Question 3 - A-Level Maths

Edexcel FP2 2010 June — Question 3 7 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2010
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Modulus function
Type	Solve inequality with reciprocal in modulus
Difficulty	Standard +0.8 This FP2 question requires solving a rational inequality, then extending to modulus cases. Part (a) is standard A-level (multiplying by (x+3)² to avoid sign cases), but part (b) requires systematic consideration of \|x+3\| cases and combining solution sets—a moderately challenging extension requiring careful algebraic manipulation and logical reasoning beyond routine exercises.
Spec	1.02g Inequalities: linear and quadratic in single variable 1.02l Modulus function: notation, relations, equations and inequalities

3. (a) Find the set of values of $x$ for which $$x + 4 > \frac { 2 } { x + 3 }$$ (b) Deduce, or otherwise find, the values of $x$ for which $$x + 4 > \frac { 2 } { | x + 3 | }$$

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Question 3:

Part (a)

Answer	Marks	Guidance
Answer	Marks	Guidance
$(x+4)(x+3)^2 - 2(x+3) = 0$, $(x+3)(x^2+7x+10) = 0$ so $(x+2)(x+3)(x+5) = 0$	M1	Or alternative method including calculator
Finds critical values $-2$ and $-5$	A1 A1
Establishes $x > -2$	A1ft
Finds and uses critical value $-3$ to give $-5 < x < -3$	M1 A1	(6 marks)

Part (b)

Answer	Marks	Guidance
Answer	Marks	Guidance
$x > -2$	B1ft	(1 mark)

## Question 3:

**Part (a)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x+4)(x+3)^2 - 2(x+3) = 0$, $(x+3)(x^2+7x+10) = 0$ so $(x+2)(x+3)(x+5) = 0$ | M1 | Or alternative method including calculator |
| Finds critical values $-2$ and $-5$ | A1 A1 | |
| Establishes $x > -2$ | A1ft | |
| Finds and uses critical value $-3$ to give $-5 < x < -3$ | M1 A1 | (6 marks) |

**Part (b)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x > -2$ | B1ft | (1 mark) |

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3. (a) Find the set of values of $x$ for which

$$x + 4 > \frac { 2 } { x + 3 }$$

(b) Deduce, or otherwise find, the values of $x$ for which

$$x + 4 > \frac { 2 } { | x + 3 | }$$

\hfill \mbox{\textit{Edexcel FP2 2010 Q3 [7]}}

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Q1 7 Q2 5 Q3 7 Q4 10 Q5 10 Q6 10 Q7 12 Q8 14