| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Show gradient expression then find coordinates |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring verification by substitution, then using the chain rule to find dy/dx, and solving a simple equation for stationary points. All steps are standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(r = \frac{a_{n+1}}{a_n}\) and \(r = \frac{a_n}{a_{n-1}}\) | M1 | For either ratio used |
| \(a_{n+1} = 2a_n + 3a_{n-1}\) | ||
| dividing through by \(a_n \Rightarrow r = 2 + \frac{3}{r}\) | M1 | |
| \(\Rightarrow r^2 - 2r - 3 = 0\) | A1 | |
| \(\Rightarrow (r-3)(r+1) = 0\) | ||
| \(\Rightarrow r = 3\) (discounting -1) | E1 [4] | SC B2 if dividing terms as far as \(a_n/a_s = a_0/a_0 = 3.00\) |
Let $r = \frac{a_{n+1}}{a_n}$ and $r = \frac{a_n}{a_{n-1}}$ | M1 | For either ratio used
$a_{n+1} = 2a_n + 3a_{n-1}$ | |
dividing through by $a_n \Rightarrow r = 2 + \frac{3}{r}$ | M1 |
$\Rightarrow r^2 - 2r - 3 = 0$ | A1 |
$\Rightarrow (r-3)(r+1) = 0$ | |
$\Rightarrow r = 3$ (discounting -1) | E1 [4] | SC B2 if dividing terms as far as $a_n/a_s = a_0/a_0 = 3.00$6 A curve has cartesian equation $y ^ { 2 } - x ^ { 2 } = 4$.\\
(i) Verify that
$$x = t - \frac { 1 } { t } , \quad y = t + \frac { 1 } { t } ,$$
are parametric equations of the curve.\\
(ii) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( t - 1 ) ( t + 1 ) } { t ^ { 2 } + 1 }$. Hence find the coordinates of the stationary points of the curve.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI C4 2005 Q6 [8]}}