FP1-7 Inequalities (Further Pure 1) — revision question pack

1 specification points · notes, questions, answers and worked methods

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.

How this checking works

FP1-7.1 · The manipulation and solution of algebraic inequalities and inequations, including those involving the modulus sign.

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

1.
Solve the inequality $x^2-1>2(x+1)$ .
(3)
(Total for Question 1 is 3 marks)
2.
Solve the inequality $|2x-1|<5$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Solve the inequality $\dfrac{1}{x-2}>3$ .
(5)
(Total for Question 1 is 5 marks)
2.
Solve the inequality $\dfrac{x}{x+1}\le2$ .
(5)
(Total for Question 2 is 5 marks)
3.
Solve the inequality $|x-3|>|x+1|$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Solve the inequality $\dfrac{1}{x-1}>\dfrac{x}{x-2}$ .
(7)
(Total for Question 1 is 7 marks)
2.
Solve the inequality $\left|x^2-4\right|\le3x$ .
(7)
(Total for Question 2 is 7 marks)
3.
Solve the inequality $\dfrac{x+1}{x-1}<\dfrac{x-2}{x+2}$ .
(7)
(Total for Question 3 is 7 marks)
4.
Solve the inequality $\dfrac{x^2-5x+6}{x+1}\ge0$ , showing your sign table.
(6)
(Total for Question 4 is 6 marks)
5.
Solve the inequality $\left|\dfrac{x}{x-2}\right|<3$ .
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FP1-7.1 · The manipulation and solution of algebraic inequalities and inequations, including those involving the modulus sign.

Tier 1 · Easy

Mark scheme for FP1-7.1 Tier 1 · Easy
Question	Scheme	Marks
1	$x^2-2x-3>0$ $(x-3)(x+1)>0$ $x<-1$ or $x>3$	3
(3 marks)		3
Notes
Expand and collect on one side, factorise, and note that a positive quadratic product lies outside its roots. Independent check: at $x=0$ the original reads $-1>2$ , which is false, confirming that the interval between the roots is excluded; at $x=4$ it reads $15>10$ , which is true.
2	$-5<2x-1<5$ $-4<2x<6$ $-2<x<3$	3
(3 marks)		3
Notes
A modulus less than a positive constant is equivalent to a double inequality. Independent check by squaring: $(2x-1)^2<25$ gives $4x^2-4x-24<0$ , that is $(x-3)(x+2)<0$ , the same interval.

Tier 2 · Standard

Mark scheme for FP1-7.1 Tier 2 · Standard
Question	Scheme	Marks
1	Multiplying by $(x-2)^2>0$ gives $x-2>3(x-2)^2$ $(x-2)\left[3(x-2)-1\right]<0$ $(x-2)(3x-7)<0$ $2<x<\dfrac73$	5
(5 marks)		5
Notes
Multiplying by the square of the denominator preserves the inequality; factorising and using a sign line gives the interval between the critical values. Independent check: at $x=2.1$ the left side is $10>3$ , which is true, while at $x=3$ it is $1>3$ , which is false, and at $x=0$ it is $-0.5>3$ , also false.
2	Multiplying by $(x+1)^2>0$ gives $x(x+1)\le2(x+1)^2$ $0\le(x+1)\left[2(x+1)-x\right]=(x+1)(x+2)$ $(x+1)(x+2)\ge0$ gives $x\le-2$ or $x\ge-1$ $x=-1$ must be excluded, so the solution is $x\le-2$ or $x>-1$	5
(5 marks)		5
Notes
Multiply by the squared denominator, collect on one side and factorise; the critical value $x=-1$ is a root of the factorised expression but makes the original fraction undefined. Independent check: at $x=-3$ the left side is $1.5\le2$ , true; at $x=-1.5$ it is $3\le2$ , false; at $x=0$ it is $0\le2$ , true.
3	Squaring both sides gives $x^2-6x+9>x^2+2x+1$ $-8x>-8$ $x<1$	4
(4 marks)		4
Notes
Both sides are non-negative, so squaring is reversible; the quadratic terms cancel, leaving a linear inequality. Independent check: $x=1$ is equidistant from $3$ and $-1$ , and points to its left are nearer $-1$ , so $\|x-3\|$ is the larger there; at $x=0$ the inequality reads $3>1$ , which is true.

