| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle equation from centre and radius |
| Difficulty | Moderate -0.8 This question requires finding a constant by substituting a known point into an equation. At point P, y=0 and x=16, giving 256=16a, so a=16. This is straightforward substitution with basic algebra, requiring fewer steps than a typical circle equation problem and no geometric insight or manipulation into standard form. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(y = 0\) and \(x = 16\) into \(x^2 + y^2 = a\sqrt{x} - y\) | M1 | |
| \(16^2 + 0^2 = a\sqrt{16} - 0 \Rightarrow 256 = 4a\) | A1 | |
| \(a = 64\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either \(2y\frac{dy}{dx}\) or \(-\frac{dy}{dx}\) seen | B1 | Differentiates implicitly |
| Differentiates any two of the four terms correctly | M1 | Can be in terms of \(a\) or with their \(a\) value |
| \(2x + 2y\frac{dy}{dx} = \frac{64}{2}x^{-\frac{1}{2}} - \frac{dy}{dx}\) | A1F | Fully correct differentiated equation; follow through their \(a\) value |
| Uses \(\frac{dy}{dx} = 0\) | M1 | |
| \(2x = \frac{32}{\sqrt{x}} \Rightarrow x^{\frac{3}{2}} = 16 \Rightarrow x = 6.3496...\) | M1 | Substitutes numerical \(x\) where \(0 < x < 16\) into model |
| \((6.3496...)^2 + y^2 = 64\sqrt{6.3496...} - y \Rightarrow y = 10.51\) | R1 | Must obtain \(y\) AWRT 10.51, state units, conclude maximum height is approximately 10.5 metres AG; CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The entrance to the cave is unlikely to be a perfectly smooth curve | E1 | Accept: the cave has dents; entrance is not perfectly smooth. Ignore comments about the floor or vertical cross section |
## Question 13(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $y = 0$ and $x = 16$ into $x^2 + y^2 = a\sqrt{x} - y$ | M1 | |
| $16^2 + 0^2 = a\sqrt{16} - 0 \Rightarrow 256 = 4a$ | A1 | |
| $a = 64$ | | |
## Question 13(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Either $2y\frac{dy}{dx}$ or $-\frac{dy}{dx}$ seen | B1 | Differentiates implicitly |
| Differentiates any two of the four terms correctly | M1 | Can be in terms of $a$ or with their $a$ value |
| $2x + 2y\frac{dy}{dx} = \frac{64}{2}x^{-\frac{1}{2}} - \frac{dy}{dx}$ | A1F | Fully correct differentiated equation; follow through their $a$ value |
| Uses $\frac{dy}{dx} = 0$ | M1 | |
| $2x = \frac{32}{\sqrt{x}} \Rightarrow x^{\frac{3}{2}} = 16 \Rightarrow x = 6.3496...$ | M1 | Substitutes numerical $x$ where $0 < x < 16$ into model |
| $(6.3496...)^2 + y^2 = 64\sqrt{6.3496...} - y \Rightarrow y = 10.51$ | R1 | Must obtain $y$ AWRT 10.51, state units, conclude maximum height is approximately 10.5 metres AG; CSO |
## Question 13(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The entrance to the cave is unlikely to be a perfectly smooth curve | E1 | Accept: the cave has dents; entrance is not perfectly smooth. Ignore comments about the floor or vertical cross section |13 Figure 2 shows the approximate shape of the vertical cross section of the entrance to a cave. The cave has a horizontal floor.
The entrance to the cave joins the floor at the points $O$ and $P$.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{22ff390e-1360-43bd-8c7f-3d2b58627e91-24_396_991_584_529}
\end{center}
\end{figure}
Garry models the shape of the cross section of the entrance to the cave using the equation
$$x ^ { 2 } + y ^ { 2 } = a \sqrt { x } - y$$
where $a$ is a constant, and $x$ and $y$ are the horizontal and vertical distances respectively, in metres, measured from $O$.
13 (a) The distance $O P$ is 16 metres.\\
Find the value of $a$ that Garry should use in the model.\\
\includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-25_2518_1723_196_148}
\hfill \mbox{\textit{AQA Paper 1 2022 Q13 [9]}}