## (i)
**Answer:** $y = -x^3 - 3x^2 + 4x - kx + k$; $\frac{dy}{dx} = -3x^2 - 6x + 4 - k$; When $x = -3$: $\frac{dy}{dx} = 0$; $-27 + 18 + 4 - k = 0$; $k = -5$

**Marks:** M1 | A1 | M1* | DM1* | A1 [7]

**Guidance:** Attempt to multiply out brackets | Can be unsimplified | Attempt to differentiate their expansion (M0 if signs have changed throughout) | Sets $\frac{dy}{dx} = 0$; Substitutes $x = -3$ into their $\frac{dy}{dx} = 0$ | www

**Additional notes:** Must have $\pm x^3$ and 5 or 6 terms; If using product rule: Clear attempt at correct rule M1*; Differentiates both parts correctly A1; Expand brackets of both parts *DM1; Then as main scheme

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## (ii)
**Answer:** $\frac{d^2y}{dx^2} = -6x - 6$; When $x = -3$, $\frac{d^2y}{dx^2}$ is positive so min point

**Marks:** M1 | A1 [2]

**Guidance:** Evaluates second derivative at $x = -3$ or other fully correct method | No incorrect working seen in this part i.e. if second derivative is evaluated, it must be 12. (Ignore errors in $k$ value)

**Additional notes:** Alternate valid methods include: 1) Evaluating gradient at either side of −3; 2) Evaluating y at either side of −3; 3) Finding other turning point and stating "negative cubic so min before max"

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## (iii)
**Answer:** $-3x^2 - 6x + 9 = 9$; $3x(x+2) = 0$; $x = 0$ or $x = -2$; When $x = 0, y = -9$ for line; $y = -5$ for curve; When $x = -2, y = -27$ for line; $y = -27$ for curve; $x = -2, y = -27$ www (Check $k$ correct)

**Marks:** M1 | A1 | M1 | M1 | M1 | A1 [5]

**Guidance:** Sets their gradient function from (i) (or from a restart) to 9 | Correct x-values | One of their x-values substituted into both curve and line/substituted into one and verified to be on the other | Conclusion that $x = -2$ is the correct value or Second x-value substituted into both curve and line/verified as above | $x = -2, y = -27$ www (Check $k$ correct)

**Additional notes:** Allow first M even if k not found but look out for correct answer from wrong working. SEE NEXT PAGE FOR ALTERNATIVE METHODS; Note: Putting a value into $x^3 + 3x^2 - 4 = 0$ (where the line and curve meet) is equivalent. If curve equated to line before differentiating: M0 A0, can get M1M1 but A0 ww; Maximum mark 2/5

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## (iii) Alternative method

**Answer:** Attempt to solve equations of curve and tangent simultaneously and uses valid method to establish at least one root of the resulting cubic ($x^3 + 3x^2 - 4 = 0$ oe) M1; All roots found A1; Either 1) States $x = -2$ is repeated root so tangent M2; (If double root found but not explicitly stated that repeated root implies tangent then M0 but B1 if $(-2, -27)$ found) Or 2) Substitutes one x value into their gradient function to determine if equal to gradient of the line M1; Substitutes other x value into their gradient function to determine if equal to gradient of the line or conclusion that −2 is the correct one M1; $x = -2, y = -27$ A1 www

**Marks for alternative:** M2 or M1 B1 (then M1, M1) and A1 www / B1 2/5

**Additional notes:** SC Trial and Improvement: Finds at least one value at which the gradient of the curve is 9 B1; Verifies on both line and curve B1 2/5