| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Session | Specimen |
| 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 standard implicit differentiation question requiring application of the product and chain rules, followed by solving a system of equations to find stationary points. While it involves multiple steps and algebraic manipulation, the techniques are routine for C4 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(4x + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\) | B1 | For correct terms \(x\frac{dy}{dx} + y\) |
| B1 | For correct term \(2y\frac{dy}{dx}\) | |
| M1 | For solving for \(\frac{dy}{dx}\) | |
| Hence \(\frac{dy}{dx} = \frac{-4x-y}{x+2y}\) | A1 | For any correct form of expression |
| 4 | ||
| (ii) \(\frac{dy}{dx} = 0 \Rightarrow y = -4x\) | M1 | For stating or using their \(\frac{dy}{dx} = 0\) |
| Hence \(2x^2 + (-4x^2) + (-4x)^2 = 14\) | M1 | For solving simultaneously with curve equation |
| i.e. \(x^2 = 1\) | A1 | For correct value of \(x^2\) (or \(y^2\)) |
| So the two points are \((1, -4)\) and \((-1, 4)\) | A1 | For both correct points identified |
| 4 |
**(i)** $4x + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$ | B1 | For correct terms $x\frac{dy}{dx} + y$
| B1 | For correct term $2y\frac{dy}{dx}$
| M1 | For solving for $\frac{dy}{dx}$
Hence $\frac{dy}{dx} = \frac{-4x-y}{x+2y}$ | A1 | For any correct form of expression
| **4** |
**(ii)** $\frac{dy}{dx} = 0 \Rightarrow y = -4x$ | M1 | For stating or using their $\frac{dy}{dx} = 0$
Hence $2x^2 + (-4x^2) + (-4x)^2 = 14$ | M1 | For solving simultaneously with curve equation
i.e. $x^2 = 1$ | A1 | For correct value of $x^2$ (or $y^2$)
So the two points are $(1, -4)$ and $(-1, 4)$ | A1 | For both correct points identified
| **4** |5 (i) For the curve $2 x ^ { 2 } + x y + y ^ { 2 } = 14$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.\\
(ii) Deduce that there are two points on the curve $2 x ^ { 2 } + x y + y ^ { 2 } = 14$ at which the tangents are parallel to the $x$-axis, and find their coordinates.
\hfill \mbox{\textit{OCR C4 Q5 [8]}}