# Question 13:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $3^{-2}\left(1+\frac{x}{3}\right)^{-2} = 3^{-2}(1+...x+...x^2)$ | M1 | Attempts binomial expansion by taking out factor of $3^{-2}$ or $\frac{1}{3^2}$ or $\frac{1}{9}$, achieves at least first 3 terms |
| $(-2)\left(\frac{x}{3}\right)$ or $\frac{(-2)(-3)}{2!}\left(\frac{x}{3}\right)^2$ | M1 | Correct method to find either the $x$ or $x^2$ term unsimplified. Award for $(-2)(kx)$ or $\frac{(-2)(-2-1)}{2!}(kx)^2$ where $k\neq 1$ |
| $\left(1+\frac{x}{3}\right)^{-2} = 1+(-2)\left(\frac{x}{3}\right)+\frac{(-2)(-3)}{2!}\left(\frac{x}{3}\right)^2$ | A1 | Correct unsimplified or simplified expansion. Condone $\left(-\frac{x}{3}\right)^2$ for $\left(\frac{x}{3}\right)^2$ |
| $3^{-2}\left(1+\frac{x}{3}\right)^{-2} = \frac{1}{9}-\frac{2x}{27}+\frac{x^2}{27}$ | A1 | cao. Allow terms to be listed. Ignore extra terms. isw once correct simplified answer seen |

**Direct expansion alternative:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(3+x)^{-2} = 3^{-2}+3^{-3}\times\alpha x+3^{-4}\times\beta x^2$ | M1 | |
| $(-2)\times 3^{-3}x$ or $\frac{(-2)(-2-1)}{2!}\times 3^{-4}x^2$ | M1 | Condone invisible brackets |
| $3^{-2}-2\times 3^{-3}\times x+\frac{-2(-2-1)}{2!}\times 3^{-4}\times x^2$ | A1 | |
| $\frac{1}{9}-\frac{2x}{27}+\frac{x^2}{27}$ | A1 | Also award for at least 2 correct simplified terms with both method marks scored |

---

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 6x\times\left(\frac{1}{9}-\frac{2x}{27}+\frac{x^2}{27}\right)dx = \int\left(\frac{2x}{3}-\frac{4x^2}{9}+\frac{2x^3}{9}\right)dx = ...$ | M1 | Attempts to multiply expansion from (a) by $6x$ or just $x$ and integrates. Look for $x^n\to x^{n+1}$ at least once. Condone copying slips |
| $\int\left(\frac{2x}{3}-\frac{4x^2}{9}+\frac{2x^3}{9}\right)dx = \frac{x^2}{3}-\frac{4x^3}{27}+\frac{x^4}{18}$ oe | A1 | Correct integration, simplified or unsimplified |
| $\left[\frac{x^2}{3}-\frac{4x^3}{27}+\frac{x^4}{18}\right]_{0.2}^{0.4} = \left(\frac{(0.4)^2}{3}-\frac{4(0.4)^3}{27}+\frac{(0.4)^4}{18}\right)-\left(\frac{(0.2)^2}{3}-\frac{4(0.2)^3}{27}+\frac{(0.2)^4}{18}\right)$ | dM1 | Substituting 0.4 and 0.2 and subtracting either way round (may be implied). Depends on first M mark |
| $= \text{awrt } 0.03304$ or $\frac{223}{6750}$ | A1 | Note actual value is $0.032865...$; answers using additional terms score A0 |

**Integration by parts alternative (taking $u=6x$, $\frac{dv}{dx}=$ expansion):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 6x\times\left(\frac{1}{9}-\frac{2x}{27}+\frac{x^2}{27}\right)dx = kx\times f(x)-k\int f(x)dx = kx\times f(x)-kg(x)$ | M1 | Full attempt at integration by parts. $f(x)$ is attempt to integrate expansion with $x^n\to x^{n+1}$ at least once, $g(x)$ is attempt to integrate $f(x)$ |
| $= 6x\left(\frac{x}{9}-\frac{x^2}{27}+\frac{x^3}{81}\right)-\left(\frac{x^2}{3}-\frac{2x^3}{27}+\frac{6x^4}{324}\right)$ | A1 | Fully correct integration, then dM1A1 as main scheme |

