Inequalities with rational expressions: critical values, sketches and sign tables

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Further Pure Mathematics FP2 (WFM02/01) is an A Level unit, and this unit, inequalities, the method of differences and further complex numbers, is half of it: on the June 2024, January 2025 and June 2025 FP2 papers it carried 114 of the 225 marks.

Each of those three papers had one rational or modulus inequality question, worth between 6 and 10 marks.

FP2 assumes P1 to P4 and FP1, and this section leans on two pieces of P1: solving a quadratic inequality, and sketching a curve with a vertical asymptote.

The idea: an inequality is a question about where

An equation such as $\frac{x+5}{x-1} = 3$ asks for particular values of $x$ .

An inequality such as

\frac{x+5}{x-1} > 3

asks for a whole set of values: all the $x$ for which the left-hand side is bigger than 3.

Its answer is usually one or more intervals.

Try a few values.

At $x = 2$ the left-hand side is $\frac{7}{1} = 7$ , which is more than 3.

At $x = 5$ it is $\frac{10}{4} = 2.5$ , which is not.

At $x = 0$ it is $\frac{5}{-1} = -5$ , which is not.

At $x = 1$ it is not defined at all.

So the answer is some interval that contains 2 but not 0 or 5, and the edges of that interval are what we need.

The edges can only be at two kinds of place:

where the two sides are equal (here $x + 5 = 3(x - 1)$ , so $x = 4$ ), because a continuous expression can only change from "bigger" to "smaller" by passing through "equal";

where an expression is undefined (here $x = 1$ , the vertical asymptote), because a rational expression can jump from very large positive to very large negative there without passing through 3.

These are the critical values.

Between neighbouring critical values the inequality is either true all the way or false all the way, so one test value in each interval decides it.

Here the intervals are $x < 1$ , $1 < x < 4$ and $x > 4$ , and the tests above give the answer $1 < x < 4$ .

Why you cannot just multiply by the denominator

With an equation you clear fractions by multiplying both sides by $x - 1$ .

With an inequality that is dangerous, because $x - 1$ is positive for some $x$ and negative for others, and multiplying by a negative number reverses an inequality.

A student who multiplies through gets $x + 5 > 3x - 3$ , so $x < 4$ , and wrongly includes every $x$ below 1, such as $x = 0$ , where the left-hand side is $-5$ .

There are two safe routes, and both are credited in full.

Route 1: multiply by the square of the denominator. A square is positive (away from the critical value itself), so multiplying by $(x - 1)^2$ keeps the inequality the right way round:

(x + 5)(x - 1) > 3(x - 1)^2.

Do not expand everything.

Collect on one side and take out the common factor:

(x - 1)\left[(x + 5) - 3(x - 1)\right] > 0 \quad\Longrightarrow\quad (x - 1)(8 - 2x) > 0.

This is a quadratic inequality, solved as in P1: the parabola $y = (x-1)(8-2x)$ opens downwards, crosses at $x = 1$ and $x = 4$ , and is positive between them.

Route 2: one side, one fraction, then a sign table. Subtract so that one side is 0, and combine:

\frac{x + 5}{x - 1} - 3 > 0 \quad\Longrightarrow\quad \frac{x + 5 - 3(x - 1)}{x - 1} > 0 \quad\Longrightarrow\quad \frac{8 - 2x}{x - 1} > 0.

The critical values are where the numerator is 0 ( $x = 4$ ) and where the denominator is 0 ( $x = 1$ ).

A fraction is positive when top and bottom have the same sign:

$x < 1$

$1 < x < 4$

$x > 4$

$8 - 2x$

$+$

$-$

$x - 1$

$-$

$+$

$\frac{8 - 2x}{x - 1}$

$-$

$+$

$-$

Either way the answer is $\{x : 1 < x < 4\}$ .

Why multiplying by a square is safe (for understanding). If $p > q$ and $m > 0$ , then $mp - mq = m(p - q)$ is a positive number times a positive number, so $mp > mq$ .

A square $(x - a)^2$ is positive for every $x \neq a$ , and $x = a$ is excluded anyway because the original fraction is undefined there.

The graphical picture

Drawing both sides makes the critical values visible.

For $\frac{x}{x+1} < \frac{2}{x-3}$ , solved in Worked example 2 below, the two curves have vertical asymptotes at $x = -1$ and $x = 3$ and cross at two more points.

The inequality holds wherever the curve $y = \frac{x}{x+1}$ is below the curve $y = \frac{2}{x-3}$ .

Sketch of y = x/(x + 1), with asymptotes x = −1 and y = 1, and y = 2/(x − 3), with asymptote x = 3 and the x-axis. The curves cross twice, at x about −0.37 and x about 5.37. The first curve is below the second just right of x = −1 up to the first crossing, and just right of x = 3 up to the second crossing.

