| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| 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 to find dy/dx, evaluate at a point, and find stationary points by setting dy/dx = 0. While it involves multiple steps, each is routine for C3 level with no novel problem-solving 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Differentiating implicitly: \((4y \pm 1)\frac{dy}{dx} = 18x\) | M1 | \((4y+1)\frac{dy}{dx} = \ldots\) allow \(4y + 1\frac{dy}{dx} = \ldots\); condone omitted bracket if intention implied by following line. \(4y\frac{dy}{dx} + 1\) M1 A0 |
| \(\Rightarrow \frac{dy}{dx} = \frac{18x}{4y+1}\) | A1 | |
| When \(x=1\), \(y=2\): \(\frac{dy}{dx} = \frac{18}{9} = 2\) | M1 A1cao [4] | Substituting \(x=1\), \(y=2\) into their derivative (provided it contains \(x\)'s and \(y\)'s). Allow unsupported answers. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 0\) when \(x = 0\) | B1 | \(x = 0\) from their numerator \(= 0\) (must have a denominator) |
| \(\Rightarrow 2y^2 + y = 1\) | M1 | Obtaining correct quadratic and attempt to factorise or use quadratic formula \(y = \frac{-1 \pm \sqrt{1-4\times -2}}{4}\) |
| \(\Rightarrow 2y^2 + y - 1 = 0\) | ||
| \(\Rightarrow (2y-1)(y+1) = 0\) | ||
| \(\Rightarrow y = \frac{1}{2}\) or \(y = -1\) | A1 A1 | cao; allow unsupported answers provided quadratic is shown |
| So coords are \((0, \frac{1}{2})\) and \((0, -1)\) | [4] |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Differentiating implicitly: $(4y \pm 1)\frac{dy}{dx} = 18x$ | M1 | $(4y+1)\frac{dy}{dx} = \ldots$ allow $4y + 1\frac{dy}{dx} = \ldots$; condone omitted bracket if intention implied by following line. $4y\frac{dy}{dx} + 1$ M1 A0 |
| $\Rightarrow \frac{dy}{dx} = \frac{18x}{4y+1}$ | A1 | |
| When $x=1$, $y=2$: $\frac{dy}{dx} = \frac{18}{9} = 2$ | M1 A1cao [4] | Substituting $x=1$, $y=2$ into their derivative (provided it contains $x$'s and $y$'s). Allow unsupported answers. |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 0$ when $x = 0$ | B1 | $x = 0$ from their numerator $= 0$ (must have a denominator) |
| $\Rightarrow 2y^2 + y = 1$ | M1 | Obtaining correct quadratic and attempt to factorise or use quadratic formula $y = \frac{-1 \pm \sqrt{1-4\times -2}}{4}$ |
| $\Rightarrow 2y^2 + y - 1 = 0$ | | |
| $\Rightarrow (2y-1)(y+1) = 0$ | | |
| $\Rightarrow y = \frac{1}{2}$ or $y = -1$ | A1 A1 | cao; allow unsupported answers provided quadratic is shown |
| So coords are $(0, \frac{1}{2})$ and $(0, -1)$ | [4] | |
---4 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 Q4 [8]}}