Harder Inequalities and Curved Regions

x in the denominator: multiply by x², not x, so the sign does not flip

In the exam for Edexcel International AS Maths · 8 real past-paper questions below

x on the bottom: times x²

You can already solve quadratic inequalities from their roots. AS questions add a trap: an inequality with $x$ on the bottom of a fraction, like $\frac{5}{x} < 2$ .

To clear the fraction you want to multiply by $x$ . But if $x$ is negative, the sign must flip, and you do not know the sign of $x$ . Multiply by $x^{2}$ instead. It is always positive, so the sign stays.

Formula

\frac{a}{x} < b \Rightarrow a x < b x^{2}

$x^{2}$ — the safe multiplier. It is positive for every $x \neq = 0$ , so < stays <.
$a$ — the number on top. It ends up times $x$ .
$b$ — the number on the right. It ends up times $x^{2}$ .

Memory trick

A square is never negative. Square = safe: times $x^{2}$ , and the sign never flips.

Worked example

1
Solve $\frac{5}{x} < 2$ .
2
$\frac{5}{x} < 2, x \neq = 0$ — You cannot divide by 0, so 0 is never in the answer.
3
$5 x < 2 x^{2}$ — Multiply both sides by $x^{2}$ . It is positive, so < stays.
4
$0 < 2 x^{2} - 5 x$ — Take $5 x$ from both sides, so 0 is on one side.
5
$0 < x (2 x - 5)$ — Factorise. Do not divide by $x$ : you would lose a root.
6
$x = 0, x = \frac{5}{2}$ — The critical values.
7
$x < 0 or x > \frac{5}{2}$ — The U is above 0 outside the roots.
8
$x = - 1 : \frac{5}{- 1} = - 5 < 2 ✓$ — Check a negative value. It works.

Rule

Marks usually go: one for both critical values, one for choosing the right region, one for the answer. With ≤ the answer is $x < 0$ or $x \geq \frac{5}{2}$ : 0 stays strict, because $\frac{5}{0}$ does not exist.

Common mistake

Multiplying by $x$ : $5 < 2 x$ , so $x > \frac{5}{2}$ . This loses every negative $x$ , and it is the most common wrong answer.

Key idea

Thomas Harriot gave us the signs < and >. They were printed in his book in 1631, ten years after he died.

Key idea

$x$ on the bottom? Multiply by $x^{2}$ , factorise, and read the answer from the critical values.

Signs from roots or a graph

A cubic works like a quadratic, with three roots instead of two. The roots cut the number line into pieces. The curve is above the axis on some pieces and below on others.

Rule

If the $x^{3}$ term is positive, the curve starts low and ends high. From left to right it goes below, above, below, above the axis, changing at each root.

Worked example

1
$f (x) = (2 x + 5) (x - 1) (x - 4)$ . State the values of $x$ for which $f (x) < 0$ .
2
$x = - \frac{5}{2}, 1, 4$ — The roots: each bracket equals 0.
3
$2 x \times x \times x = 2 x^{3}$ — The $x^{3}$ term is positive, so the curve starts low.
4
$below: x < - \frac{5}{2}, and between 1 and 4$ — Below, above, below, above.
5
$x < - \frac{5}{2} or 1 < x < 4$ — The answer has two pieces.
6
$x = 2 : (9) (1) (- 2) = - 18 < 0 ✓$ — Check a value inside the middle piece.

“State” means no long working. Use the brackets you are given: do not expand first. Often one mark for one piece, two for both.

Example

From a graph: a positive cubic $y = g (x)$ meets the line $y = 2$ at $x = - 3$ , $0$ and $5$ . For $g (x) > 2$ , take where the curve is above the line: $- 3 < x < 0$ or $x > 5$ .

Common mistake

Writing the middle piece as two loose inequalities, $x > 1$ , $x < 4$ . Write it as one: $1 < x < 4$ .

Key idea

Find the roots, decide the start from the sign of the top power, then take the pieces you need.

Two conditions, one range

Some questions give two conditions that must both be true. Solve each one, then keep only the numbers that pass both.

