# Question 8 (Direct Expansion / Binomial):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{4-9x} = (4-9x)^{\frac{1}{2}} = 4^{\frac{1}{2}} + \left(\frac{1}{2}\right)4^{-\frac{1}{2}} \times (-9x)^1 + \left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)4^{-\frac{3}{2}} \times \frac{(-9x)^2}{2!} + \left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)4^{-\frac{5}{2}} \times \frac{(-9x)^3}{3!}$ | B1 | For 2 or $\sqrt{4}$ or $4^{\frac{1}{2}}$ as constant term |
| Correct form for term 3 or term 4 e.g. $\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right) \times \frac{(\ldots x)^2}{2!}$ or $\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \times \frac{(\ldots x)^3}{3!}$ where $\ldots \neq 1$ | M1 | Condone missing brackets around $x$ terms; binomial coefficients must be correct. Allow 2! and/or 3! or 2 and/or 6 |
| $\sqrt{4-9x} = 2 - \frac{9x}{4} - \frac{81x^2}{64} - \frac{729x^3}{512}$ | A1 | OR at least 2 correct simplified terms from $-\frac{9x}{4}$, $-\frac{81x^2}{64}$, $-\frac{729x^3}{512}$; condone e.g. $2 + \frac{-9x}{4} - \frac{81x^2}{64} - \frac{729x^3}{512}$ |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| States approximation will be an **overestimate** because all terms after the first one in the expansion are negative | B1 | e.g. "overestimate because the terms are negative"; "overestimate as terms are being taken away (from 2)"; condone "overestimate as every term is negative". Mark depends on having obtained expansion of form $k - px - qx^2 - rx^3$, $k,p,q,r > 0$ |

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# Question 8 (Integration — direct method):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{(x-2)(x-4)}{4\sqrt{x}} = \frac{x^2 - 6x + 8}{4\sqrt{x}} = \frac{1}{4}x^{\frac{3}{2}} - \frac{3}{2}x^{\frac{1}{2}} + 2x^{-\frac{1}{2}}$ | M1, A1 | Correct attempt to write $\frac{(x-2)(x-4)}{4\sqrt{x}}$ as sum of terms with indices; look for at least two different terms with correct index |
| $\int \frac{1}{4}x^{\frac{3}{2}} - \frac{3}{2}x^{\frac{1}{2}} + 2x^{-\frac{1}{2}}\,dx = \frac{1}{10}x^{\frac{5}{2}} - x^{\frac{3}{2}} + 4x^{\frac{1}{2}} (+c)$ | dM1, A1 | Integrates $x^n \to x^{n+1}$ for at least 2 correct indices; dependent on first M |
| Deduces limits are 2 and 4; applies to $\frac{1}{10}x^{\frac{5}{2}} - x^{\frac{3}{2}} + 4x^{\frac{1}{2}}$ | M1 | Independent mark; limits must be applied to a changed function |
| $\left(\frac{32}{10} - 8 + 8\right) - \left(\frac{2}{5}\sqrt{2} - 2\sqrt{2} + 4\sqrt{2}\right) = \frac{16}{5} - \frac{12}{5}\sqrt{2}$; Area $R = \frac{12}{5}\sqrt{2} - \frac{16}{5}$ | A1 | Allow $\frac{16}{5} - \frac{12}{5}\sqrt{2}$; correct working leading to $\frac{12}{5}\sqrt{2} - \frac{16}{5}$ |

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# Question 8 (Integration by parts):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{(x-2)(x-4)}{4\sqrt{x}}\,dx = \frac{1}{2}(x-2)(x-4)x^{\frac{1}{2}} - \int \frac{1}{2}(2x-6)x^{\frac{1}{2}}\,dx$ | M1, A1 | Reaches form $\alpha(x-2)(x-4)x^{\frac{1}{2}} \pm \int(px+q)x^{\frac{1}{2}}\,dx$, $\alpha,p \neq 0$ |
| $= \frac{1}{2}(x-2)(x-4)x^{\frac{1}{2}} - \frac{2}{5}x^{\frac{5}{2}} + 2x^{\frac{3}{2}}$ | dM1, A1 | Attempts $\int(px+q)x^{\frac{1}{2}}\,dx$ by expanding and integrating or parts again |
| Applies limits 2 and 4 | M1 | Limits must be applied to a changed function |
| Area $R = \frac{12}{5}\sqrt{2} - \frac{16}{5}$ | A1 | Allow $\frac{16}{5} - \frac{12}{5}\sqrt{2}$ |

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# Question 8 (Integration by substitution):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \sqrt{x}\ (x=u^2) \Rightarrow \int\frac{(x-2)(x-4)}{4\sqrt{x}}\,dx = \int\frac{(u^2-2)(u^2-4)}{4u}\cdot\frac{dx}{du}\,du = \int\frac{(u^2-2)(u^2-4)}{4u}\cdot 2u\,du$ | M1, A1 | Applies substitution $u=\sqrt{x}$; attempts $k\int\frac{(u^2-2)(u^2-4)}{u}\frac{dx}{du}\,du$ |
| $= \frac{1}{2}\int(u^4 - 6u^2 + 8)\,du = \frac{1}{2}\left(\frac{u^5}{5} - \frac{6u^3}{3} + 8u\right)(+c)$ | dM1, A1 | Expands and integrates $u^n \to u^{n+1}$ for at least 2 correct indices |
| Deduces limits $\sqrt{2}$ and 2; applies to $\frac{1}{2}\left(\frac{u^5}{5} - \frac{6u^3}{3} + 8u\right)$ | M1 | 2 and $\sqrt{2}$ must be substituted either way round with evidence of subtraction |
| $\frac{1}{2}\left(\frac{32}{5} - 16 + 16 - \left(\frac{4\sqrt{2}}{5} - 4\sqrt{2} + 8\sqrt{2}\right)\right)$; Area $R = \frac{12}{5}\sqrt{2} - \frac{16}{5}$ | A1 | Allow $\frac{16}{5} - \frac{12}{5}\sqrt{2}$ |