Tier 3 · Hard

Mark scheme for FP1-7.1 Tier 3 · Hard
Question	Scheme	Marks
1	Multiplying by $(x-1)^2(x-2)^2>0$ gives $(x-1)(x-2)^2>x(x-1)^2(x-2)$ $(x-1)(x-2)\left[(x-2)-x(x-1)\right]>0$ $(x-1)(x-2)\left(-x^2+2x-2\right)>0$ $x^2-2x+2=(x-1)^2+1>0$ for all $x$ , so the inequality becomes $(x-1)(x-2)<0$ $1<x<2$	7
(7 marks)		7
Notes
Multiplying by the product of both squared denominators avoids any sign assumption; after factorising, the quadratic factor is always positive and can be divided out, reversing the inequality because of its leading minus sign. Independent check: at $x=1.5$ the two sides are $2$ and $-3$ , so the inequality holds; at $x=3$ they are $0.5$ and $3$ , so it fails; at $x=0$ they are $-1$ and $0$ , so it fails.
2	A modulus is never negative, so $3x\ge0$ and hence $x\ge0$ The inequality is equivalent to $-3x\le x^2-4\le3x$ $x^2-3x-4\le0$ gives $(x-4)(x+1)\le0$ , that is $-1\le x\le4$ $x^2+3x-4\ge0$ gives $(x+4)(x-1)\ge0$ , that is $x\le-4$ or $x\ge1$ Intersecting all three conditions gives $1\le x\le4$	7
(7 marks)		7
Notes
State the necessary condition $3x\ge0$ before removing the modulus, then solve the two resulting quadratic inequalities and intersect all three solution sets. Independent check: at $x=2$ the inequality reads $0\le6$ , true; at $x=0.5$ it reads $3.75\le1.5$ , false; at $x=5$ it reads $21\le15$ , false.
3	Multiplying by $(x-1)^2(x+2)^2>0$ gives $(x+1)(x-1)(x+2)^2<(x-2)(x+2)(x-1)^2$ $(x-1)(x+2)\left[(x+1)(x+2)-(x-2)(x-1)\right]<0$ $(x+1)(x+2)-(x-2)(x-1)=\left(x^2+3x+2\right)-\left(x^2-3x+2\right)=6x$ $6x(x-1)(x+2)<0$ , that is $x(x-1)(x+2)<0$ $x<-2$ or $0<x<1$	7
(7 marks)		7
Notes
Multiply by both squared denominators, take out the common factor $(x-1)(x+2)$ and simplify the bracket, which collapses to $6x$ . A sign line with critical values $-2$ , $0$ and $1$ gives the two intervals where the cubic is negative. Independent check: at $x=-3$ the two sides are $0.5$ and $5$ , so the inequality holds; at $x=0.5$ they are $-3$ and $-0.6$ , so it holds; at $x=-1$ they are $0$ and $-3$ , so it fails; at $x=2$ they are $3$ and $0$ , so it fails.
4	The numerator factorises as $(x-2)(x-3)$ The critical values are $x=-1$ , $x=2$ and $x=3$ For $x<-1$ the expression is negative; for $-1<x<2$ it is positive; for $2<x<3$ it is negative; for $x>3$ it is positive The numerator vanishes at $x=2$ and $x=3$ , which are included $-1<x\le2$ or $x\ge3$	6
(6 marks)		6
Notes
Factorise the numerator, mark the three critical values on a sign line and record the sign of each factor in every interval; the quotient is non-negative where the signs multiply to a positive value, together with the zeros of the numerator. Independent check: at $x=0$ the expression is $6>0$ , true; at $x=2.5$ it is $\tfrac{-0.25}{3.5}<0$ , false; at $x=-2$ it is $\tfrac{20}{-1}<0$ , false; at $x=4$ it is $\tfrac{2}{5}>0$ , true.
5	Both sides are non-negative, so squaring gives $x^2<9(x-2)^2$ $9(x-2)^2-x^2>0$ $\left[3(x-2)-x\right]\left[3(x-2)+x\right]>0$ , that is $(2x-6)(4x-6)>0$ $(x-3)(2x-3)>0$ $x<\dfrac32$ or $x>3$	7
(7 marks)		7
Notes
Squaring is valid because both sides are non-negative, and the resulting difference of two squares factorises directly. The excluded value $x=2$ already lies between the two critical values, so it is not in the solution set. Independent check: at $x=0$ the left side is $0<3$ , true; at $x=2.5$ it is $5<3$ , false; at $x=4$ it is $2<3$ , true; at $x=1.5$ it is exactly $3$ , so the strict inequality correctly excludes it.