Linear & quadratic inequalities, regions on graphs – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator3 marks

The diagram shows an inequality on a number line. The circle at $-2$ is empty and the circle at $3$ is filled in.

(a) Write down the inequality shown, using $x$. [2]

(b) Write down all the integers that satisfy this inequality. [1]

(a) one end correct: $x > -2$ or $x \le 3$ (or $-2 < x$ seen in a double inequality with the other end wrong)B1
(a) $-2 < x \le 3$ (accept $3 \ge x > -2$; do not accept $-2 \le x \le 3$ or $-2 < x < 3$)B1
(b) $-1, 0, 1, 2, 3$ and no others (ignore order)B1
Answer: (a) $-2 < x \le 3$ (b) $-1, 0, 1, 2, 3$
Examiner tip: An empty circle means the end value is NOT included (strict $<$); a filled circle means it is included ($\le$), so $-2$ is not in the list but $3$ is.

2Grade 4–5Non-calculator3 marks

Kofi solves the inequality $7 - 3x < 1$. His working is shown.

$-3x < -6$
$x < 2$

(a) Explain the mistake Kofi has made. [1]

(b) Solve $7 - 3x < 1$ correctly. [2]

(a) explanation that when dividing (both sides) by a negative number ($-3$) the inequality sign must be reversed (accept e.g. 'should be $>$ not $<$ because he divided by $-3$')C1
(b) $-3x < -6$ or $6 < 3x$ or $7 - 1 < 3x$ (a correct first step)M1
(b) $x > 2$ (accept $2 < x$; $x = 2$ or $x \ge 2$ scores M1 A0)A1
Answer: (a) He divided by $-3$ without reversing the sign. (b) $x > 2$
Examiner tip: Avoid dividing by a negative altogether: add $3x$ to both sides first ($6 < 3x$) so the sign never needs flipping.

3Grade 4–5Non-calculator3 marks

$n$ is an integer.

$$-7 \le 3n - 1 < 11$$

Write down all the possible values of $n$. [3]

adding 1 to all parts: $-6 \le 3n < 12$, or one correct critical value $-2$ or $4$ from workingM1
$-2 \le n < 4$ (both ends correct, inequality signs may be missing or wrong)A1
$-2, -1, 0, 1, 2, 3$ and no others (ignore order); allow one omission or one extra for M1 A1 A0 onlyA1
Answer: $-2, -1, 0, 1, 2, 3$
Examiner tip: Do the same operation to all three parts, then check the ends: $-2$ is included ($\le$) but $4$ is not ($<$).

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4Grade 4–5Non-calculator3 marks

A rectangle has length $(2x + 3)$ cm and width $(x - 1)$ cm.

The perimeter of the rectangle is less than 40 cm.

Find the range of possible values of $x$. Give your answer as an inequality. [3]

forms a correct inequality for the perimeter, e.g. $2(2x + 3) + 2(x - 1) < 40$ or $6x + 4 < 40$M1
$x < 6$A1
$1 < x < 6$ (uses width $x - 1 > 0$; accept $x > 1$ and $x < 6$ stated separately)B1
Answer: $1 < x < 6$
Examiner tip: Lengths must be positive, so the width $x - 1 > 0$ gives the lower limit; many students stop at $x < 6$.

5Grade 4–5FoundationNon-calculator3 marks

The region $\mathbf{R}$ is shaded on the grid. It is bounded by the lines $x = 1$, $y = 2$ and $x + y = 6$.

Write down the three inequalities that define the region $\mathbf{R}$. [3]

$x \ge 1$ (accept $x > 1$)B1
$y \ge 2$ (accept $y > 2$)B1
$x + y \le 6$ (accept $x + y < 6$, or equivalent such as $y \le 6 - x$)B1
Answer: $x \ge 1$, $y \ge 2$, $x + y \le 6$
Examiner tip: Test a point inside the region, e.g. $(2, 3)$: $2 + 3 = 5$, which is less than 6, so the sign is $\le$.