Worked example

1
Find the values of $x$ for which $2 x + 1 > 7$ and $2 x^{2} - 9 x - 5 < 0$ .
2
$2 x > 6 \Rightarrow x > 3$ — Solve the linear one.
3
$(2 x + 1) (x - 5) < 0$ — Factorise the quadratic.
4
$- \frac{1}{2} < x < 5$ — Less than 0, so between the roots.
5
$3 < x < 5$ — Keep the overlap: above 3 and below 5.

Rule

Draw both answers on one number line. “And” keeps the overlap. “Or” keeps everything in either.

“Hence state the range of values of $x$ ” is the same job. It joins your two earlier answers, often from a shape where lengths must be positive.

Common mistake

Writing $x > 3$ or $- \frac{1}{2} < x < 5$ . “Or” keeps numbers like 10, which fail the quadratic. The answer needs both, so write $3 < x < 5$ , or $x > 3$ and $x < 5$ .

Key idea

Two conditions, one number line, keep the overlap.

Regions with curves

You can already describe a region bounded by straight lines. AS regions also use curves and axes. Each edge gives one inequality, and you test a point inside to pick each sign.

Rule

Curve or line $y = f (x)$ : write $y \leq f (x)$ or $y \geq f (x)$ .
The x-axis: $y \geq 0$ (or $y \leq 0$ below it).
A vertical edge, for example through a turning point at $x = h$ : $x \geq h$ or $x \leq h$ .

Worked example

1
$C$ : $y = x^{2} - x - 6$ and $l$ : $y = x + 9$ . Region R lies left of the y-axis, bounded by $C$ , $l$ and the x-axis. Its corners are $(- 9, 0)$ , $(- 3, 6)$ and $(- 2, 0)$ . Define R.
2
$y \geq 0$ — The x-axis is R's floor, so R is above it.
3
$(- 4, 1) is inside R$ — Pick a test point that is not on any edge.
4
$1 \leq - 4 + 9 = 5 \Rightarrow y \leq x + 9$ — R is below $l$ .
5
$1 \leq (- 4)^{2} - (- 4) - 6 = 14 \Rightarrow y \leq x^{2} - x - 6$ — R is below $C$ : outside the U.
6
$(4, 1) passes all three too$ — But it is in a second piece, right of the root $x = 3$ .
7
$x \leq - 2$ — Add an x-limit to cut that piece off.
8
$y \geq 0, y \leq x + 9, y \leq x^{2} - x - 6, x \leq - 2$ — The answer: four inequalities.

Any limit $x \leq k$ with $- 2 \leq k < 3$ works. Between the roots, $C$ is below the axis, so no point there passes. But $k = 3$ fails: the point $(3, 0)$ passes all four and is not in R. The x-values come from where the edges meet, which earlier parts usually make you find.

Rule

Marks usually go: one for two correct inequalities, then marks for three and for all of them. The last mark often needs all < or all ≤, not a mix. In set notation, join with ∩, not ∪.

Common mistake

Writing $R \leq x + 9$ or $C \leq \dots$ . Every inequality must be in $x$ and $y$ . Using R scores nothing.

Common mistake

Stopping at three inequalities when they also fit a second piece. Always ask: is there anywhere else these are all true?

Key idea

One inequality per edge, a test point for each sign, then an x-limit if R could be confused with another piece.

Which method? Read the question

The wording tells you which tool to use.

You see	You do
$x$ on the bottom of a fraction	Multiply by $x^{2}$ ; 0 is never in the answer
A factorised cubic and “ $f (x) > 0$ ”	Roots, start from the top power, pick the pieces
A graph and “ $f (x) < k$ ”	Where the curve is under the line $y = k$
“and”, or “Hence state the range”	Keep the overlap on one number line
“Use inequalities to define R”	One inequality per edge, in $y$ ; test a point
Your inequalities fit two pieces	Add an x-limit between them
R bounded by $C$ and $l$ only, in one quadrant	$y$ between them, plus an x-limit for that side

Common mistake

Mixing < and ≤ in one region answer. Pick one kind and use it for every inequality.

Key idea

Keep the sign safe, read signs from roots, keep the overlap, and give every edge its own inequality.

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