Read the sketch from left to right: below $x = -1$ the first curve is above the second; just to the right of $x = -1$ the first curve has come up from very large negative values, so it is below; and so on.

Each change happens at an asymptote or at a crossing point, and nowhere else.

A sketch like this, or a sign table, is the most reliable way to read off the final set.

Strict and non-strict, and "inequations"

In a strict inequality ( $<$ or $>$ ) no critical value is included.

In a non-strict inequality ( $\leqslant$ or $\geqslant$ ) the values where the two sides are equal are included, but the values that make a denominator zero are never included: the expression does not exist there.

An inequation is an older word for a statement with $<$ , $>$ , $\leqslant$ , $\geqslant$ or $\neq$ between two expressions. Solving one means finding the set of values that make it true, exactly as here.

Write the answer as intervals, in set notation or in words.

$\{x : -3 < x \leqslant 2\} \cup \{x : x > 5\}$ and " $-3 < x \leqslant 2$ or $x > 5$ " both score.

What loses the final mark is a wrong end, a square bracket or $\leqslant$ at an asymptote, or "and" where "or" is meant: $x < -2$ and $x > 1$ describes no numbers at all.

Connections

This builds on P1 quadratic inequalities, and the sketch is the P1 and P3 skill of drawing rational functions with asymptotes.

The next section adds the modulus.

In the method of differences (the third section) an inequality in $n$ decides the smallest number of terms for a sum to pass a target, and in the loci section a region of the Argand diagram is just an inequality in two variables.

Beyond this course, inequalities are the language of optimisation and of error bounds: "the approximation is within 0.001 for every $n > N$ " is an inequality to be solved.

How the marks work

M1 is a method mark for a correct method applied to this question. dM1 and ddM1 need one or two earlier M marks. A1 is an accuracy mark and needs its M mark. B1 is an independent mark for a correct fact or step.

A1* marks a printed answer ("show that"): every step must be on the page, with the line before the result, and no slip anywhere. A1cso means correct solution only. ft (follow through) lets a later mark use your own earlier wrong value.

awrt means any answer which rounds to the value given; isw means later work is ignored once a correct answer is seen. A misread costs two of the A or B marks in that part.

A "Hence" part must use the part before it: any other route scores 0, not half. Under the banner saying solutions relying entirely on calculator technology are not acceptable, a bare answer scores 0.

A quadratic earns its method mark only if it is solved by factorising, by the formula with the values in, or by completing the square, and a formula must be quoted or correctly used. Exact answers lose credit when given only as decimals.

Key terms

Inequation: a statement that two expressions are related by $<$ , $>$ , $\leqslant$ , $\geqslant$ or $\neq$ ; you need to know the manipulation and solution of algebraic inequalities and inequations, including those involving the modulus sign.

Critical values: the values of $x$ at which the two sides are equal, together with the values at which an expression is undefined (the vertical asymptotes). The truth of the inequality can change only at these values.

Solution set: all the values of $x$ for which the inequality is true, written as intervals in set notation or with "or" between them.

Formulae: none to look up. Solving a quadratic by the formula uses $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , which is on the P1 list of formulae to remember: learn it (it is not in the booklet).

Worked example 1: one side, one fraction

Use algebra to determine the values of $x$ for which

\frac{x + 5}{x - 1} > 3. \qquad (4)

1. Critical value from the denominator: $x = 1$ B1

2. $\dfrac{x + 5}{x - 1} - 3 > 0 \;\Rightarrow\; \dfrac{8 - 2x}{x - 1} > 0$ M1 (one side zero, combined into one fraction)

3. Other critical value: $8 - 2x = 0$ , $x = 4$ A1

4. Sign table: positive only for $1 < x < 4$ , so $\{x : 1 < x < 4\}$ A1

Answer: $1 < x < 4$ .

Check with $x = 2$ : $\frac{7}{1} = 7 > 3$ , true; with $x = 0$ : $-5 > 3$ , false.

The answer is a single interval, closed off on the left by the asymptote.

