# Question 6:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\sin x\times-\sin x-\cos x\times\cos x}{\sin^2 x}$ may be implied by $\frac{-\sin^2 x-\cos^2 x}{\sin^2 x}$ | M1 | Or $-\sin x\times\frac{1}{\sin x}+\cos x\times(-\sin x)^{-2}\times\cos x$ oe; allow sign errors only; if **M0**, **SC1** for just $\frac{-\sin^2 x-\cos^2 x}{\sin^2 x}=\frac{-1}{\sin^2 x}$ |
| e.g. $=\frac{-(\sin^2 x+\cos^2 x)}{\sin^2 x}$ and completion to $\frac{-1}{\sin^2 x}$ **AG** | A1 | Need to see at least two correct constructive steps and statement of given answer; NB $\sin^2x+\cos^2x=1$ seen may be a constructive intermediate step |

## Question 6(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos 2x = 2\cos^2 x - 1$ substituted in numerator | M1 | Or alternative form of double angle formula plus Pythagoras leading to no term in $\sin^2 x$ in numerator; may be awarded if not seen as part of fraction |
| $\sin 2x = 2\sin x \cos x$ substituted in denominator | M1 | |
| $\frac{\sqrt{2}\cos x}{2\sin^2 x \cos x}$ | A1 | NB $\int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \frac{1}{\sqrt{2}\sin^2 x}\,dx$ |
| $F[x] = \pm k\frac{\cos x}{\sin x}$ | M1* | $k$ must not be obtained from square rooting a negative number; NB $-\frac{\cos x}{\sqrt{2}\sin x}$ |
| $F\left[\frac{1}{4}\pi\right] - F\left[\frac{1}{6}\pi\right]$ | M1dep* | e.g. $\frac{-\cos\frac{\pi}{4}}{\sqrt{2}\times\sin\frac{\pi}{4}} - \frac{-\cos\frac{\pi}{6}}{\sqrt{2}\times\sin\frac{\pi}{6}}$; at least one correct intermediate step following substitution needed as well as statement of given result; e.g. $-\frac{\sqrt{2}}{2}(1-\sqrt{3})$ |
| $= \frac{1}{2}(\sqrt{6}-\sqrt{2})$ www | A1 **[6]** | |

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