## Question 9(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply the form $\frac{Ax+B}{2+3x^2}+\frac{C}{2-x}$ | B1 | If incorrect partial fractions e.g. $A=0$ or $Ax^2+B$ then M1, A1, A0 for correct $C$. Only allow single A1 even if other coefficients correct. B1 recoverable by correct form end statement. |
| Use a correct method for finding a coefficient | M1 | e.g. $(Ax+B)(2-x)+C(2+3x^2) = (3C-A)x^2+(2A-B)x+(2B+2C) = 17x^2-7x+16$ |
| Obtain one of $A=-2$, $B=3$ and $C=5$ | A1 | If error present in above still allow A1 for $C$ |
| Obtain a second value | A1 | |
| Obtain the third value | A1 | Extra term in partial fractions, $D/(2+3x^2)$, that is 4 unknowns $A$, $B$, $C$ and $D$ then B0 unless recover at end. If $B$ or $D=0$ and all coefficients correctly found then allow all A marks but still B0. |

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## Question 9(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct method to find first two terms of expansion: $(2-x)^{-1}=2^{-1}+(-1)\,2^{-2}(-x)+[(-1)(-2)2^{-3}(-x)^2/2!]$, $\left(1+\frac{3x^2}{2}\right)^{-1}=1-\frac{3x^2}{2}$ or $\left(1-\frac{x}{2}\right)^{-1}=1-\left(\frac{-x}{2}\right)$ | M1 | Symbolic coefficients are not sufficient for the M1 |
| $\frac{Ax+B}{2}\left[1+(-1)\frac{3x^2}{2}\cdots\right]$, $A=-2$, $B=3$; and $\frac{C}{2}\left[1+(-1)\left(\frac{-x}{2}\right)+\frac{(-1)(-2)}{2}\left(\frac{-x}{2}\right)^2+\frac{(-1)(-2)(-3)}{6}\left(\frac{-x}{2}\right)^3\cdots\right]$, $C=5$ | A1 FT | Obtain correct un-simplified expansions up to the term in $x^3$ of each partial fraction |
| $=\frac{3-2x}{2}\left(1-\frac{3x^2}{2}\right)+\frac{5}{2}\left(1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}\right)$ $=\left(\frac{3}{2}+\frac{5}{2}\right)+\left(-1+\frac{5}{4}\right)x+\left(-\frac{9}{4}+\frac{5}{8}\right)x^2+\left(\frac{3}{2}+\frac{5}{16}\right)x^3$ | A1 FT | Unsimplified $(2-x)^{-1}$ expanded correctly; error in simplifying before $C$ is involved, allow A1FT when $C$ is introduced. FT is on $A$, $B$, $C$ |
| Multiply expansion of $\left(1+\frac{3x^2}{2}\right)^{-1}$ (must reach $1\pm\frac{3x^2}{2}$) by $Ax+B$, where $AB\neq 0$, up to term in $x^3$ | M1 | Allow either $\pm 2$ or $\pm 2^{-1}$ outside bracket or missing. Allow one error in actual multiplication to acquire the 4 terms. Ignore errors in higher powers. |
| Obtain final answer $4+\frac{1}{4}x-\frac{13}{8}x^2+\frac{29}{16}x^3$, or equivalent | A1 | Maclaurin's Series: $f(0)=4$ B1, $f'(0)=\frac{1}{4}$ B1, $f''(0)=-\frac{13}{4}$ and $f'''(0)=\frac{87}{8}$ B1. If final answer multiplied throughout e.g. by 16 then A0. |

**Total: 5 marks**

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## Question 9(c):

| Answer | Mark | Guidance |
|--------|------|----------|
| State answer $|x|<\sqrt{\frac{2}{3}}$ or $-\sqrt{\frac{2}{3}}<x<\sqrt{\frac{2}{3}}$, clear conclusion required | B1 | Or exact equivalent. Strict inequality. |

**Total: 1 mark**

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