C4 June 2013 Q7

Notes

M1: Differentiates implicitly to include either \(4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\pm ky\dfrac{\mathrm{d}y}{\mathrm{d}x}\). (Ignore \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right)\)).

A1: \(\left(x^2\right) \to \left(\underline{2x}\right)\) and \(\left(\ldots + y^2 + 27 = 0 \to \underline{+ 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0}\right)\).
Note: If an extra term appears then award A0.
Note: The "\(= 0\)" can be implied by rearrangement of their equation.
i.e.: \(2x + 4y + 4x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) leading to \(4x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = -2x - 4y\) will get A1 (implied).

B1: \(4y + 4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(4\left(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\) or equivalent

dM1: An attempt to factorise out \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) as long as there are at least two terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
ie. \(\ldots + (4x + 2y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) or \(\ldots + 2(2x + y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\)
Note: This mark is dependent on the previous method mark being awarded.

A1: For \(\dfrac{-2x - 4y}{4x + 2y}\) or equivalent. Eg: \(\dfrac{+2x + 4y}{-4x - 2y}\) or \(\dfrac{-2(x + 2y)}{4x + 2y}\) or \(\dfrac{-x - 2y}{2x + y}\)
cso: If the candidate’s solution is not completely correct, then do not give this mark.

Notes

M1: Sets the denominator of their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to zero (or the numerator of their \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) equal to zero) oe.

A1: Rearranges to give either \(y = -2x\) or \(x = -\dfrac{1}{2}y\). (correct solution only).
The first two marks can be implied from later working, i.e. for a correct substitution of either \(y = -2x\) into \(y^2\) or for \(x = -\dfrac{1}{2}y\) into \(4xy\).

M1*: Substitutes \(y = \pm\lambda x\) or or \(x = \pm\mu y\) or \(y = \pm\lambda x \pm a\) or \(x = \pm\mu y \pm b\) \((\lambda \neq 0,\ \mu \neq 0)\) into
\(x^2 + 4xy + y^2 + 27 = 0\) to form an equation in one variable.

dM1*: leading to at least either \(x^2 = A,\ A > 0\) or \(y^2 = B,\ B > 0\)
Note: This mark is dependent on the previous method mark (M1*) being awarded.

A1: For \(x = -3\) (ignore \(x = 3\)) or if \(y\) was found first, \(y = 6\) (ignore \(y = -6\)) (correct solution only).

ddM1*: Substitutes their value of \(x\) into \(y = \pm\lambda x\) to give \(y = \) value
or substitutes their value of \(x\) into \(x^2 + 4xy + y^2 + 27 = 0\) to give \(y = \) value.
Alternatively, substitutes their value of \(y\) into \(x = \pm\mu y\) to give \(x = \) value
or substitutes their value of \(y\) into \(x^2 + 4xy + y^2 + 27 = 0\) to give \(x = \) value
Note: This mark is dependent on the two previous method marks (M1* and dM1*) being awarded.

A1: \((-3, 6)\) cso.

Note: If a candidate offers two sets of coordinates without either rejecting the incorrect set or accepting the correct set then award A0. DO NOT APPLY ISW ON THIS OCCASION.

Note: \(x = -3\) followed later in working by \(y = 6\) is fine for A1.

Note: \(y = 6\) followed later in working by \(x = -3\) is fine for A1.

Note: \(x = -3, 3\) followed later in working by \(y = 6\) is A0, unless candidate indicates that they
are rejecting \(x = 3\)

Note: Candidates who set the numerator of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to 0 (or the denominator of their \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) equal to zero) can only achieve a maximum of 3 marks in this part. They can only achieve the 2^nd, 3^rd and 4^th Method marks to give a maximum marking profile of M0A0M1M1A0M1A0. They will usually find \((-6, 3)\) { or even \((6, -3)\) }.

Note: Candidates who set the numerator or the denominator of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to \(\pm k\) (usually \(k = 1\)) can only achieve a maximum of 3 marks in this part. They can only achieve the 2^nd, 3^rd and 4^th Method marks to give a marking profile of M0A0M1M1A0M1A0.

Special Case: It is possible for a candidate who does not achieve full marks in part (a), (but has a correct denominator for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)) to gain all 7 marks in part (b).
Eg: An incorrect part (a) answer of \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x - 4y}{4x + 2y}\) can lead to a correct \((-3, 6)\) in part (b) and 7 marks.