**(a)** Sets $f(3) = 0 \to$ equation in $k$. Eg. $27k - 135 - 96 - 12 = 0$ | M1 | Attempts to set $f(3) = 0 \to$ equation in $k$. Eg. $27k - 135 - 96 - 12 = 0$. Condone slips.

$\Rightarrow 27k = 243 \Rightarrow k = 9$ * ($= 0$ must be seen) | A1* | Completes proof with at least one intermediate "solvable" line namely $27k = 243 \Rightarrow k = 9$ or $27k - 243 = 0 \Rightarrow k = 9$. This is a given answer so there should be no errors. It is a "show that" question so expect to see (i) Either $f(3) = 0$ explicitly stated or implied by sight of $27k - 135 - 96 - 12 = 0$ or $27k - 243 = 0$ (ii) One solvable intermediate line followed by $k = 9$

| (2 marks)

**(b)** $9x^3 - 15x^2 - 32x - 12 = (x - 3)(9x^2 + 12x + 4)$ | M1 A1 | M1 A1: Attempt to divide or factorise out $(x - 3)$. Condone students who use a different value of $k$.

$= (x - 3)(3x + 2)^2$ | dM1 A1 | dM1: Attempt at factorising their $9x^2 + 12x + 4$ Apply the usual rules for factorising $(x - 3)(3x + 2)(3x + 2)$ on one line.

Accept $9(x - 3)\left(x + \frac{2}{3}\right)$ oe. It must be seen as a product

Remember to isw for candidates who go on to give roots $f(x) = (x - 3)(3x + 2)^2 \Rightarrow x = ...$

| (4 marks)

**(c)** Attempts $\cos\theta = -\frac{2}{3}$ | M1 | A correct attempt to find one value of $\theta$ in the given range for their $\cos\theta = -\frac{2}{3}$ (You may have to use a calculator). So if (b) is factorised correctly the mark is for one of the values.

$\theta = 131.8°, 228.2°$ (awrt) | A1 | CSO awrt $\theta = 131.8°, 228.2°$ with no additional solutions within the range $0 < \theta < 360°$. Watch for correct solutions appearing from $3\cos\theta - 2 = 0 \Rightarrow \cos\theta = \frac{2}{3}$. This is M0 A0.

| (2 marks, 8 marks total)

**Additional notes:**

(a) M1: Attempts to set $f(3) = 0 \to$ equation in $k$. Eg. $27k - 135 - 96 - 12 = 0$. Condone slips. Score when you see embedded values within the equation or two correct terms on the lhs of the equation. It is implied by sight of $27k - 243 = 0$ or $27k = 135 + 96 + 12$.

A1*: Completes proof with at least one intermediate "solvable" line namely $27k = 243 \Rightarrow k = 9$ or $27k - 243 = 0 \Rightarrow k = 9$. This is a given answer so there should be no errors. It is a "show that" question so expect to see (i) Either $f(3) = 0$ explicitly stated or implied by sight of $27k - 135 - 96 - 12 = 0$ or $27k - 243 = 0$ (ii) One solvable intermediate line followed by $k = 9$

A candidate could use $k = 9$ and start with $f(x) = 9x^3 - 15x^2 - 32x - 12$. Alt attempts to divide $f(x)$ by (x-3). See below on how to score such an attempt.

M1: For attempting $f(3) = 9 \times 3^3 - 15 \times 3^2 - 32 \times 3 - 12$. Alt attempts to divide $f(x)$ by (x-3). See below on how to score such an attempt.

A1*: Shows that $f(3) = 0$ and makes a minimal statement to the effect that "so $k = 9$". If division is attempted it must be correct and a statement is required to the effect that there is no remainder, " so $k = 9$"

If candidates have divided (correctly) in part (a) they can be awarded the first two marks in (b) when they start factorising the $9x^2 + 12x + 4$ term.

(b) M1: Attempt to divide or factorise out $(x - 3)$. Condone students who use a different value of $k$. For factorisation look for first and last terms $9x^3 - 15x^2 - 32x - 12 = (x - 3)(\pm 9x^2 .......\pm 4)$. For division look for the following line $x - 3) 9x^3 - 15x^2 - 32x - 12 \quad 9x^2 + ...........$

A1: Correct quadratic factor $9x^2 + 12x + 4$. You may condone division attempts that don't quite work as long as the correct factor is seen.

dM1: Attempt at factorising their $9x^2 + 12x + 4$ Apply the usual rules for factorising $(x - 3)(3x + 2)(3x + 2)$ on one line. Accept $9(x - 3)\left(x + \frac{2}{3}\right)$ oe. It must be seen as a product.

Remember to isw for candidates who go on to give roots $f(x) = (x - 3)(3x + 2)^2 \Rightarrow x = ....$

Note: Part (b) is "Hence" so take care when students write down the answer to (b) without method.

If candidates state $x = -\frac{2}{3}, 3 \Rightarrow f(x) = (x + \frac{2}{3})(x + \frac{2}{3})(x - 3)$ score 0 0 0.

If candidates state $x = -\frac{2}{3}, 3 \Rightarrow f(x) = (3x + 2)(3x + 2)(x - 3)$ they score SC 1010.

If candidate writes down $f(x) = (x - 3)(3x + 2)$ are factors it is 0000.

If a candidate writes down $f(x) = (3x + 2)(3x + 2)(x - 3)$ with no working they score SC 1010.

(c) M1: A correct attempt to find one value of $\theta$ in the given range for their $\cos\theta = -\frac{2}{3}$ (You may have to use a calculator). So if (b) is factorised correctly the mark is for one of the values. This can be implied by sight of awrt 132 or 228 in degrees or awrt 2.3 which is the radian solution.

A1: CSO awrt $\theta = 131.8°, 228.2°$ with no additional solutions within the range $0 < \theta < 360°$. Watch for correct solutions appearing from $3\cos\theta - 2 = 0 \Rightarrow \cos\theta = \frac{2}{3}$. This is M0 A0.

Answers without working are acceptable.

M1 For one correct answer

M1 A1 For two correct answers with no additional solutions within the range.

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