6Grade 6–7HigherNon-calculator3 marks

A point $(x, y)$ has whole-number coordinates and satisfies all three inequalities

$$x \ge 0 \qquad y > 1 \qquad x + 2y \le 8$$

Work out how many such points there are. [3]

uses $y > 1$ as $y \ge 2$ for integers, or finds a correct set of points for one value of $y$ (e.g. $y = 2$: $x = 0, 1, 2, 3, 4$)M1
a systematic method covering $y = 2, 3, 4$, e.g. counts 5, 3, 1 (or a correct sketch of the region with points marked), with at most one errorM1
9 caoA1
Answer: 9 points
Examiner tip: Work row by row and remember $y > 1$ excludes $y = 1$ but $x \ge 0$ includes points on the $y$-axis.

7Grade 6–7HigherNon-calculator3 marks

Solve $2x^2 + 5x - 12 \ge 0$ [3]

factorises correctly: $(2x - 3)(x + 4)$, or uses a correct formula substitutionM1
critical values $\tfrac{3}{2}$ and $-4$A1
$x \le -4$ or $x \ge \tfrac{3}{2}$ (accept 1.5; accept in set notation; do not accept $-4 \ge x \ge 1.5$)A1
Answer: $x \le -4$ or $x \ge \tfrac{3}{2}$
Examiner tip: Sketch the U-shaped parabola: '$\ge 0$' means the parts above the $x$-axis, which is the two OUTSIDE regions, not the bit between the roots.

8Grade 8–9HigherCalculator4 marks

Find all the integer values of $x$ for which

$$x^2 \le 4x + 7$$

Show your working clearly. [4]

rearranges to $x^2 - 4x - 7 \le 0$ (or $= 0$)M1
correct substitution into the quadratic formula, e.g. $\dfrac{4 \pm \sqrt{(-4)^2 - 4(1)(-7)}}{2}$, or completing the square $(x - 2)^2 \le 11$M1
critical values $2 \pm \sqrt{11}$, i.e. $-1.32$ and $5.32$ (awrt)A1
$-1, 0, 1, 2, 3, 4, 5$ and no othersA1
Answer: $-1, 0, 1, 2, 3, 4, 5$
Examiner tip: The roots are not whole numbers, so find them accurately and then pick the integers that lie between them; $-1$ is often missed.

9Grade 6–7HigherNon-calculator4 marks

The region $\mathbf{R}$ is shaded on the grid. It is bounded by three straight lines.

Write down the three inequalities that define the region $\mathbf{R}$. [4]

$y \ge 1$ (accept $y > 1$)B1
method to find the equation of either sloping line, e.g. gradient 2 through the origin, or gradient $-\tfrac{1}{2}$ with $y$-intercept 4M1
$y \le 2x$ (accept $y < 2x$)A1
$x + 2y \le 8$ (accept $y \le -\tfrac{1}{2}x + 4$, $<$ instead of $\le$, or any equivalent)A1
Answer: $y \ge 1$, $y \le 2x$, $x + 2y \le 8$
Examiner tip: Find each line's equation first, then test a point inside $\mathbf{R}$ such as $(2, 2)$ to choose the direction of each sign.

10Grade 6–7HigherNon-calculator4 marks

$x$ is an integer such that

$$x^2 < 3x + 10 \quad \text{and} \quad 4x - 3 \ge 2x + 3$$

Find all the possible values of $x$. [4]

solves the linear inequality: $x \ge 3$M1
rearranges and factorises: $(x - 5)(x + 2) < 0$ or finds critical values $5$ and $-2$M1
$-2 < x < 5$A1
$3, 4$ and no othersA1
Answer: $3, 4$
Examiner tip: Solve each inequality separately, then take only the values that satisfy BOTH; $5$ is excluded because the quadratic inequality is strict.

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