# Question 5:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^3 - 4y^2 = 12xy$; substituting $x=-8$: $-512 - 4y^2 = 12(-8)y$ | M1 | Substitutes $x=-8$ into equation to obtain three-term quadratic in $y$; condone loss of $=0$ |
| $4y^2 - 96y + 512 = 0 \Rightarrow y^2 - 24y + 128 = 0$ | | |
| $(y-16)(y-8)=0$ | dM1 | Attempt to solve quadratic by factorising, formula, or completing the square |
| $y=16$ or $y=8$, i.e. $(-8,8)$ and $(-8,16)$ | A1 | Both $y=16$ and $y=8$ |

**[3 marks]**

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $3x^2 - 8y\frac{dy}{dx} = 12y + 12x\frac{dy}{dx}$ | M1 | Differentiates implicitly to include either $\pm ky\frac{dy}{dx}$ or $12x\frac{dy}{dx}$; ignore $\frac{dy}{dx}=\ldots$ |
| Correct LHS: $3x^2 - 8y\frac{dy}{dx}$ | A1 | Correct LHS equation |
| Correct product rule application on RHS | (B1) | |
| $\frac{dy}{dx} = \frac{3x^2-12y}{12x+8y}$ | | Not necessarily required |
| At $(-8,8)$: $\frac{dy}{dx} = \frac{3(64)-12(8)}{12(-8)+8(8)} = \frac{96}{-32} = -3$ | dM1 | Substitutes $x=-8$ and at least one $y$-value to find any $\frac{dy}{dx}$ |
| At $(-8,16)$: $\frac{dy}{dx} = \frac{3(64)-12(16)}{12(-8)+8(16)} = \frac{0}{32} = 0$ | A1 | One gradient found |
| Both gradients $-3$ and $0$ correctly found | A1 **cso** | Both gradients correct |

**[6 marks]**

**Total: 9 marks**

## Question 5(b) – Way 2 (Aliter):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2\frac{dx}{dy} - 8y = 12y\frac{dx}{dy} + 12x$ | M1 | Differentiates implicitly to include either $\pm kx^2\frac{dx}{dy}$ or $12y\frac{dx}{dy}$. Ignore $\frac{dx}{dy} = \ldots$ |
| Correct LHS equation | A1 | |
| Correct application of product rule | (B1) | |
| $\frac{dy}{dx} = \frac{3x^2 - 12y}{12x + 8y}$ | — | *not necessarily required* |
| @ $(-8, 8)$: $\frac{dy}{dx} = \frac{3(64)-12(8)}{12(-8)+8(8)} = \frac{96}{-32} = -3$ | dM1 | Substitutes $x = -8$ and *at least one* of their $y$-values to attempt to find one of $\frac{dy}{dx}$ or $\frac{dx}{dy}$ |
| @ $(-8, 16)$: $\frac{dy}{dx} = \frac{3(64)-12(16)}{12(-8)+8(16)} = \frac{0}{32} = 0$ | A1 | One gradient found |
| Both gradients of $-3$ and $0$ **correctly** found | A1 cso | |

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## Question 5(b) – Way 3 (Aliter):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = -\frac{3}{2}x \pm \frac{1}{2}(9x^2 + x^3)^{\frac{1}{2}}$ | M1 | A credible attempt to make $y$ the subject and an attempt to differentiate either $-\frac{3}{2}x$ or $\frac{1}{2}(9x^2+x^3)^{\frac{1}{2}}$ |
| $\frac{dy}{dx} = -\frac{3}{2} \pm \frac{1}{2}\left(\frac{1}{2}\right)(9x^2+x^3)^{-\frac{1}{2}}\cdot(18x+3x^2)$ | A1 | $\frac{dy}{dx} = -\frac{3}{2} \pm k(9x^2+x^3)^{-\frac{1}{2}}(g(x))$ |
| $\frac{dy}{dx} = -\frac{3}{2} \pm \frac{18x+3x^2}{4(9x^2+x^3)^{\frac{1}{2}}}$ | A1 | |
| @ $x=-8$: $\frac{dy}{dx} = -\frac{3}{2} \pm \frac{18(-8)+3(64)}{4(9(64)+(-512))^{\frac{1}{2}}} = -\frac{3}{2} \pm \frac{48}{4\sqrt{64}} = -\frac{3}{2} \pm \frac{48}{32}$ | dM1 | Substitutes $x = -8$, find any one of $\frac{dy}{dx}$ |
| $\frac{dy}{dx} = -\frac{3}{2} \pm \frac{3}{2} = -3, 0$ | A1 | One gradient correctly found |
| Both gradients of $-3$ and $0$ correctly found | A1 | |

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