| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question with routine steps: substituting a point to find a constant, applying product and chain rules systematically, then using the derivative at a point. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\sqrt{3})^3 \sin\frac{\pi}{6} + \cos\frac{\pi}{6} = A\sqrt{3}\) | M1 | Substitutes \(x = \sqrt{3}\) and \(y = \frac{\pi}{6}\) to obtain an equation or expression for \(A\) |
| \(\frac{3\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = A\sqrt{3}\), \(\frac{3}{2} + \frac{1}{2} = A\), \(A = 2\) | R1 | Completes argument to show \(A = 2\); must clearly show use of \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(\sin\frac{\pi}{6} = \frac{1}{2}\); AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3x^2 \sin y + x^3 \cos y \frac{dy}{dx} - \sin y \frac{dy}{dx} = 2\) | M1 | Uses implicit differentiation correctly at least once with sight of \(\sin y \frac{dy}{dx}\) or \(\cos y \frac{dy}{dx}\); condone sign error |
| \(\frac{dy}{dx}(x^3 \cos y - \sin y) = 2 - 3x^2 \sin y\) | M1 | Uses product rule with sight of \(Px^2 \sin y \pm x^3 \cos y \frac{dy}{dx}\); condone omission of \(\frac{dy}{dx}\) |
| \(Px^2 \sin y \pm x^3 \cos y \frac{dy}{dx} \pm \sin y \frac{dy}{dx} = 2\) | A1 | Obtains equation of this form |
| Completely correct equation | A1 | |
| \(\frac{dy}{dx} = \frac{2 - 3x^2 \sin y}{x^3 \cos y - \sin y}\) | R1 | Isolates \(\frac{dy}{dx}\) terms and factorises to complete rigorous argument with no slips to show the given result; AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{2 - 3(\sqrt{3})^2 \sin\frac{\pi}{6}}{(\sqrt{3})^3 \cos\frac{\pi}{6} - \sin\frac{\pi}{6}}\) | M1 | Substitutes \(x = \sqrt{3}\) and \(y = \frac{\pi}{6}\) to obtain expression for gradient |
| \(= -\frac{5}{8}\) | A1 | Obtains correct gradient; OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y - \frac{\pi}{6} = -\frac{5}{8}(x - \sqrt{3})\) | M1 | Forms equation for tangent (condone normal) at P using their gradient and \(\left(\sqrt{3}, \frac{\pi}{6}\right)\); ACF; or writes equation as \(y = mx + c\) using their gradient (condone normal) and substitutes \(\left(\sqrt{3}, \frac{\pi}{6}\right)\) to obtain equation in \(c\); PI by correct exact value for \(x\) |
| Fully correct tangent equation, note \(c = \frac{5\sqrt{3}}{8} + \frac{\pi}{6}\) or \(c = 1.606\ldots\) | A1F | Follow through their gradient of tangent from 12(b)(ii); must be to at least 3 dp |
| \(0 - \frac{\pi}{6} = -\frac{5}{8}(x - \sqrt{3})\) | M1 | Substitutes \(y = 0\) into their tangent (condone normal) equation and solves to find \(x\) coordinate of Q; accept decimals |
| \(x = \sqrt{3} + \frac{4\pi}{15}\) | A1 | OE must be exact form; e.g. \(x = \frac{8}{5}\left(\frac{5\sqrt{3}}{8} + \frac{\pi}{6}\right)\) |
## Question 12(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\sqrt{3})^3 \sin\frac{\pi}{6} + \cos\frac{\pi}{6} = A\sqrt{3}$ | M1 | Substitutes $x = \sqrt{3}$ and $y = \frac{\pi}{6}$ to obtain an equation or expression for $A$ |
| $\frac{3\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = A\sqrt{3}$, $\frac{3}{2} + \frac{1}{2} = A$, $A = 2$ | R1 | Completes argument to show $A = 2$; must clearly show use of $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6} = \frac{1}{2}$; AG |
---
## Question 12(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2 \sin y + x^3 \cos y \frac{dy}{dx} - \sin y \frac{dy}{dx} = 2$ | M1 | Uses implicit differentiation correctly at least once with sight of $\sin y \frac{dy}{dx}$ or $\cos y \frac{dy}{dx}$; condone sign error |
| $\frac{dy}{dx}(x^3 \cos y - \sin y) = 2 - 3x^2 \sin y$ | M1 | Uses product rule with sight of $Px^2 \sin y \pm x^3 \cos y \frac{dy}{dx}$; condone omission of $\frac{dy}{dx}$ |
| $Px^2 \sin y \pm x^3 \cos y \frac{dy}{dx} \pm \sin y \frac{dy}{dx} = 2$ | A1 | Obtains equation of this form |
| Completely correct equation | A1 | |
| $\frac{dy}{dx} = \frac{2 - 3x^2 \sin y}{x^3 \cos y - \sin y}$ | R1 | Isolates $\frac{dy}{dx}$ terms and factorises to complete rigorous argument with no slips to show the given result; AG |
---
## Question 12(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{2 - 3(\sqrt{3})^2 \sin\frac{\pi}{6}}{(\sqrt{3})^3 \cos\frac{\pi}{6} - \sin\frac{\pi}{6}}$ | M1 | Substitutes $x = \sqrt{3}$ and $y = \frac{\pi}{6}$ to obtain expression for gradient |
| $= -\frac{5}{8}$ | A1 | Obtains correct gradient; OE |
---
## Question 12(b)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y - \frac{\pi}{6} = -\frac{5}{8}(x - \sqrt{3})$ | M1 | Forms equation for tangent (condone normal) at P using their gradient and $\left(\sqrt{3}, \frac{\pi}{6}\right)$; ACF; or writes equation as $y = mx + c$ using their gradient (condone normal) and substitutes $\left(\sqrt{3}, \frac{\pi}{6}\right)$ to obtain equation in $c$; PI by correct exact value for $x$ |
| Fully correct tangent equation, note $c = \frac{5\sqrt{3}}{8} + \frac{\pi}{6}$ or $c = 1.606\ldots$ | A1F | Follow through their gradient of tangent from 12(b)(ii); must be to at least 3 dp |
| $0 - \frac{\pi}{6} = -\frac{5}{8}(x - \sqrt{3})$ | M1 | Substitutes $y = 0$ into their tangent (condone normal) equation and solves to find $x$ coordinate of Q; accept decimals |
| $x = \sqrt{3} + \frac{4\pi}{15}$ | A1 | OE must be exact form; e.g. $x = \frac{8}{5}\left(\frac{5\sqrt{3}}{8} + \frac{\pi}{6}\right)$ |
---12 A curve $C$ has equation
$$x ^ { 3 } \sin y + \cos y = A x$$
where $A$ is a constant.\\
$C$ passes through the point $P \left( \sqrt { 3 } , \frac { \pi } { 6 } \right)$\\
12
\begin{enumerate}[label=(\alph*)]
\item Show that $A = 2$\\
12
\item (i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 - 3 x ^ { 2 } \sin y } { x ^ { 3 } \cos y - \sin y }$\\
12 (b) (ii) Hence, find the gradient of the curve at $P$.\\
12 (b) (iii) The tangent to $C$ at $P$ intersects the $x$-axis at $Q$.\\
Find the exact $x$-coordinate of $Q$.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2020 Q12 [13]}}