Edexcel A-level Further Maths revision notes

Inequalities (Further Pure 1)

Section FP1-7

Both years

1 specification point

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section FP1-7

Checked against Edexcel 9FM0 section FP1-7. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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FP1-7.1

The manipulation and solution of algebraic inequalities and inequations, including those involving the modulus sign.

Notes

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Explanation

Never multiply an inequality by an expression whose sign you do not know, because a negative multiplier reverses the inequality. Two safe techniques cover everything in this topic.
For a rational inequality, multiply both sides by the square of every denominator, which is positive, then collect everything on one side, factorise and read the solution from a sign table or sketch; remember to exclude the values that make a denominator zero.
For a modulus inequality between two moduli, square both sides, since $|A|<|B|$ is equivalent to $A^2<B^2$ ; for $|A|<k$ with $k>0$ , the equivalent statement is $-k<A<k$ .
A modulus set equal to a linear expression, as in $|f(x)|\le g(x)$ , needs the extra condition $g(x)\ge0$ before squaring, and that condition is a mark in its own right.
Always finish by testing one value from each candidate interval against the original inequality.

Worked example

Solve $\dfrac{1}{x-2}>3$ .

1.Multiply both sides by $(x-2)^2$ , which is positive for $x\ne2$ : $x-2>3(x-2)^2$ .
2.Rearranging, $3(x-2)^2-(x-2)<0$ , that is $(x-2)\left[3(x-2)-1\right]<0$ .
3.So $(x-2)(3x-7)<0$ , with critical values $2$ and $\dfrac73$ .
4.The product is negative between the roots.

Answer: $2<x<\dfrac73$ .

Common mistakes

Don't fall into the trap of multiplying by $x-2$ rather than $(x-2)^2$ , which silently assumes $x>2$ .
Don't fall into the trap of including a value that makes a denominator zero in the final solution set.
Don't fall into the trap of squaring $|f(x)|\le g(x)$ without first requiring $g(x)\ge0$ .

Exam tip

Set the inequality to zero, factorise fully and draw a sign line with every critical value marked; then test one point in each interval before writing the answer.

Tier 1 · Easy

ORIGINAL

Solve the inequality

x^2-1>2(x+1)

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

Solve the inequality

\dfrac{1}{x-2}>3

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

Solve the inequality

\dfrac{1}{x-1}>\dfrac{x}{x-2}

(7)

(Total for Question 1 is 7 marks)

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