| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show root in interval |
| Difficulty | Standard +0.3 This is a standard C3 numerical methods question involving sign change verification, iterative formula application, and algebraic manipulation using trig identities. Part (i) is routine substitution, part (ii) is straightforward iteration (given the formula), and part (iii) requires recognizing the relationship between the equations using sec²(2x) = 1 + tan²(2x). While it tests multiple techniques, all are standard C3 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt calculations involving 1.0 and 1.1 | M1 | using radians or values to 1 dp (rounded or truncated); or equivs (where eqn rearranged) |
| Obtain \(-0.57\) and \(0.76\) | A1 | — |
| Refer to sign change (or equiv for rearranged eqn) | A1 3 AG | following correct work only |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain correct first iterate | B1 | using value \(x_1\) such that \(1.0 \le x_1 \le 1.1\) |
| Carry out iteration process | M1 | obtaining at least 3 iterates in all so far |
| Obtain at least 3 correct iterates | A1 | showing at least 3 dp |
| Obtain 1.050830 | A1 4 | answer required to exactly 5 d.p. |
| Answer | Marks | Guidance |
|---|---|---|
| State or imply \(\sec^2 2x = 1 + \tan^2 2x\) | B1 | — |
| Relate to earlier equation | M1 | by halving or doubling answer to (ii) or carrying out equivalent iteration process |
| Deduce \(2x = 1.050830\) and hence \(0.525\) | A1\N 3 | following their answer to (ii); or greater accuracy |
| [SC: Rearrange to obtain \(x = \frac{1}{2}\cos^{-1}(2x + 3)^{-1}\) | B1 | — |
| Use iterative process to obtain 0.525 | B1 2 or greater accuracy] |
## (i)
Attempt calculations involving 1.0 and 1.1 | M1 | using radians or values to 1 dp (rounded or truncated); or equivs (where eqn rearranged)
Obtain $-0.57$ and $0.76$ | A1 | —
Refer to sign change (or equiv for rearranged eqn) | A1 3 AG | following correct work only
## (ii)
Obtain correct first iterate | B1 | using value $x_1$ such that $1.0 \le x_1 \le 1.1$
Carry out iteration process | M1 | obtaining at least 3 iterates in all so far
Obtain at least 3 correct iterates | A1 | showing at least 3 dp
Obtain 1.050830 | A1 4 | answer required to exactly 5 d.p.
[1 → 1.047198 → 1.050571 → 1.050809 → 1.050826 → 1.050827;
1.05 → 1.050769 → 1.050823 → 1.050827 → 1.050827;
1.1 → 1.054268 → 1.051070 → 1.050844 → 1.050829 → 1.050827]
## (iii)
State or imply $\sec^2 2x = 1 + \tan^2 2x$ | B1 | —
Relate to earlier equation | M1 | by halving or doubling answer to (ii) or carrying out equivalent iteration process
Deduce $2x = 1.050830$ and hence $0.525$ | A1\N 3 | following their answer to (ii); or greater accuracy
[SC: Rearrange to obtain $x = \frac{1}{2}\cos^{-1}(2x + 3)^{-1}$ | B1 | —
Use iterative process to obtain 0.525 | B1 2 or greater accuracy]
---\begin{enumerate}[label=(\roman*)]
\item Show by calculation that the equation
$$\tan^2 x - x - 2 = 0,$$
where $x$ is measured in radians, has a root between 1.0 and 1.1. [3]
\item Use the iteration formula $x_{n+1} = \tan^{-1}\sqrt{2 + x_n}$ with a suitable starting value to find this root correct to 5 decimal places. You should show the outcome of each step of the process. [4]
\item Deduce a root of the equation
$$\sec^2 2x - 2x - 3 = 0.$$ [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2010 Q6 [10]}}