Question 11 - A-Level Maths

AQA Paper 1 2021 June — Question 11 8 marks

Exam Board	AQA
Module	Paper 1 (Paper 1)
Year	2021
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Differential equations
Type	Geometric curve properties
Difficulty	Challenging +1.8 This is a challenging A-level question requiring separation of variables to solve a non-linear differential equation, then analyzing the resulting implicit/explicit function to determine axis intersections. It demands careful algebraic manipulation, consideration of domain restrictions, and rigorous justification of uniqueness—going beyond routine differential equation solving to require geometric insight and proof.
Spec	1.02q Use intersection points: of graphs to solve equations 1.08k Separable differential equations: dy/dx = f(x)g(y)

11 A curve, $C$, passes through the point with coordinates $( 1,6 )$ The gradient of $C$ is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 6 } ( x y ) ^ { 2 }$$ Show that $C$ intersects the coordinate axes at exactly one point and state the coordinates of this point. Fully justify your answer.

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Question 11:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Separates variables: $\int \frac{1}{y^2}\, dy = \int \frac{1}{6}x^2\, dx$	M1 (AO 3.1a)	Must have integral signs and consistent $dy$ and $dx$. Condone $x$ instead of $x^2$
Integrates one integral correctly	M1 (AO 1.1a)
$-y^{-1} = \frac{x^3}{18} + c$	A1 (AO 1.1b)	Condone missing $+c$
Substitutes $(1, 6)$: $-\frac{1}{6} = \frac{1}{18} + c \Rightarrow c = -\frac{2}{9}$	M1 (AO 1.1a)
$y$ cannot equal zero as $y^{-1}$ is undefined at $y = 0$, therefore $C$ does not intersect the $x$-axis	E1F (AO 2.4)	Explains why $y$ cannot equal zero, following through their equation
Substitutes $x = 0$: $-y^{-1} = -\frac{2}{9} \Rightarrow y = \frac{9}{2}$	M1 (AO 3.1a)
$y = 4.5$; curve crosses $y$-axis at $(0, 4.5)$	A1, R1 (AO 1.1b, 2.2a)	Must state $C$ intersects axes at (exactly/only) one point, state coordinate $(0, 4.5)$, AND state $C$ does not intersect $x$-axis. CSO

Total: 8 marks

## Question 11:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Separates variables: $\int \frac{1}{y^2}\, dy = \int \frac{1}{6}x^2\, dx$ | M1 (AO 3.1a) | Must have integral signs and consistent $dy$ and $dx$. Condone $x$ instead of $x^2$ |
| Integrates one integral correctly | M1 (AO 1.1a) | |
| $-y^{-1} = \frac{x^3}{18} + c$ | A1 (AO 1.1b) | Condone missing $+c$ |
| Substitutes $(1, 6)$: $-\frac{1}{6} = \frac{1}{18} + c \Rightarrow c = -\frac{2}{9}$ | M1 (AO 1.1a) | |
| $y$ cannot equal zero as $y^{-1}$ is undefined at $y = 0$, therefore $C$ does not intersect the $x$-axis | E1F (AO 2.4) | Explains why $y$ cannot equal zero, following through their equation |
| Substitutes $x = 0$: $-y^{-1} = -\frac{2}{9} \Rightarrow y = \frac{9}{2}$ | M1 (AO 3.1a) | |
| $y = 4.5$; curve crosses $y$-axis at $(0, 4.5)$ | A1, R1 (AO 1.1b, 2.2a) | Must state $C$ intersects axes at (exactly/only) one point, state coordinate $(0, 4.5)$, AND state $C$ does not intersect $x$-axis. CSO |

**Total: 8 marks**

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11 A curve, $C$, passes through the point with coordinates $( 1,6 )$

The gradient of $C$ is given by

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 6 } ( x y ) ^ { 2 }$$

Show that $C$ intersects the coordinate axes at exactly one point and state the coordinates of this point.

Fully justify your answer.\\

\hfill \mbox{\textit{AQA Paper 1 2021 Q11 [8]}}

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