C4 June 2018 Q2

Notes

M1: Differentiates implicitly to include either \(x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(y^2 \to 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(-5y \to -5\dfrac{\mathrm{d}y}{\mathrm{d}x}\). \(\left(\text{Ignore } \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\right)\)

A1: \(x^2 \to 2x\) and \(y^2 - 4x - 5y + 1 = 0 \to 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 4 - 5\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)

B1: \(xy \to y + x\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

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

Note: \(2x + y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 4 - 5\dfrac{\mathrm{d}y}{\mathrm{d}x} \to 2x + y - 4 = -x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} + 5\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
will get 1^st A1 (implied) as the "\(= 0\)" can be implied the rearrangement of their equation.

dM1: dependent on the previous M mark
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 cso: \(\dfrac{2x + y - 4}{5 - x - 2y}\) or \(\dfrac{4 - 2x - y}{x + 2y - 5}\)
If the candidate’s solution is not completely correct, then do not give the final A mark

Alt 1 for part (a)

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

A1: \(x^2 \to 2x\dfrac{\mathrm{d}x}{\mathrm{d}y}\) and \(y^2 - 4x - 5y + 1 = 0 \to 2y - 4\dfrac{\mathrm{d}x}{\mathrm{d}y} - 5 = 0\)

B1: \(xy \to y\dfrac{\mathrm{d}x}{\mathrm{d}y} + x\)

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

Note: \(2x\dfrac{\mathrm{d}x}{\mathrm{d}y} + y\dfrac{\mathrm{d}x}{\mathrm{d}y} + x + 2y - 4\dfrac{\mathrm{d}x}{\mathrm{d}y} - 5 \to x + 2y - 5 = -2x\dfrac{\mathrm{d}x}{\mathrm{d}y} - y\dfrac{\mathrm{d}x}{\mathrm{d}y} + 4\dfrac{\mathrm{d}x}{\mathrm{d}y}\)
will get 1^st A1 (implied) as the "\(= 0\)" can be implied the rearrangement of their equation.

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

A1 cso: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x + y - 4}{5 - x - 2y}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4 - 2x - y}{x + 2y - 5}\)
If the candidate’s solution is not completely correct, then do not give the final A mark

Further notes for part (a)

Note: Writing down from no working

• \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x + y - 4}{5 - x - 2y}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4 - 2x - y}{x + 2y - 5}\) scores M1 A1 B1 M1 A1
• \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4 - 2x - y}{5 - x - 2y}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x + y - 4}{x + 2y - 5}\) scores M1 A0 B1 M1 A0

Note: Writing \(2x\,\mathrm{d}x + y\,\mathrm{d}x + x\,\mathrm{d}y + 2y\,\mathrm{d}y - 4\,\mathrm{d}x - 5\,\mathrm{d}y = 0\) scores M1 A1 B1

Notes

Alt 1 for part (b)

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) o.e.

Note: This mark can also be gained by setting \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to zero in their differentiated equation from (a)

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

dM1: dependent on the previous M mark
Substitutes their \(x\) or their \(y\) (from their numerator = 0) into the printed equation to give an equation in one variable only

A1: For obtaining the correct 3TQ. E.g.: either \(3x^2 - 6x - 3\ \{= 0\}\) or \(-3x^2 + 6x + 3\ \{= 0\}\)

Note: This mark can also be awarded for a correct 3 term equation. E.g. either \(3x^2 - 6x = 3\)
\(x^2 - 2x - 1 = 0\) or \(x^2 = 2x + 1\) are all fine for A1

ddM1: dependent on the previous 2 M marks
See page 6: Method mark for solving THEIR 3-term quadratic in one variable
Quadratic Equation to solve: \(3x^2 - 6x - 3 = 0\)
Way 1: \(x = \dfrac{6 \pm \sqrt{(-6)^2 - 4(3)(-3)}}{2(3)}\)
Way 2: \(x^2 - 2x - 1 = 0 \Rightarrow (x - 1)^2 - 1 - 1 = 0 \Rightarrow x = \ldots\)
Way 3: Or writes down at least one exact correct \(x\)-root (or one correct \(x\)-root to 2 dp) from their quadratic equation. This is usually found on their calculator.
Way 4: (Only allowed if their 3TQ can be factorised)

• \((x^2 + bx + c) = (x + p)(x + q)\), where \(|pq| = |c|\), leading to \(x = \ldots\)
• \((ax^2 + bx + c) = (mx + p)(nx + q)\), where \(|pq| = |c|\) and \(|mn| = a\), leading to \(x = \ldots\)

Note: If a candidate applies the alternative method then they also need to use their \(x = \dfrac{4 - y}{2}\) to find at least one value for \(x\) in order to gain the final M mark.

A1: Exact values of \(x = 1 + \sqrt{2},\ 1 - \sqrt{2}\) (or \(1 \pm \sqrt{2}\)), cao. Apply isw if \(y\)-values are also found.

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)