Edexcel FP2 2002 June — Question 5 7 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2002
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypeRational inequality algebraically
DifficultyModerate -0.3 This is a straightforward rational inequality requiring standard algebraic manipulation: multiply by x² (always positive), rearrange to 2x²-5x-3>0, factor or use quadratic formula, then determine sign regions. While it's Further Maths (FP2), it's a routine single-part question testing a well-practiced technique with no conceptual surprises, making it slightly easier than average.
Spec1.02g Inequalities: linear and quadratic in single variable
5. Using algebra, find the set of values of \(x\) for which \(2 x - 5 > \frac { 3 } { x }\).
Question 5:
AnswerMarks Guidance
AnswerMarks Guidance
\((x>0)\ 2x^2-5x>3\) or \(2x^2-5x=3\)M1
\((2x+1)(x-3)\), critical values \(-\frac{1}{2}\) and \(3\)A1, A1
\(x>3\)A1 ft
\(x<0\quad 2x^2-5x<3\)M1
Using critical value 0: \(-\frac{1}{2}M1, A1 ft
Alt: \(2x-5-\frac{3}{x}<0\) or \((2x-5)x^2>3x\)M1
\(\frac{(2x+1)(x-3)}{x}>0\) or \(x(2x+1)(x-3)>0\)M1, A1
Critical values \(-\frac{1}{2}\) and \(3\), \(x>3\)A1, A1 ft
Using critical value 0: \(-\frac{1}{2}M1, A1 ft (7 marks) 
# Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x>0)\ 2x^2-5x>3$ or $2x^2-5x=3$ | M1 | |
| $(2x+1)(x-3)$, critical values $-\frac{1}{2}$ and $3$ | A1, A1 | |
| $x>3$ | A1 ft | |
| $x<0\quad 2x^2-5x<3$ | M1 | |
| Using critical value 0: $-\frac{1}{2}<x<0$ | M1, A1 ft | |
| **Alt:** $2x-5-\frac{3}{x}<0$ or $(2x-5)x^2>3x$ | M1 | |
| $\frac{(2x+1)(x-3)}{x}>0$ or $x(2x+1)(x-3)>0$ | M1, A1 | |
| Critical values $-\frac{1}{2}$ and $3$, $x>3$ | A1, A1 ft | |
| Using critical value 0: $-\frac{1}{2}<x<0$ | M1, A1 ft | **(7 marks)** |

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5. Using algebra, find the set of values of $x$ for which $2 x - 5 > \frac { 3 } { x }$.\\

\hfill \mbox{\textit{Edexcel FP2 2002 Q5 [7]}}
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Q1 5 Q2 10 Q3 13 Q4 18 Q5 7 Q6 11 Q7 14 Q8 15 Q9 7 Q10 12