## Question 10(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply the form $\dfrac{A}{1+2x} + \dfrac{B}{3-x} + \dfrac{C}{(3-x)^2}$ | **B1** | Alternative form: $\dfrac{A}{1+2x} + \dfrac{Dx+E}{(3-x)^2}$ |
| Use a correct method to find a constant | **M1** | Incorrect format for partial fractions: Allow M1 and a possible A1 if obtain one of these correct values. Max 2/5. Allow M1 even if multiply up by $(1+2x)(3-x)^3$. |
| Obtain one of $A = 2$, $B = 2$ and $C = -3$ | **A1** | Alternative form: obtain one of $A = 2$, $D = -2$ and $E = 3$ |
| Obtain a second value | **A1** | |
| Obtain the third value | **A1** | Do not need to substitute values back into original form. If $\dfrac{A}{1+2x} + \dfrac{B}{3-x} + \dfrac{Cx+D}{(3-x)^2}$ B0 but M1 A1 for $A$, A1 for $B$ and A1 for $C$ and $D$. If $C = 0$ then recovers B1 from above. |

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## Question 10(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| Use a correct method to obtain the first two terms of one of the unsimplified expansions $(1+2x)^{-1}$, $\left(1-\frac{1}{3}x\right)^{-1}$, $\left(1-\frac{1}{3}x\right)^{-2}$, $(3-x)^{-1}$, $(3-x)^{-2}$ | **M1** | $(1+2x)^{-1} = 1 + (-1)(2x) + \ldots$ $\left(1-\frac{1}{3}x\right)^{-1} = 1 + (-1)(-x/3) + \ldots$ $\left(1-\frac{1}{3}x\right)^{-2} = 1 + (-2)(-x/3) + \ldots$ $(3-x)^{-1} = 3^{-1} + (-1)3^{-2}(-x)\ldots$ $(3-x)^{-2} = 3^{-2} + (-2)3^{-3}(-x) + \ldots$ |
| Obtain the correct unsimplified expansions up to the term in $x^2$ for each partial fraction. If correct, should be working with $\dfrac{2}{1+2x} + \dfrac{2}{3-x} - \dfrac{3}{(3-x)^2}$ or $\dfrac{2}{1+2x} + \dfrac{-2x+3}{(3-x)^2}$ | **A1 FT** | Follow through on their $A$, $B$, $C$: $A(1+(-1)(2x)+((-1)(-2)/2)(2x)^2+\ldots)$ $\frac{B}{3}(1+(-1)(-x/3)+((-1)(-2)/2)(-x/3)^2+\ldots)$ $\frac{C}{3^2}(1+(-2)(-x/3)+((-2)(-3)/2)(-x/3)^2+\ldots)$ Must be their coefficients from (a) but may be unsimplified expansions for FT marks. If correct, expect to see $2(1-2x+(2x)^2)$ or $2-4x+8x^2$ $\frac{2}{3}\left(1+\frac{x}{3}+\left(\frac{x}{3}\right)^2\right)$ or $\frac{2}{3}+\frac{2}{9}x+\frac{2}{27}x^2$ $-\frac{1}{3}\left(1+\frac{2x}{3}+(3)\left(\frac{x}{3}\right)^2\right)$ or $-\frac{1}{3}-\frac{2}{9}x-\frac{x^2}{9}$ |
| **A1 FT** (second partial fraction) | **A1 FT** | |
| **A1 FT** (third partial fraction) | **A1 FT** | |
| Obtain final answer $\dfrac{7}{3} - 4x + \dfrac{215}{27}x^2$ | **A1** | Accept $2\frac{1}{3} - 4x + 7\frac{26}{27}x^2$. No ISW. |

## Question 10(b) – Alternative Method (Partial Fractions):

| Answer | Marks | Guidance |
|--------|-------|----------|
| For the form $\frac{A}{1+2x} + \frac{Dx+E}{(3-x)^2}$ | M1* | For the first two terms of an expanded partial fraction, following their $A$, $D$, $E$ |
| $A(1+(-1)(2x)+((-1)(-2)/2)(2x)^2+\ldots) + (Dx+E)\frac{1}{3^2}(1+(-2)(-x/3)+((-2)(-3)/2)(-x/3)^2+\ldots)$, i.e. $2(1-2x+(2x)^2+\ldots) + \frac{-2x+3}{3^2}(1+\frac{2x}{3}+(3)(\frac{x}{3})^2+\ldots)$ | A1FT | |
| Multiply out fully | DM1 | Provided $DE\neq 0$. Ignore cubic terms and above. Allow error in one term but all terms must be present. If correct, expect to see $2-4x+8x^2-\frac{2}{9}x-\frac{4}{27}x^2+\frac{1}{3}+\frac{2}{9}x+\frac{1}{9}x^2$ |
| Obtain final answer $\frac{7}{3}-4x+\frac{215}{27}x^2$ | A1 | Accept $2\frac{1}{3}-4x+7\frac{26}{27}x^2$. No ISW |

## Question 10(b) – Alternative Method (Maclaurin's Series):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct derivatives for $A(1+2x)^{-1}$, $B(3-x)^{-1}$ and $C(3-x)^{-2}$: $(-1)(2)A(1+2x)^{-2}$, $(-1)(-1)B(3-x)^{-2}$ and $(-2)(-1)C(3-x)^{-3}$ | B1FT | |
| One of: $(-2)(2)(-1)(2)A(1+2x)^{-3}$, $(-2)(-1)(-1)(-1)B(3-x)^{-3}$ and $(-3)(-1)(-2)(-1)C(3-x)^{-4}$ | B1FT | |
| All correct | B1FT | |
| Substitute in $f(0)+xf'(0)+\frac{x^2}{2}f''(0)$ | M1 | |
| Obtain final answer $\frac{7}{3}-4x+\frac{215}{27}x^2$ | A1 | Accept $2\frac{1}{3}-4x+7\frac{26}{27}x^2$. No ISW |
| | **5** | |