# Question 10:

## Part (i)(a):
| Working | Mark | Guidance |
|---------|------|----------|
| $(3+2x)^6$: any correct form for any term of ${}^6C_1 3^5(2x)^1$ or ${}^6C_2 3^4(2x)^2$ or ${}^6C_4 3^2(2x)^4$ | M1 | Attempt at binomial expansion to find one required term; condone omission of brackets on $2x$ terms |
| Any two correct: $t(2)=6\times3^5\times(2x)^1$, $t(3)=15\times3^4\times(2x)^2$, $t(5)=15\times3^2\times(2x)^4$ | A1 | Any two correct terms simplified or not; ${}^6C_2$ must be numerical |
| All 3 terms correct: 2nd term $= 2916x$, 3rd term $= 4860x^2$, 5th term $= 2160x^4$ | A1 | All three terms correct and simplified |

## Part (i)(b):
| Working | Mark | Guidance |
|---------|------|----------|
| Method using consecutive terms of a GP: $\frac{2160x^4}{4860x^2} = \frac{4860x^2}{2916x}$ | M1 | Method using ratio of consecutive terms of GP |
| $\Rightarrow x = ...$ | dM1 | Proceeds to solve for $x$ |
| $\Rightarrow x = \frac{15}{4}$ | A1 | Correct value |

## Part (ii)(a):
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts to use $S_\infty = \frac{a}{1-r} \Rightarrow \frac{8}{5} = \frac{\sin^2\theta}{1-2\cos\theta}$ | M1 | |
| $\Rightarrow \frac{8}{5} = \frac{1-\cos^2\theta}{1-2\cos\theta}$ | dM1 | Uses $\sin^2\theta = 1-\cos^2\theta$ |
| $\Rightarrow 8 - 16\cos\theta = 5 - 5\cos^2\theta$ | | |
| $\Rightarrow 5\cos^2\theta - 16\cos\theta + 3 = 0$ | A1* | Shown with sufficient working |

## Part (ii)(b):
| Working | Mark | Guidance |
|---------|------|----------|
| Solves $5\cos^2\theta - 16\cos\theta + 3 = 0 \Rightarrow \cos\theta = \frac{1}{5}$ | B1 | |
| Attempts $ar = \sin^2\theta \times 2\cos\theta = \left(1-\left(\frac{1}{5}\right)^2\right) \times 2 \times \frac{1}{5}$ | M1 | |
| $= \frac{48}{125}$ | A1 | |

## Question (i)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses 2nd, 3rd and 5th terms of binomial expansion as consecutive GP terms to set up equation in $x$ | M1 | Common ways: $\frac{"2160x^4"}{" 4860x^2"} = \frac{"4860x^2"}{" 2916x"}$, or $\frac{"4860x^2"}{" 2160x^4"} = \frac{"2916x"}{" 4860x^2"}$, or $"2160x^4" = \frac{"4860x^2"}{" 2916x"} \times "4860x^2"$. Condone slips e.g. 2196 for 2916. Must use 2nd, 3rd and 5th terms only. |
| Proceeds to equation of form $...x = ...$ | dM1 | Dependent on first M1; must have correct index on powers of $x$ |
| $x = \dfrac{15}{4}$ or exact equivalent | A1 | Condone extra $x = 0$ |

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## Question (ii)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $S_\infty = \dfrac{a}{1-r} \Rightarrow \dfrac{8}{5} = \dfrac{\sin^2\theta}{1 - 2\cos\theta}$ | M1 | Condone slips only; formula must be correct. Allow different parameter e.g. $\theta \leftrightarrow x$ |
| Uses identity $\sin^2\theta = 1 - \cos^2\theta$ to form equation in just $\cos\theta$ | dM1 | |
| Completes proof with sufficient working, final line correct and includes $= 0$ | A1* | Condone one slip in notation e.g. $\sin\theta^2 \leftrightarrow \sin^2\theta$, $\cos \leftrightarrow \cos\theta$, but final line must be correct |

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## Question (ii)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $\cos\theta = \dfrac{1}{5}$ | B1 | |
| Attempts $ar = \sin^2\theta \times 2\cos\theta = \left(1 - \left(\dfrac{1}{5}\right)^{"2"}\right) \times 2 \times "\dfrac{1}{5}"$ | M1 | Alternatively: finds $\theta$ from $\cos\theta = \dfrac{1}{5}$ and substitutes into $ar = \sin^2\theta \times 2\cos\theta$ |
| $\dfrac{48}{125}$ or exact equivalent, e.g. $0.384$ | A1 | Cannot score by rounding a non-exact answer or from a decimal value of $\theta$. Answers of $\dfrac{48}{125}$ achieved using $\theta = 78.46...$ do not score this A1 |