# Question 8(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cot 2x\{+\tan x\} = \frac{\cos 2x}{\sin 2x}\left\{+\frac{\sin x}{\cos x}\right\}$ | M1 | Uses $\cot 2x = \frac{\cos 2x}{\sin 2x}$ or e.g. $\frac{\cos 2x}{2\sin x\cos x}$ |
| $\frac{\cos 2x + 2\sin^2 x}{2\sin x\cos x}$; using e.g. $\cos 2x = 1-2\sin^2 x$ or $\cos^2 x - \sin^2 x + 2\sin^2 x$; reaching $\frac{2\cos^2 x - 1 + 2\sin^2 x}{\sin 2x}$ or $\frac{\cos 2x+1-\cos 2x}{\sin 2x}$ | A1 (M1 on ePen) | Uses sufficient correct identities to obtain correct single fraction with numerator in terms of $\sin x$ and/or $\cos x$. Qualifying fraction must be seen before $\frac{1}{2\sin x\cos x}$ or $\frac{1}{\sin 2x}$. Condone poor notation. |
| $= \frac{1}{2\sin x\cos x}$ or $\frac{1}{\sin 2x} = \csc 2x$ * | A1* | Fully correct proof with one of the two intermediate fractions seen. All notation correct — no mixed or missing arguments. |

**Alt method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cot 2x\{+\tan x\} = \frac{1-\tan^2 x}{2\tan x}\{+\tan x\}$ | M1 | Uses $\cot 2x = \frac{1-\tan^2 x}{2\tan x}$ |
| $\frac{1-\tan^2 x+2\tan^2 x}{2\tan x}$; e.g. $\frac{\tan^2 x+1}{2\tan x} \Rightarrow \frac{\cos x(\sin^2 x+\cos^2 x)}{2\cos^2 x\sin x}$ or $\frac{\sec^2 x}{2\tan x}$ | A1 (M1 on ePen) | Uses correct identities. Allow $\frac{\sec^2 x}{2\tan x}$ following $\sec^2 x=1+\tan^2 x$. Qualifying fraction must be seen. |
| $= \frac{1}{2\sin x\cos x}$ or $\frac{1}{\sin 2x} = \csc 2x$ * | A1* | Fully correct proof with intermediate fraction seen. All notation correct. |

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# Question 8(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| [Examples shown in markscheme] | M1 A1 | |

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# Question 8(b) [differentiation part]:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y^2 = w\sin 2x \Rightarrow 2y\frac{dy}{dx} = \frac{dw}{dx}\sin 2x + 2w\cos 2x$ | M1 | Attempts differentiation of substitution using product/quotient and chain rules **and** obtains equation in $\frac{dy}{dx}$ and $\frac{dw}{dx}$ of correct form (sign/coefficient errors only). Not available for work in $\frac{dy}{dw}$ or $\frac{dw}{dy}$ unless appropriate work follows. A1: Correct differentiation |
| $y\frac{dy}{dx}+y^2\tan x = \sin x$; eliminates $y$ to obtain equation in $\frac{dw}{dx}$, $w$ and $x$ only | M1 | Recognisable attempt to eliminate $y$ from original equation |
| $\frac{dw}{dx}+2w(\cot 2x+\tan x) = \frac{2\sin x}{\sin 2x}$ $\Rightarrow \frac{dw}{dx}+2w\csc 2x = \sec x$ * | A1* | Fully correct work leading to given equation. Allow $\cot 2x$ written as $\frac{1}{\tan 2x}$ or $\frac{\cos 2x}{\sin 2x}$ and/or $\tan x$ written as $\frac{\sin x}{\cos x}$. If result in (a) not clearly used, must be full equivalent work. |

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# Question 8(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dw}{dx}+2w\csc 2x = \sec x \Rightarrow \text{IF} = e^{2\int\csc 2x\,dx} = \tan x$ | M1 A1 | M1: $e^{2\int\csc 2x\,dx}$ condoning omission of one or both 2's. A1: $\tan x$ oe. Allow $k\tan x$. Not just $e^{\ln(\tan x)}$ |
| $\Rightarrow w\tan x = \int \tan x \sec x\,dx$ | M1 | Correctly applies integrating factor: $\text{IF}\times w = \int\text{IF}\times\sec x\,dx$. Allow equivalents for $\sec x$. Condone "$y$" for "$w$". |
| $\Rightarrow w\tan x = \sec x (+c)$ | A1 | Correct equation with or without constant |
| $y^2 = w\sin 2x$ and $w\tan x = \sec x + c \Rightarrow \frac{y^2}{\sin 2x}\tan x = \sec x+c$ $\Rightarrow y^2 = \frac{\sin 2x}{\tan x}(\sec x+c) = \frac{2\sin x\cos^2 x}{\sin x}\left(\frac{1}{\cos x}+c\right)$ $\Rightarrow y^2 = 2\cos x + A\cos^2 x$ | ddM1 | Substitutes for $w$ correctly and reaches $y^2=\ldots$. Must be consistent with their equation in $w$ and $x$. **Requires both previous M marks and attempt at integration including "$+c$"** |
| $y^2 = 2\cos x + A\cos^2 x$ | A1 | Any correct $y^2=\ldots$ with RHS fully in terms of $\cos x$. Accept e.g. $y^2=2\cos x+2c\cos^2 x$ or $y^2=\cos x(2+A\cos x)$. Ignore inconsistencies with constant. |

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