C4 June 2015 Q2

Notes

Alternative method for part (a)

General

Note: Writing down \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x - 3y}{3x + 8y}\) or \(\dfrac{3y - 2x}{-3x - 8y}\) from no working is full marks

Note: Writing down \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x - 3y}{-3x - 8y}\) or \(\dfrac{3y - 2x}{3x + 8y}\) from no working is M1A0B1M1A0

Note: Few candidates will write \(2x\,\mathrm{d}x - 3y\,\mathrm{d}x - 3x\,\mathrm{d}y - 8y\,\mathrm{d}y = 0\) leading to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x - 3y}{3x + 8y}\), o.e. This should get full marks.

Marks

M1: Differentiates implicitly to include either \(\pm 3x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(-4y^2 \to \pm ky\dfrac{\mathrm{d}y}{\mathrm{d}x}\). (Ignore \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right)\)).

A1: Both \(x^2 \to \underline{2x}\) and \(\ldots - 4y^2 + 64 = 0 \to \underline{-8y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0}\)

Note: If an extra term appears then award A0.

M1: \(-3xy \to -3x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3y\) or \(-3x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y\) or \(3x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3y\) or \(3x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y\)

Note: \(2x - 3y - 3x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 8y\dfrac{\mathrm{d}y}{\mathrm{d}x} \to 2x - 3y = 3x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 8y\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
will get 1^st A1 (implied) as the "\(= 0\)" can be implied by the 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 \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
i.e. \(\ldots + (-3x - 8y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) or \(\ldots = (3x + 8y)\dfrac{\mathrm{d}y}{\mathrm{d}x}\). (Allow combining in 1 variable).

A1: \(\dfrac{2x - 3y}{3x + 8y}\) or \(\dfrac{3y - 2x}{-3x - 8y}\) or equivalent.

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

Note: You cannot recover work for part (a) in part (b).

Notes

M1: Sets their 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) o.e.

Note: 1^st M1 can also be gained by setting \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to zero in their “\(2x - 3y - 3x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 8y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)”

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

Note: If their numerator is a constant then no marks are available in part (b)

Note: If their numerator is in the form \(\pm ax^2 \pm by = 0\) or \(\pm ax \pm by^2 = 0\) then the first 3 marks are possible in part (b).

Note: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x - 3y}{3x + 8y} = 0\) is not sufficient for M1.

A1ft: Either

• Sets \(2x - 3y\) to zero and obtains either \(y = \dfrac{2}{3}x\) or \(x = \dfrac{3}{2}y\)
• the follow through result of making either \(y\) or \(x\) the subject from setting their numerator of their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to zero

dM1: dependent on the first method mark being awarded.
Substitutes either their \(y = \dfrac{2}{3}x\) or their \(x = \dfrac{3}{2}y\) into the original equation to give an equation in one variable only.

A1: Obtains either \(x = \dfrac{24}{5}\) or \(-\dfrac{24}{5}\) or \(y = \dfrac{16}{5}\) or \(-\dfrac{16}{5}\), (or equivalent) by correct solution only.
i.e. You can allow for example \(x = \dfrac{48}{10}\) or 4.8, etc.

Note: \(x = \sqrt{\dfrac{576}{25}}\) (not simplified) or \(y = \sqrt{\dfrac{256}{25}}\) (not simplified) is not sufficient for A1.

ddM1: dependent on both previous method marks being awarded in this part.

Method 1
Either:

• substitutes their \(x\) into their \(y = \dfrac{2}{3}x\) or substitutes their \(y\) into their \(x = \dfrac{3}{2}y\), or
• substitutes the other of their \(y = \dfrac{2}{3}x\) or their \(x = \dfrac{3}{2}y\) into the original equation,

and achieves either:

• exactly two sets of two coordinates or
• exactly two distinct values for \(x\) and exactly two distinct values for \(y\).

Method 2
Either:

• substitutes their first \(x\)-value, \(x_1\) into \(x^2 - 3xy - 4y^2 + 64 = 0\) to obtain one \(y\)-value, \(y_1\) and substitutes their second \(x\)-value, \(x_2\) into \(x^2 - 3xy - 4y^2 + 64 = 0\) to obtain 1 \(y\)-value \(y_2\) or
• substitutes their first \(y\)-value, \(y_1\) into \(x^2 - 3xy - 4y^2 + 64 = 0\) to obtain one \(x\)-value \(x_1\) and substitutes their second \(y\)-value, \(y_2\) into \(x^2 - 3xy - 4y^2 + 64 = 0\) to obtain one \(x\)-value \(x_2\).

Note: Three or more sets of coordinates given (without identification of two sets of coordinates) is ddM0.

A1: Both \(\left(\dfrac{24}{5}, \dfrac{16}{5}\right)\) and \(\left(-\dfrac{24}{5}, -\dfrac{16}{5}\right)\), only by cso. Note that decimal equivalents are fine.

Note: Also allow \(x = \dfrac{24}{5}, y = \dfrac{16}{5}\) and \(x = -\dfrac{24}{5}, y = -\dfrac{16}{5}\) all seen in their working to part (b).

Note: Allow \(x = \pm\dfrac{24}{5}, y = \pm\dfrac{16}{5}\) for 3^rd A1.

Note: \(x = \pm\dfrac{24}{5}, y = \pm\dfrac{16}{5}\) followed by eg. \(\left(\dfrac{16}{5}, \dfrac{24}{5}\right)\) and \(\left(-\dfrac{16}{5}, -\dfrac{24}{5}\right)\) (eg. coordinates stated the wrong way round) is 3^rd A0.

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 6 marks in part (b).

Note: Decimal equivalents to fractions are fine in part (b). i.e. \((4.8, 3.2)\) and \((-4.8, -3.2)\).

Note: \(\left(\dfrac{24}{5}, \dfrac{16}{5}\right)\) and \(\left(-\dfrac{24}{5}, -\dfrac{16}{5}\right)\) from no working is M0A0M0A0M0A0.

Note: Candidates could potentially lose the final 2 marks for setting both their numerator and denominator to zero.

Note: No credit in this part can be gained by only setting the denominator to zero.