C4 June 2017 Q4

Notes

Note: For the first four marks
Writing down from no working

• \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4y - 8x}{-3y^2 - 4x + 2^y\ln 2}\) or \(\dfrac{8x - 4y}{3y^2 + 4x - 2^y\ln 2}\) scores M1A1M1B1
• \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8x - 4y}{-3y^2 - 4x + 2^y\ln 2}\) or \(\dfrac{4y - 8x}{3y^2 + 4x - 2^y\ln 2}\) scores M1A0M1B1

Writing \(8x\,\mathrm{d}x - 3y^2\,\mathrm{d}y - 4y\,\mathrm{d}x - 4x\,\mathrm{d}y + 2^y\ln 2\,\mathrm{d}y = 0\) scores M1A1M1B1

1^st M1: Differentiates implicitly to include either \(\pm 4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(-y^3 \to \pm\lambda y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(2^y \to \pm\mu 2^y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (Ignore \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \right)\)). \(\lambda, \mu\) are constants which can be 1

1^st A1: Both \(4x^2 - y^3 \to 8x - 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(= 0 \to = 0\)

Note: e.g. \(8x - 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} - 4y - 4x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2^y\ln 2\dfrac{\mathrm{d}y}{\mathrm{d}x} \to -3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} - 4x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2^y\ln 2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4y - 8x\)
or e.g. \(-16 - 48\dfrac{\mathrm{d}y}{\mathrm{d}x} - 16 + 8\dfrac{\mathrm{d}y}{\mathrm{d}x} + 16\ln 2\dfrac{\mathrm{d}y}{\mathrm{d}x} \to -48\dfrac{\mathrm{d}y}{\mathrm{d}x} + 8\dfrac{\mathrm{d}y}{\mathrm{d}x} + 16\ln 2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 32\)
will get 1^st A1 (implied) as the "\(= 0\)" can be implied by the rearrangement of their equation.

2^nd M1: \(-4xy \to -4y - 4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(4y - 4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(-4y + 4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(4y + 4x\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

\(\overline{\overline{\text{B1}}}\): \(2^y \to 2^y\ln 2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(2^y \to \mathrm{e}^{y\ln 2}\ln 2\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

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

3^rd dM1: dependent on the first M mark
For substituting \(x = -2\) and \(y = 4\) into an equation involving \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

Note: M1 can be gained by seeing at least one example of substituting \(x = -2\) and at least one example of substituting \(y = 4\) unless it is clear that they are instead applying \(x = 4\) and \(y = -2\)
Otherwise, you will NEED to check (with your calculator) that \(x = -2,\ y = 4\) that has been substituted into their equation involving \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

Note: A1 cso: If the candidate’s solution is not completely correct, then do not give this mark.

Note: isw: You can, however, ignore subsequent working following on from correct solution.

Way 2

Way 2 notes

1^st M1: Differentiates implicitly to include either \(\pm 4y\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(4x^2 \to \pm\lambda x\dfrac{\mathrm{d}x}{\mathrm{d}y}\) (Ignore \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \right)\)). \(\lambda\) is a constant which can be 1

1^st A1: Both \(4x^2 - y^3 \to 8x\dfrac{\mathrm{d}x}{\mathrm{d}y} - 3y^2\) and \(= 0 \to = 0\)

2^nd M1: \(-4xy \to -4y\dfrac{\mathrm{d}x}{\mathrm{d}y} - 4x\) or \(4y\dfrac{\mathrm{d}x}{\mathrm{d}y} - 4x\) or \(-4y\dfrac{\mathrm{d}x}{\mathrm{d}y} + 4x\) or \(4y\dfrac{\mathrm{d}x}{\mathrm{d}y} + 4x\)

\(\overline{\overline{\text{B1}}}\): \(2^y \to 2^y\ln 2\)

3^rd dM1: dependent on the first M mark
For substituting \(x = -2\) and \(y = 4\) into an equation involving \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\)

Notes

Note: The 2^nd M1 mark can be implied by later working.
Eg. Award 1^st M1 and 2^nd M1 for \(\dfrac{y - 4}{2} = \dfrac{-1}{\text{their } m_T \text{ evaluated at } x = -2 \text{ and } y = 4}\)

Note: A1: Allow the alternative answer \(\{y =\}\ \ln\left(\dfrac{1}{2}\right) + \dfrac{13}{2\ln 2}(\ln 2)\) which is in the form \(p + q\ln 2\)