| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question involving standard hyperbolic function manipulation, sketching familiar curves, algebraic rearrangement using the definition of sech x, and interpreting given iteration values. All parts are routine applications of FP2 techniques with no novel problem-solving required—slightly easier than average even for Further Maths. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams4.07b Hyperbolic graphs: sketch and properties |
| Answer | Marks |
|---|---|
| (i) For (either point) + (difference between vectors) | M1 |
| \(\mathbf{r} = (3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})\) or i + \(\mathbf{j}\) + 0\(\mathbf{k}\) or \(2\mathbf{i}\) – \(\mathbf{j}\) – \(\mathbf{k}\) | A1 |
| 'r' can be '+', '×' etc. Check other formats, e.g. \(ta + (1 - t)b\) | |
| 'r' must be 'r' but need not be bold | |
| (ii) State/imply their \(\mathbf{r}\) and their \(-2\mathbf{i} + \mathbf{j} + \mathbf{k}\) are perpendicular | *M1 |
| dep*M1 | dep*M1 |
| Consider scalar product = 0 | |
| Obtain \(t = -\frac{1}{6}\) or \(-\frac{1}{6}\) or \(-\frac{5}{6}\) or \(\frac{5}{6}\) | A1 |
| Subst their \(t\) into their equation of AB | M1 |
| Obtain \(\frac{1}{6}(16\mathbf{i} + 13\mathbf{j} + 19\mathbf{k})\) |
| | |
|---|---|
| (i) For (either point) + (difference between vectors) | M1 |
| $\mathbf{r} = (3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})$ or i + $\mathbf{j}$ + 0$\mathbf{k}$ or $2\mathbf{i}$ – $\mathbf{j}$ – $\mathbf{k}$ | A1 |
| 'r' can be '+', '×' etc. Check other formats, e.g. $ta + (1 - t)b$ | |
| 'r' must be 'r' but need not be bold | |
| (ii) State/imply their $\mathbf{r}$ and their $-2\mathbf{i} + \mathbf{j} + \mathbf{k}$ are perpendicular | *M1 |
| dep*M1 | dep*M1 |
| Consider scalar product = 0 | |
| Obtain $t = -\frac{1}{6}$ or $-\frac{1}{6}$ or $-\frac{5}{6}$ or $\frac{5}{6}$ | A1 |
| Subst their $t$ into their equation of AB | M1 |
| Obtain $\frac{1}{6}(16\mathbf{i} + 13\mathbf{j} + 19\mathbf{k})$4 (i) Sketch, on the same diagram, the curves with equations $y = \operatorname { sech } x$ and $y = x ^ { 2 }$.\\
(ii) By using the definition of $\operatorname { sech } x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, show that the $x$-coordinates of the points at which these curves meet are solutions of the equation
$$x ^ { 2 } = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 } .$$
(iii) The iteration
$$x _ { n + 1 } = \sqrt { \frac { 2 \mathrm { e } ^ { x _ { n } } } { \mathrm { e } ^ { 2 x _ { n } } + 1 } }$$
can be used to find the positive root of the equation in part (ii). With initial value $x _ { 1 } = 1$, the approximations $x _ { 2 } = 0.8050 , x _ { 3 } = 0.8633 , x _ { 4 } = 0.8463$ and $x _ { 5 } = 0.8513$ are obtained, correct to 4 decimal places. State with a reason whether, in this case, the iteration produces a 'staircase' or a 'cobweb' diagram.
\hfill \mbox{\textit{OCR FP2 2008 Q4 [8]}}