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Date	November Example questions	Marks available	1	Reference code	EXN.2.AHL.TZ0.7
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term		Question number	7	Adapted from	N/A

Question

Consider the identity $\frac{2 + 7 x}{(1 + 2 x) (1 - x)} \equiv \frac{A}{1 + 2 x} + \frac{B}{1 - x}$ , where $A, B \in ℤ$ .

Find the value of $A$ and the value of $B$ .

[3]

Hence, expand $\frac{2 + 7 x}{(1 + 2 x) (1 - x)}$ in ascending powers of $x$ , up to and including the term in $x^{2}$ .

[4]

Give a reason why the series expansion found in part (b) is not valid for $x = \frac{3}{4}$ .

[1]

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

$2 + 7 x \equiv A (1 - x) + B (1 + 2 x)$

EITHER

substitutes $x = 1$ and attempts to solve for $B$ and substitutes $x = - \frac{1}{2}$ and attempts to solve for $A$ (M1)

$9 = 3 B \Rightarrow B = 3; \frac{3 A}{2} = - \frac{3}{2} \Rightarrow A = - 1$

forms $A + B = 2$ and $- A + 2 B = 7$ and attempts to solve for $A$ and $B$ (M1)

THEN

$A = - 1$ and $B = 3$ A1A1

[3 marks]

uses the binomial expansion on either $3 {(1 - x)}^{- 1}$ or ${(1 + 2 x)}^{- 1}$ M1

$3 {(1 - x)}^{- 1} = 3 (1 + x + x^{2} + \dots)$ A1

${(1 + 2 x)}^{- 1} = (1 - 2 x + \frac{(- 1) (- 2)}{2!} {(2 x)}^{2} + \dots) (= 1 - 2 x + 4 x^{2} + \dots)$ A1

$3 + 3 x + 3 x^{2} - (1 - 2 x + 4 x^{2})$

so the expansion is $2 + 5 x - x^{2}$ (in ascending powers of $x$ ) A1

[4 marks]

${(1 + 2 x)}^{- 1}$ (is convergent) requires $|x| < \frac{1}{2}$ and $x = \frac{3}{4}$ is outside this so the expansion is not valid R1

[1 mark]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » AHL 1.10—Perms and combs, binomial with negative and fractional indices

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22M.2.AHL.TZ2.9b:
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19M.1.AHL.TZ1.H_11c:
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EXM.3.AHL.TZ0.1d:
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EXM.3.AHL.TZ0.1f:
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Topic 1—Number and algebra » AHL 1.11—Partial fractions

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