| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard technique application. Part (i) involves routine differentiation and substitution, while part (ii) requires solving simultaneous equations to find stationary points—slightly above average due to the algebraic manipulation needed, but still a standard C3 exercise with no novel insight required. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(4y\frac{dy}{dx} + \frac{dy}{dx} = 18x\) | M1 | Implicit differentiation |
| \(\frac{dy}{dx}(4y+1) = 18x\) | A1 | Correct differentiation |
| \(\frac{dy}{dx} = \frac{18x}{4y+1}\) | A1 | Correct expression |
| At \((1,2)\): \(\frac{dy}{dx} = \frac{18}{9} = 2\) | A1 | [4] Correct gradient |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 0 \Rightarrow 18x = 0 \Rightarrow x = 0\) | M1 | Setting numerator to zero |
| Substituting \(x=0\): \(2y^2 + y = 1\) | M1 | Substituting back into curve equation |
| \(2y^2 + y - 1 = 0 \Rightarrow (2y-1)(y+1)=0 \Rightarrow y = \frac{1}{2}\) or \(y = -1\) | A1 | Correct values of \(y\) |
| \((0, \frac{1}{2})\) and \((0, -1)\) | A1 | [4] Both coordinates correct |
# Question 3:
**Part (i):**
| $4y\frac{dy}{dx} + \frac{dy}{dx} = 18x$ | M1 | Implicit differentiation |
| $\frac{dy}{dx}(4y+1) = 18x$ | A1 | Correct differentiation |
| $\frac{dy}{dx} = \frac{18x}{4y+1}$ | A1 | Correct expression |
| At $(1,2)$: $\frac{dy}{dx} = \frac{18}{9} = 2$ | A1 | [4] Correct gradient |
**Part (ii):**
| $\frac{dy}{dx} = 0 \Rightarrow 18x = 0 \Rightarrow x = 0$ | M1 | Setting numerator to zero |
| Substituting $x=0$: $2y^2 + y = 1$ | M1 | Substituting back into curve equation |
| $2y^2 + y - 1 = 0 \Rightarrow (2y-1)(y+1)=0 \Rightarrow y = \frac{1}{2}$ or $y = -1$ | A1 | Correct values of $y$ |
| $(0, \frac{1}{2})$ and $(0, -1)$ | A1 | [4] Both coordinates correct |
---3 A curve has equation $2 y ^ { 2 } + y = 9 x ^ { 2 } + 1$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$. Hence find the gradient of the curve at the point $\mathrm { A } ( 1,2 )$.\\
(ii) Find the coordinates of the points on the curve at which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$.
\hfill \mbox{\textit{OCR MEI C3 2007 Q3 [8]}}