# Question 7 (Implicit differentiation):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $6x^2 + 6y + 6x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0$ | M1 | 1.1a | Attempt implicit differentiation. Either of the two $\frac{dy}{dx}$ terms correct, allowing sign errors. Condone $6x^2dx + 6ydx + 6xdy - 6ydy$ |
| | B1 | 1.1a | Use product rule correctly on middle term. Both terms correct. Must now be $6y + 6x\frac{dy}{dx}$, or implied in correct expression for $\frac{dy}{dx}$ |
| $6x^2 + 6y + 6x - 6y = 0$ | M1 | 3.1a | Use $\frac{dy}{dx} = 1$ in their equation. Must now be equation, but RHS could be incorrect (e.g. $= 2$) |
| $x^2 + x = 0$ | B1 | 1.1a | 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 |
| $x = 0,\ x = -1$ | | |
| $x = 0$ gives $3y^2 = -2$, but $y^2$ has to be $\geq 0$, so no solutions | B1 | 2.3 | Explicitly reject $x = 0$ with reasoning. $x = 0$ must come from $x^2 + x = 0$. e.g. negative numbers cannot be square rooted or $y^2 \neq -\frac{2}{3}$ as $y$ is real. (just $y^2 \neq -\frac{2}{3}$ is insufficient). Must be sensible reason, not just 'math error' or 'not possible'. Could say there are only imaginary (or not real) roots – condone 'complex' roots |
| $x = -1$ gives $3y^2 + 6y + 4 = 0$ | M1 | 2.1 | Attempt to determine the number of real roots of their 3 term quadratic in $y$. Consider discriminant, or use quadratic formula, or attempt minimum value of function |
| $b^2 - 4ac = 36 - 48 = -12$ | | |
| $-12 < 0$ hence no (real) roots | A1 | 2.4 | Obtain correct discriminant from correct quadratic and conclude appropriately. $x = -1$ must come from $x^2 + x = 0$. If using quadratic formula then must be fully correct and attention drawn to why there are no real roots |
| | **[8]** | |

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