June 2022 Paper 1 Q7

Notes

M1: Either of the two \(\frac{\mathrm{d}y}{\mathrm{d}x}\) terms correct, allowing sign errors
Condone \(6x^2\mathrm{d}x + 6y\mathrm{d}x + 6x\mathrm{d}y - 6y\mathrm{d}y\)

B1: Both terms correct
Must now be \(6y + 6x\frac{\mathrm{d}y}{\mathrm{d}x}\), or implied in a correct expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\)

A1: Condone missing or incorrect RHS
Must now have \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and not just \(\mathrm{d}y\) or \(\mathrm{d}x\) in terms

M1: Use \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1\) in their equation
Must now be equation, but RHS could be incorrect (eg ‘= 2’)

B1: Solve correct quadratic in \(x\) to obtain two correct roots (possibly BC)
Quadratic must come from correct implicit differentiation
B0 if \(x\) ‘cancelled’ in quadratic to give \(x = -1\) as only root, but M1A1 still available

B1: Explicitly reject \(x = 0\), with reasoning
\(x = 0\) must come from \(x^2 + x = 0\)
eg negative numbers cannot be square rooted or \(y^2 \ne -\frac{2}{3}\) as \(y\) is real
(just \(y^2 \ne -\frac{2}{3}\) is insufficient)
Must be sensible reason and not just ‘math error’ or ‘not possible’
Could say that there are only imaginary (or not real) roots – condone ‘complex’ roots

M1: Attempt to determine the number of real roots of their 3 term quadratic in \(y\)
From substituting their \(x\) value into the equation of the curve
Consider discriminant, or use quadratic formula, or attempt minimum value of function

A1: Obtain correct discriminant from correct quadratic and conclude appropriately
\(x = -1\) must come from \(x^2 + x = 0\)
If using quadratic formula then it must be fully correct and attention drawn to why there are no real roots