C4 June 2014 (R) Q3

Notes

M1: Differentiates implicitly to include either \(\pm 4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(y^2 \to 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(2y \to 2\dfrac{\mathrm{d}y}{\mathrm{d}x}\). (Ignore \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right)\)).

A1: \(x^2 + y^2 + 10x + 2y \to 2x + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} + 10 + 2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(10 \to 0\)

M1: \(-4xy \to \pm 4y \pm 4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

Note: If an extra term appears then award 1^st A0.

Note: \(2x + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} + 10 + 2\dfrac{\mathrm{d}y}{\mathrm{d}x} - 4y - 4x\dfrac{\mathrm{d}y}{\mathrm{d}x} \ \to\ 2x + 10 - 4y = -2y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
will get 1^st A1 (implied) as the “\(= 0\)” can be implied by rearrangement of their equation.

dM1: dependent on the first method mark being awarded.
An attempt to factorise out all the terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) as long as there are at least two terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).

A1: \(\dfrac{x + 5 - 2y}{2x - y - 1}\) or \(\dfrac{-x - 5 + 2y}{-2x + y + 1}\) (must be simplified).

cso: If the candidate’s solution is not completely correct, then do not give this mark.

Notes

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

NOTE: If the numerator involves one variable only then only the 1^st M1 mark is possible in part (b).

M1: Substitutes their \(x\) or their \(y\) into the printed equation to give an equation in one variable only.

A1: For obtaining either \(-3y^2 + 22y - 35\ \{= 0\}\) or \(3y^2 - 22y + 35\ \{= 0\}\)

Note: This mark can also awarded for a correct three term equation, eg. either \(-3y^2 + 22y = 35\),
\(3y^2 - 22y = -35\) or \(3y^2 + 35 = 22y\) are all fine for A1.

ddM1: Dependent on the previous 2 M marks.
See notes at the beginning of the mark scheme: Method mark for solving a 3 term quadratic

• \((3y - 7)(y - 5) = 0 \Rightarrow y = \ldots\)
• \(y = \dfrac{22 \pm \sqrt{(-22)^2 - 4(3)(35)}}{2(3)}\)
• \(y^2 - \dfrac{22}{3}y + \dfrac{35}{3} = 0 \Rightarrow \left(y - \dfrac{11}{3}\right)^2 - \dfrac{121}{9} + \dfrac{35}{3} = 0 \Rightarrow y = \ldots\) (corrected from the printed mark scheme: \(y^2 - \dfrac{22}{3}y - \dfrac{35}{3} = 0\))
• Or writes down at least one correct \(y\)- root from their quadratic equation. This is usually found from their calculator.

Note: If a candidate applies the alternative method then they also need to use their \(y = \dfrac{x + 5}{2}\) in order to find at least one value for \(y\) in order to gain the final M1.

A1: \(y = \dfrac{7}{3},\ 5\). cao. (2.33 or 2.3 without reference to \(\dfrac{7}{3}\) or \(2\dfrac{1}{3}\) is not allowed for this mark.)

Note: It is possible for a candidate who does not achieve full marks in part (a), (but has a correct numerator for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)) to gain all 5 marks in part (b).