(6 marks)

**Part i.** | (3 marks)

**M1** | Attempts the quotient rule to obtain an expression of the form $\frac{(x + k)(pe^x + 2xe^x) - (2xe^x)q}{(x + k)^2}$, $p, q > 0$ condoning bracketing errors. If the quotient rule formula is stated it must be correct. May also use product rule to $2xe^x(x + k)^{-1}$ to obtain an expression of the form $pxe^x(x + k)^{-2} + q(x + k)^{-1}e^x(x + 1)$, $p < 0$, $q > 0$ condoning bracketing errors. If the quotient rule formula is stated it must be correct

**A1** | Correct differentiation in any form with correct bracketing which may be implied by subsequent work. Example: $\frac{(x + k)(2e^x + 2xe^x) - (2xe^x)(1)}{(x + k)^2}$ or $-2xe^x(x + k)^{-2} + 2(x + k)^{-1}e^x(x + 1)$

**A1** | Obtains $\frac{dy}{dx} = \frac{e^x(2x^2 + 2kx + 2k)}{(x + k)^2}$

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**Part ii.** | (3 marks)

**B1** | States that $\frac{dy}{dx} = 0$ and then $2x^2 + 2kx + 2k = 0$

**M1** | Attempts $b^2 - 4ac \cdots 0$ with $a = 2$, $b = 2k$, $c = 2k$ where $\cdots$ is either "=", "<, >"

Example: $(2k)^2 - 4 \times 2 \times (2k) \cdots 0$ $\Rightarrow 4k^2 - 16k \cdots 0$

Alternatively attempts to complete the square and sets rhs $\cdots 0$

Example: $2[x^2 + kx + k] = 0$ $\Rightarrow \left(x + \frac{k}{2}\right)^2 - \left(\frac{k}{2}\right)^2 + k = 0$ $\Rightarrow \left(x + \frac{k}{2}\right)^2 = \frac{k^2}{4} - k$

leading to $\frac{k^2}{4} - k = 0$

**A1** | $k = 4$

Example: $4k^2 - 16k = 0 \Rightarrow 4k(k - 4) = 0 \Rightarrow k = 4$

**OR** $\frac{k^2}{4} - k = 0 \Rightarrow k^2 - 4k = 0 \Rightarrow k(k - 4) = 0 \Rightarrow k = 4$

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