---

## Part (c):

**Main method (substitution $u=3+x$):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Overall problem-solving mark | M1 | Using appropriate integration technique (substitution, parts, or partial fractions); integrates one term to natural logarithm e.g. $\frac{a}{3+x}\to b\ln(3+x)$; substitutes correct limits and subtracts |
| $u=3+x \Rightarrow \int_{3.2}^{3.4}f(u)\,du \Rightarrow \int_{3.2}^{3.4}\frac{6(u-3)}{u^2}\,du = \int_{3.2}^{3.4}\frac{6}{u}-\frac{18}{u^2}\,du \Rightarrow ...\ln u+...u^{-1}$ | M1 | Integrates to achieve $\pm\frac{a}{u}\pm b\ln u$ |
| $\int_{3.2}^{3.4}\frac{6(u-3)}{u^2}\,du = \int_{3.2}^{3.4}\frac{6}{u}-\frac{18}{u^2}\,du \Rightarrow 6\ln u+18u^{-1}$ | A1 | |
| $\left[6\ln u+18u^{-1}\right]_{3.2}^{3.4} = \left(6\ln 3.4+\frac{18}{3.4}\right)-\left(6\ln 3.2+\frac{18}{3.2}\right)=...$ | ddM1 | |
| $6\ln\left(\frac{17}{16}\right)-\frac{45}{136}$ oe | A1 | |

**Alt 1 (integration by parts):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Overall problem-solving mark | M1 | |
| $\int 6x(3+x)^{-2}\,dx = \frac{...x}{3+x}\pm...\int(3+x)^{-1}dx = \frac{...x}{3+x}\pm...\ln(3+x)$ oe | M1 | Condone missing brackets e.g. $...\ln x+3$ for $...\ln(3+x)$ |
| $= 6\ln(3+x)-\frac{6x}{3+x}$ oe | A1 | |
| $\left(6\ln(3.4)-\frac{6(0.4)}{3.4}\right)-\left(6\ln(3.2)-\frac{6(0.2)}{3.2}\right)=...$ | ddM1 | |
| $6\ln\left(\frac{17}{16}\right)-\frac{45}{136}$ oe | A1 | |

**Alt 2 (partial fractions):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Overall problem-solving mark | M1 | |
| $\int 6x(3+x)^{-2}\,dx = \int\left(\frac{...}{(3+x)}+\frac{...}{(3+x)^2}\right)dx = ...\ln(3+x)+\frac{...}{3+x}$ oe | M1 | Condone missing brackets |
| $= 6\ln(3+x)+\frac{18}{3+x}$ oe | A1 | |
| $\left(6\ln(3.4)+\frac{18}{3.4}\right)-\left(6\ln(3.2)+\frac{18}{3.2}\right)=...$ | ddM1 | |
| $6\ln\left(\frac{17}{16}\right)-\frac{45}{136}$ oe | A1 | |

---

## Part (c) — Set notation answer guidance:

| Answer | Score | Notes |
|---|---|---|
| $t...\text{awrt }1.7$ or $t...\text{awrt }4.3$ (any inequality/equation) | 11000 | |
| $t...\frac{5}{3}$ or $t...\frac{13}{3}$ (any inequality/equation) | 11100 | one correct c.v. |
| Both $t<\text{awrt }1.7$ and $t>b$ where $\left\{b>\frac{5}{3}\right\}$ | 11010 | outside region |
| Both $t<a$ and $t>\text{awrt }4.3$ where $\left\{a<\frac{13}{3}\right\}$ | 11010 | outside region |
| Both $t<\frac{5}{3}$ and $t>b$ where $\left\{b>\frac{5}{3}\right\}$ | 11110 | outside region, one correct |
| Both $t<a$ and $t>\frac{13}{3}$ where $\left\{a<\frac{13}{3}\right\}$ | 11110 | outside region, one correct |
| Both $t<\frac{5}{3}$ and $t>\frac{13}{3}$ | 11110 | outside region, one correct |
| $\left\{t:t<\frac{5}{3}\right\}\cup\left\{t:t>\frac{13}{3}\right\}$ | 11111 | fully correct |

## Integration Question (A1 notes):

**Correct integration forms:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitution: $u = x+3 \to 6\ln u + 18u^{-1}$ or $u=(x+3)^2 \to 3\ln u + \frac{18}{\sqrt{u}}$ | A1 | Correct integration for their method |
| Parts: $6\ln(3+x) - \frac{6x}{3+x}$ | A1 | |
| Partial fractions: $6\ln(3+x) + \frac{18}{3+x}$ | A1 | oe e.g. $3\ln(9+6x+x^2)+\frac{18}{3+x}$ |
| Substitutes correct limits, subtracts either way | ddM1 | Depends on both previous method marks |
| $6\ln\!\left(\frac{17}{16}\right) - \frac{45}{136}$ or $3\ln\!\left(\frac{289}{256}\right)-\frac{45}{136}$ or $-6\ln\!\left(\frac{16}{17}\right)-\frac{45}{136}$ | A1 | Exact equivalents allowed e.g. $1.0625$ or $1\frac{1}{16}$; $\frac{45}{136}$ must be exact |

---