | needs each curve to be (more or less) correct in the first quadrant and on curves being related to each other correctly there
**Total: [3]**
## (ii)
Refer, in some way, to graphs crossing $x$-axis at $x = 1$ and $x = 2$ and that intersection is between these values | B1 | AG; the values 1 and 2 may be assumed from part (i) if clearly marked there; dependent on curves being (more or less) correct in first quadrant; carrying out the sign-change routine is B0
**Total: [1]**
## (iii)
Obtain correct first iterate | B1 | to at least 3 dp (except in the case of starting value 1 leading to 2)
Show correct iterative process | M1 | involving at least 3 iterates in all; may be implied by plausible converging values
Obtain at least 3 correct iterates | A1 | allowing recovery after error; iterates given to at least 3 dp; values may be rounded or truncated
Conclude with 1.917 | A1 | answer required to exactly 3 dp; answer only with no evidence of process is 0/4
| 1 → 2 → 1.91139 → 1.91731… → 1.91690… → 1.91693… |
| 1.5 → 1.94865… → 1.91479… → 1.91707… → 1.91692… |
| 2 → 1.91139… → 1.91731… → 1.91690… → 1.91693… |
**Total: [4]**
## (iv)
Obtain 3.92 or greater accuracy | B1√ | following their answer to part (iii)
Attempt $4 \times \ln(\text{part (iii) answer})$ | M1 |
Obtain $y$-coordinate 2.60 | A1 | value required to exactly 2 dp (so A0 for 2.6 and 2.603)
**Total: [3]**
---\begin{enumerate}[label=(\roman*)]
\item By sketching the curves $y = \ln x$ and $y = 8 - 2x^2$ on a single diagram, show that the equation
$$\ln x = 8 - 2x^2$$
has exactly one real root. [3]
\item Explain how your diagram shows that the root is between 1 and 2. [1]
\item Use the iterative formula
$$x_{n+1} = \sqrt{4 - \frac{1}{2}\ln x_n},$$
with a suitable starting value, to find the root. Show all your working and give the root correct to 3 decimal places. [4]
\item The curves $y = \ln x$ and $y = 8 - 2x^2$ are each translated by 2 units in the positive $x$-direction and then stretched by scale factor 4 in the $y$-direction. Find the coordinates of the point where the new curves intersect, giving each coordinate correct to 2 decimal places. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2013 Q6 [11]}}