| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.8 This question requires implicit differentiation (part a is routine), but part (b) demands solving a system where dy/dx=0 AND the original implicit equation holds simultaneously. Students must recognize that dy/dx=0 means the numerator equals zero, then substitute back into the constraint equation to find coordinates—a multi-step problem requiring strategic thinking beyond standard textbook exercises. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4x + 3x\frac{dy}{dx} + 3y + 2y\frac{dy}{dx} = 0\) | M1 | Attempt at implicit differentiation. Either \(3x\frac{dy}{dx} + 3y\) or \(2y\frac{dy}{dx}\) correct. Condone \(\frac{dy}{dx} =\) for M1 |
| A1 | All correct | |
| \(3x\frac{dy}{dx} + 2y\frac{dy}{dx} = -4x - 3y\) | ||
| \(\frac{dy}{dx} = -\frac{4x+3y}{3x+2y}\) | E1 | AG. At least one step of working needed to achieve convincing completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 0 \Rightarrow y = -\frac{4x}{3}\) or \(x = -\frac{3y}{4}\) | M1 | Finding \(y\) in terms of \(x\) (or vice versa). Condone sign errors for M1 |
| \(2x^2 - 4x^2 + \frac{16x^2}{9} + 2 = 0\) or \(\frac{9}{8}y^2 - \frac{9}{4}y^2 + y^2 + 2 = 0\) | M1 | Substitution into equation of curve to get equation in \(x\) (or \(y\)) |
| \(\frac{2x^2}{9} = 2 \Rightarrow x = \pm 3\) or \(-\frac{1}{8}y^2 = -2 \Rightarrow y = \pm 4\) | A1 | Both values – some working needed; nfww (eg sign error in first line) |
| \((3, -4)\) and \((-3, 4)\) | A1 | Both points as coordinates. Dep on M2A1. A0 if extra values |
# Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4x + 3x\frac{dy}{dx} + 3y + 2y\frac{dy}{dx} = 0$ | M1 | Attempt at implicit differentiation. Either $3x\frac{dy}{dx} + 3y$ or $2y\frac{dy}{dx}$ correct. Condone $\frac{dy}{dx} =$ for M1 |
| | A1 | All correct |
| $3x\frac{dy}{dx} + 2y\frac{dy}{dx} = -4x - 3y$ | | |
| $\frac{dy}{dx} = -\frac{4x+3y}{3x+2y}$ | E1 | AG. At least one step of working needed to achieve convincing completion |
---
# Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 0 \Rightarrow y = -\frac{4x}{3}$ or $x = -\frac{3y}{4}$ | M1 | Finding $y$ in terms of $x$ (or vice versa). Condone sign errors for M1 |
| $2x^2 - 4x^2 + \frac{16x^2}{9} + 2 = 0$ or $\frac{9}{8}y^2 - \frac{9}{4}y^2 + y^2 + 2 = 0$ | M1 | Substitution into equation of curve to get equation in $x$ (or $y$) |
| $\frac{2x^2}{9} = 2 \Rightarrow x = \pm 3$ or $-\frac{1}{8}y^2 = -2 \Rightarrow y = \pm 4$ | A1 | Both values – some working needed; nfww (eg sign error in first line) |
| $(3, -4)$ and $(-3, 4)$ | A1 | Both points as coordinates. Dep on M2A1. A0 if extra values |5 A curve is defined implicitly by the equation $2 x ^ { 2 } + 3 x y + y ^ { 2 } + 2 = 0$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { d y } { d x } = - \frac { 4 x + 3 y } { 3 x + 2 y }$.
\item In this question you must show detailed reasoning.
Find the coordinates of the stationary points of the curve.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2022 Q5 [7]}}