| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find normal equation at point |
| Difficulty | Moderate -0.3 This question requires implicit differentiation (a standard A-level technique) and finding a normal equation, but follows a routine procedure: differentiate implicitly, substitute the point P (found in part a), take negative reciprocal for normal gradient, and write in required form. The implicit differentiation of (x+y)² is straightforward, and part (a) has already established the gradient. Slightly easier than average due to being a direct application of learned techniques with no problem-solving insight required. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| 12 | Find the equation of the normal to the curve at \(P\), giving your answer in the form \(a x + b y = c\), where \(a , b\) and \(c\) are integers. |
| [2 marks] | |
| \(\_\_\_\_\) | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitutes \(y = 0\): \(x^2 = 8 + 2x \Rightarrow x = 4\) or \(x = -2\); \(x = 4\) at P | M1, A1 (AO 3.1a, 1.1b) | PI \(x = 4\). Obtain \(x = 4\), ignore any other value |
| Differentiates implicitly: \(2x + 2y + 2x\frac{dy}{dx} + 2y\frac{dy}{dx} = 4\frac{dy}{dx} + 2\) | M1 (AO 3.1a) | Expands and uses product rule for \(Axy\) term; or uses chain rule to obtain \(2(x+y)\left(1+\frac{dy}{dx}\right)\). Condone missing brackets |
| Correctly differentiates \(4y\) or \(y^2\) terms | B1 (AO 1.1b) | |
| Correct equation from correct differentiation | A1 (AO 1.1b) | |
| Substitutes \(x = 4\), \(y = 0\): \(8 + 8\frac{dy}{dx} = 4\frac{dy}{dx} + 2 \Rightarrow 4\frac{dy}{dx} = -6 \Rightarrow \frac{dy}{dx} = -\frac{3}{2}\) | R1 (AO 2.1) | Must substitute \(x=4\) and \(y=0\) into \(\frac{dy}{dx} = \frac{2-2x-2y}{2x+2y-4}\) and obtain \(-\frac{3}{2}\). AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses gradient \(\frac{2}{3}\) (negative reciprocal) and \(y = 0\), \(x = 4\): \(y = \frac{2}{3}(x-4)\) | M1 (AO 1.1a) | |
| \(2x - 3y = 8\) | A1F (AO 1.1b) | Obtains equation in correct form |
## Question 12(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitutes $y = 0$: $x^2 = 8 + 2x \Rightarrow x = 4$ or $x = -2$; $x = 4$ at P | M1, A1 (AO 3.1a, 1.1b) | PI $x = 4$. Obtain $x = 4$, ignore any other value |
| Differentiates implicitly: $2x + 2y + 2x\frac{dy}{dx} + 2y\frac{dy}{dx} = 4\frac{dy}{dx} + 2$ | M1 (AO 3.1a) | Expands and uses product rule for $Axy$ term; or uses chain rule to obtain $2(x+y)\left(1+\frac{dy}{dx}\right)$. Condone missing brackets |
| Correctly differentiates $4y$ or $y^2$ terms | B1 (AO 1.1b) | |
| Correct equation from correct differentiation | A1 (AO 1.1b) | |
| Substitutes $x = 4$, $y = 0$: $8 + 8\frac{dy}{dx} = 4\frac{dy}{dx} + 2 \Rightarrow 4\frac{dy}{dx} = -6 \Rightarrow \frac{dy}{dx} = -\frac{3}{2}$ | R1 (AO 2.1) | Must substitute $x=4$ and $y=0$ into $\frac{dy}{dx} = \frac{2-2x-2y}{2x+2y-4}$ and obtain $-\frac{3}{2}$. AG |
**Subtotal: 6 marks**
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## Question 12(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses gradient $\frac{2}{3}$ (negative reciprocal) and $y = 0$, $x = 4$: $y = \frac{2}{3}(x-4)$ | M1 (AO 1.1a) | |
| $2x - 3y = 8$ | A1F (AO 1.1b) | Obtains equation in correct form |
**Subtotal: 2 marks**
---12 The equation of a curve is
$$( x + y ) ^ { 2 } = 4 y + 2 x + 8$$
The curve intersects the positive $x$-axis at the point $P$.\\
12
\begin{enumerate}[label=(\alph*)]
\item Show that the gradient of the curve at $P$ is $- \frac { 3 } { 2 }$\\
\begin{center}
\begin{tabular}{|l|l|}
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12
\item & Find the equation of the normal to the curve at $P$, giving your answer in the form $a x + b y = c$, where $a , b$ and $c$ are integers. \\
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& [2 marks] \\
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\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2021 Q12 [8]}}