June 2023 Paper 1 Q15

AQACurrent spec9 marksDifferentiation

15 The curve with equation

\[x^2 + 2y^3 - 4xy = 0\]

has a single stationary point at \(P\) as shown in the diagram below.

(a) Show that the \(y\)-coordinate of \(P\) satisfies the equation\[y^2(y - 2) = 0\] [7 marks]

(b) Hence, find the coordinates of \(P\) [2 marks]

Mark scheme (a)

Scheme	Marks	AO
Differentiates \(x^2\) to obtain \(2x\)	B1	1.1b
Uses implicit differentiation and obtains either \(Ay^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(Bx\dfrac{\mathrm{d}y}{\mathrm{d}x}\) terms	M1	3.1a
Uses product rule to obtain \(\pm 4y \pm 4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) condone sign errors	M1	1.1a
Obtains \(2x + 6y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} - 4y - 4x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) OE	A1	1.1b
Substitutes \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) into their differentiated expression which contains either \(Ay^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(Bx\dfrac{\mathrm{d}y}{\mathrm{d}x}\) PI by later work.	M1	1.1a
Deduces \(x = 2y\) or \(2x = 4y\) or \(-x = -2y\) or \(-2x = -4y\) Must have scored A1 with no incorrect rearrangement of \(2x + 6y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} - 4y - 4x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) used.	R1	2.2a
Eliminates \(x\) in given equation and completes reasoned argument with at least one intermediate step to show the given result. Must have scored first 6 marks with no incorrect rearrangement of \(2x + 6y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} - 4y - 4x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) used.	R1	2.1
	(7)

\[x^2 + 2y^3 - 4xy = 0\]\[2x + 6y^2\frac{\mathrm{d}y}{\mathrm{d}x} - 4y - 4x\frac{\mathrm{d}y}{\mathrm{d}x} = 0\]

At stationary point \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)

\[2x - 4y = 0\]\[x = 2y\]\[(2y)^2 + 2y^3 - 4(2y)y = 0\]\[4y^2 + 2y^3 - 8y^2 = 0\]\[2y^3 - 4y^2 = 0\]\[y^2(y - 2) = 0\]

Scheme	Marks	AO
Obtains \(y\)-coordinate of 2 Accept \(y = 2\)	B1	1.1b
Obtains \(x\)-coordinate of 4 Accept \(x = 4\)	B1	1.1b
	(2)
	(9 marks)