Worked example 2: two asymptotes and surd critical values

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Use algebra to determine the values of $x$ for which

\frac{x}{x + 1} < \frac{2}{x - 3}. \qquad (6)

1. Critical values from the denominators: $x = -1$ , $x = 3$ B1

2. Multiply by $(x + 1)^2(x - 3)^2$ : $x(x + 1)(x - 3)^2 < 2(x - 3)(x + 1)^2$ , so $(x + 1)(x - 3)\left[x(x - 3) - 2(x + 1)\right] < 0$ , giving $(x + 1)(x - 3)(x^2 - 5x - 2) < 0$ M1 (reduced to a quadratic for the remaining critical values)

3. $x^2 - 5x - 2 = 0$ : $x = \dfrac{5 \pm \sqrt{25 + 8}}{2}$ M1 (formula with values in)

4. $x = \dfrac{5 - \sqrt{33}}{2}$ , $x = \dfrac{5 + \sqrt{33}}{2}$ A1

5. Order: $-1 < \frac{5 - \sqrt{33}}{2} < 3 < \frac{5 + \sqrt{33}}{2}$ , and the quartic $(x + 1)(x - 3)(x^2 - 5x - 2)$ is negative between the first and second and between the third and fourth M1 (correct shape of set: two bounded intervals)

6. $\left\{x : -1 < x < \dfrac{5 - \sqrt{33}}{2}\right\} \cup \left\{x : 3 < x < \dfrac{5 + \sqrt{33}}{2}\right\}$ A1

Answer: $-1 < x < \frac{5 - \sqrt{33}}{2}$ or $3 < x < \frac{5 + \sqrt{33}}{2}$ .

As decimals the ends are about $-0.372$ and $5.372$ , which agrees with the sketch above: the first curve is below the second just to the right of each asymptote.

Decimal critical values alone would lose the A marks here, because exact values are expected.

Worked example 3: intersections first, then the set

The curves $C_1$ and $C_2$ have equations

C_1: y = \frac{x + 2}{x - 1}, \quad x \neq 1, \qquad\qquad C_2: y = \frac{2x}{x + 3}, \quad x \neq -3.

(a) Use algebra to determine the exact $x$ coordinates of the points where $C_1$ and $C_2$ meet. (3)

(b) Hence determine the values of $x$ for which $\dfrac{x + 2}{x - 1} \geqslant \dfrac{2x}{x + 3}$ . (3)

1. (a) $(x + 2)(x + 3) = 2x(x - 1)$ , so $x^2 + 5x + 6 = 2x^2 - 2x$ M1 (fractions cleared)

2. $x^2 - 7x - 6 = 0$ , so $x = \dfrac{7 \pm \sqrt{49 + 24}}{2}$ M1

3. $x = \dfrac{7 - \sqrt{73}}{2}$ and $x = \dfrac{7 + \sqrt{73}}{2}$ A1

4. (b) Critical values $-3$ , $\frac{7 - \sqrt{73}}{2}$ (about $-0.772$ ), $1$ , $\frac{7 + \sqrt{73}}{2}$ (about $7.772$ ) B1 (asymptotes added to the answers from (a))

5. $\dfrac{x + 2}{x - 1} - \dfrac{2x}{x + 3} = \dfrac{-(x^2 - 7x - 6)}{(x - 1)(x + 3)}$ , which is $\geqslant 0$ on the second and fourth intervals, including the points where the curves meet M1

6. $-3 < x \leqslant \dfrac{7 - \sqrt{73}}{2}$ or $1 < x \leqslant \dfrac{7 + \sqrt{73}}{2}$ A1

Answer: $\left\{x : -3 < x \leqslant \frac{7 - \sqrt{73}}{2}\right\} \cup \left\{x : 1 < x \leqslant \frac{7 + \sqrt{73}}{2}\right\}$ .

The ends at the asymptotes are strict and the ends at the meeting points are included, because the inequality is non-strict.

Part (b) uses the two values found in (a): they are two of the four critical values.

Exam tip

Before you do any algebra, write down the critical values from the denominators: they are worth a mark of their own and they are never in the answer.

Then reduce to a quadratic (factorise out the common brackets; do not expand to a quartic), and finish with a sketch or a sign table.

Expanding to a quartic and solving it on a calculator was capped in the June 2024 scheme, and under the calculator banner the quadratic must be solved by a method you show.

Common mistakes

Multiplying by a denominator that can be negative. It reverses the inequality for half the values of $x$ . Multiply by the square, or bring everything to one side.

Losing an asymptote value. The June 2024 examiners' report notes candidates who turned $(x + 2)$ into $(x - 2)$ and so lost a critical value. Copy the denominators carefully and list their zeros first.

Multiplying by the squares and then expanding everything. The January 2025 and June 2025 reports describe cubics and quartics, spurious values and lost marks. Take out $(x - a)(x - b)$ as a common factor before expanding anything.

Including an asymptote value. Even with $\leqslant$ , a value that makes a denominator zero is excluded. Only the values where the two sides are equal take the closed end.

Getting the final set wrong. The reports on all three series say this is where the marks went. Check one value in every interval, or use the sketch, and write "or" (or $\cup$ ) between separate intervals, never "and".

