# Question 4:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3ye^{-2x} = 4x^2 + y^2 + 2 \Rightarrow 3e^{-2x}\frac{dy}{dx} - 6ye^{-2x} = 8x + 2y\frac{dy}{dx}$ | $\overline{\overline{\text{M1}}}$ | Attempts product rule on $3ye^{-2x}$ to give $pe^{-2x}\frac{dy}{dx} \pm qye^{-2x}$. If product rule quoted it must be correct with the $+$. |
| | $\underline{\text{M1}}$ | Attempts chain rule on $y^2$ to give $Ay\frac{dy}{dx}$ |
| $3e^{-2x}\frac{dy}{dx} - 6ye^{-2x} = 8x + 2y\frac{dy}{dx}$ | A1 | Correct differentiation |
| $\left(3e^{-2x} - 2y\right)\frac{dy}{dx} = 8x + 6ye^{-2x}$ $\Rightarrow \frac{dy}{dx} = \ldots$ | M1 | Collects terms in $\frac{dy}{dx}$ (must be two — one from product and one from $2y\frac{dy}{dx}$) and makes $\frac{dy}{dx}$ subject of formula |
| $\frac{dy}{dx} = \frac{8x + 6ye^{-2x}}{3e^{-2x} - 2y}$ | A1 | Correct expression (allow equivalent correct forms) |

## Question 4(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=0, y=2 \Rightarrow \frac{dy}{dx} = \frac{8(0)+6(2)e^{-2(0)}}{3e^{-2(0)}-2(2)} = (-12)$ | M1 | Substitutes $x=0, y=2$ into their $\frac{dy}{dx}$ or into their differentiated equation and makes $\frac{dy}{dx}$ the subject. May be implied by their value for $\frac{dy}{dx}$ |
| $y-2 = -\frac{1}{"-12"}(x-0)$ | dM1 | Uses correct form of equation of the normal. Look for $y-2 = -\frac{1}{\text{their } dy/dx\big\|_{(0,2)}}(x-0)$ where their $dy/dx$ is non-zero or not undefined. **Dependent on the first method mark.** |
| $y = \frac{1}{12}x + 2$ | A1 | Cao cso |

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