| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| 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 to find dy/dx, setting it to zero for stationary points, then substituting back into the original equation. While the techniques are standard A-level, the algebraic manipulation needed to arrive at the given cubic form and the multi-step nature (differentiate implicitly, solve for dy/dx=0, substitute back) makes this moderately challenging, above average difficulty. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Differentiates \(x^2\) to obtain \(2x\) | B1 (1.1b) | |
| Uses implicit differentiation obtaining either \(Ay^2\frac{dy}{dx}\) or \(Bx\frac{dy}{dx}\) terms | M1 (3.1a) | |
| Uses product rule to obtain \(\pm 4y \pm 4x\frac{dy}{dx}\) | M1 (1.1a) | Condone sign errors |
| \(2x + 6y^2\frac{dy}{dx} - 4y - 4x\frac{dy}{dx} = 0\) | A1 (1.1b) | OE |
| Substitutes \(\frac{dy}{dx} = 0\) into differentiated expression | M1 (1.1a) | Expression must contain either \(Ay^2\frac{dy}{dx}\) or \(Bx\frac{dy}{dx}\); PI by later work |
| Deduces \(x = 2y\) or \(2x = 4y\) or \(-x = -2y\) or \(-2x = -4y\) | R1 (2.2a) | Must have scored A1 with no incorrect rearrangement of \(2x + 6y^2\frac{dy}{dx} - 4y - 4x\frac{dy}{dx} = 0\) used |
| Eliminates \(x\): \((2y)^2 + 2y^3 - 4(2y)y = 0 \Rightarrow 4y^2 + 2y^3 - 8y^2 = 0 \Rightarrow 2y^3 - 4y^2 = 0 \Rightarrow y^2(y-2) = 0\) | R1 (2.1) | Eliminates \(x\) and completes reasoned argument with at least one intermediate step; must have scored first 6 marks with no incorrect rearrangement used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 2\) | B1 (1.1b) | Obtains \(y\)-coordinate of 2; accept \(y = 2\) |
| \(x = 4\) | B1 (1.1b) | Obtains \(x\)-coordinate of 4; accept \(x = 4\) |
## Question 15(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Differentiates $x^2$ to obtain $2x$ | B1 (1.1b) | |
| Uses implicit differentiation obtaining either $Ay^2\frac{dy}{dx}$ or $Bx\frac{dy}{dx}$ terms | M1 (3.1a) | |
| Uses product rule to obtain $\pm 4y \pm 4x\frac{dy}{dx}$ | M1 (1.1a) | Condone sign errors |
| $2x + 6y^2\frac{dy}{dx} - 4y - 4x\frac{dy}{dx} = 0$ | A1 (1.1b) | OE |
| Substitutes $\frac{dy}{dx} = 0$ into differentiated expression | M1 (1.1a) | Expression must contain either $Ay^2\frac{dy}{dx}$ or $Bx\frac{dy}{dx}$; PI by later work |
| Deduces $x = 2y$ or $2x = 4y$ or $-x = -2y$ or $-2x = -4y$ | R1 (2.2a) | Must have scored A1 with no incorrect rearrangement of $2x + 6y^2\frac{dy}{dx} - 4y - 4x\frac{dy}{dx} = 0$ used |
| Eliminates $x$: $(2y)^2 + 2y^3 - 4(2y)y = 0 \Rightarrow 4y^2 + 2y^3 - 8y^2 = 0 \Rightarrow 2y^3 - 4y^2 = 0 \Rightarrow y^2(y-2) = 0$ | R1 (2.1) | Eliminates $x$ and completes reasoned argument with at least one intermediate step; must have scored first 6 marks with no incorrect rearrangement used |
**Subtotal: 7 marks**
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## Question 15(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 2$ | B1 (1.1b) | Obtains $y$-coordinate of 2; accept $y = 2$ |
| $x = 4$ | B1 (1.1b) | Obtains $x$-coordinate of 4; accept $x = 4$ |
**Subtotal: 2 marks**
**Question 15 Total: 9 marks**
---15 The curve with equation
$$x ^ { 2 } + 2 y ^ { 3 } - 4 x y = 0$$
has a single stationary point at $P$ as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-26_656_1138_548_450}
15
\begin{enumerate}[label=(\alph*)]
\item Show that the $y$-coordinate of $P$ satisfies the equation
$$y ^ { 2 } ( y - 2 ) = 0$$
15
\item Hence, find the coordinates of $P$\\[0pt]
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2023 Q15 [9]}}