Check your understanding

1. Use algebra to determine the values of $x$ for which

\frac{2x - 1}{x + 2} < 1. \qquad (4)

Total [4]

2. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Use algebra to determine the values of $x$ for which

\frac{4}{x - 2} \geqslant \frac{x}{x + 3}. \qquad (6)

Total [6]

3. A student solves $\dfrac{x + 1}{x - 2} > 2$ by multiplying both sides by $x - 2$ , obtaining $x + 1 > 2x - 4$ and so $x < 5$ .
(a) Show that the value $x = 0$ satisfies the student's answer but not the original inequality. (2)
(b) Use algebra to determine the correct set of values of $x$ . (3) Total [5]

4. The curves $C_1$ and $C_2$ have equations $y = \dfrac{x + 4}{x - 2}$ , $x \neq 2$ , and $y = \dfrac{x}{x + 2}$ , $x \neq -2$ .
(a) Use algebra to determine the $x$ coordinate of the point where $C_1$ and $C_2$ meet. (2)
(b) Hence determine the values of $x$ for which $\dfrac{x + 4}{x - 2} > \dfrac{x}{x + 2}$ . (3) Total [5]

5. Use algebra to determine the values of $x$ for which

\frac{x^2 + x - 2}{x - 3} \leqslant x + 5. \qquad (4)

Total [4]

6. The constant $a$ is positive.
(a) Show that the set of values of $x$ for which $\dfrac{x}{x - a} > 2$ is $\{x : a < x < 2a\}$ . (3)
(b) Hence determine the set of values of $x$ for which $\dfrac{x}{x - a} \leqslant 2$ . (2) Total [5]

Answers

1. Critical value $x = -2$ B1; $\dfrac{2x - 1}{x + 2} - 1 = \dfrac{x - 3}{x + 2} < 0$ M1; $x = 3$ A1; $-2 < x < 3$ A1

2. Critical values $x = 2$ , $x = -3$ B1; multiply by $(x - 2)^2(x + 3)^2$ : $(x - 2)(x + 3)\left[4(x + 3) - x(x - 2)\right] \geqslant 0$ , so $(x - 2)(x + 3)(x^2 - 6x - 12) \leqslant 0$ M1; $x = \dfrac{6 \pm \sqrt{36 + 48}}{2}$ M1; $x = 3 \pm \sqrt{21}$ A1; order $-3 < 3 - \sqrt{21} < 2 < 3 + \sqrt{21}$ , closed only at the surds M1; $-3 < x \leqslant 3 - \sqrt{21}$ or $2 < x \leqslant 3 + \sqrt{21}$ A1

3. (a) $x = 0$ satisfies $x < 5$ M1; but $\dfrac{0 + 1}{0 - 2} = -\dfrac{1}{2}$ , which is not greater than 2, so the student's set contains values that do not satisfy the inequality; the step that fails is multiplying by $x - 2$ , which is negative for $x < 2$ A1*
(b) $\dfrac{x + 1}{x - 2} - 2 = \dfrac{5 - x}{x - 2} > 0$ M1; critical values 2 and 5 A1; $2 < x < 5$ A1

4. (a) $(x + 4)(x + 2) = x(x - 2)$ , so $x^2 + 6x + 8 = x^2 - 2x$ M1; $x = -1$ A1
(b) Critical values $-2$ , $-1$ , $2$ B1; $\dfrac{x + 4}{x - 2} - \dfrac{x}{x + 2} = \dfrac{8(x + 1)}{(x - 2)(x + 2)} > 0$ M1; $-2 < x < -1$ or $x > 2$ A1

5. $\dfrac{x^2 + x - 2 - (x + 5)(x - 3)}{x - 3} \leqslant 0$ M1; $\dfrac{13 - x}{x - 3} \leqslant 0$ A1; critical values 3 and 13 M1; $x < 3$ or $x \geqslant 13$ A1

6. (a) $\dfrac{x}{x - a} - 2 = \dfrac{x - 2x + 2a}{x - a} = \dfrac{2a - x}{x - a}$ M1; critical values $a$ and $2a$ , with $a < 2a$ because $a > 0$ M1; the fraction is positive only when $x - a$ and $2a - x$ are both positive, that is $a < x < 2a$ , so the set is $\{x : a < x < 2a\}$ for every positive $a$ A1*
(b) Every $x$ except $x = a$ (where the fraction is undefined) satisfies exactly one of the two inequalities, so the set is the rest of the real line from (a) with $x = a$ removed; at $x = 2a$ the fraction equals 2, so $2a$ is included M1; $\{x : x < a\} \cup \{x : x \geqslant 2a\}$ A1

Inequalities with rational expressions: critical values, sketches and sign tables

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