IGCSE · Practice questions · Modulus inequalities

Between or beyond.

Six original Cambridge-style questions on modulus inequalities: |X| less than or greater than a constant, squaring both sides, critical values with a sketch, and the modulus of a quadratic.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Divide, decide, then check with a sketch.

Divide by \(k\) first, so the modulus stands alone.
Less than is between: \(|X| < c \iff -c < X < c\). Greater than is outside: \(X < -c\) or \(X > c\).
Modulus against modulus: square both sides (square \(k\) too) and solve the quadratic inequality.
Modulus against a line: solve the \(+\) and \(-\) equations, check each critical value, then sketch to choose the region.
Write the answer in standard form: \(a < x < b\), or \(x < a\) or \(x > b\).

Recall

[2 marks]

Solve the inequality \(|3x - 2| \le 7\).

\(-7 \le 3x - 2 \le 7\), so \(-5 \le 3x \le 9\). ✓
\(-\dfrac{5}{3} \le x \le 3\). ✓

Calculation

[3 marks]

Solve the inequality \(4|x + 2| > 12\).

Divide by 4: \(|x + 2| > 3\). ✓
\(x + 2 < -3\) or \(x + 2 > 3\). ✓
\(x < -5\) or \(x > 1\). ✓

Calculation

[4 marks]

Solve the inequality \(|3x + 1| > |x - 5|\).

Square both sides: \(9x^{2} + 6x + 1 > x^{2} - 10x + 25\). ✓
\(8x^{2} + 16x - 24 > 0\), so \(x^{2} + 2x - 3 > 0\). ✓
\((x + 3)(x - 1) > 0\): critical values \(-3\) and \(1\), and '\(> 0\)' means outside. ✓
\(x < -3\) or \(x > 1\) (check \(x = 0\): \(1 > 5\) is false, so the middle is excluded). ✓

Analysis

[5 marks]

(a) On the same axes, sketch the graphs of \(y = |2x - 4|\) and \(y = x + 1\), stating the coordinates of the points where each graph meets the axes.

(b) Hence solve the inequality \(|2x - 4| < x + 1\).

(a) \(y = |2x - 4|\) is a V with vertex \((2, 0)\) on the \(x\)-axis, meeting the \(y\)-axis at \((0, 4)\). \(y = x + 1\) is a straight line through \((-1, 0)\) and \((0, 1)\). ✓
(b) Case \(+\): \(2x - 4 = x + 1\) gives \(x = 5\); check \(|6| = 6\). ✓
Case \(-\): \(-(2x - 4) = x + 1\) gives \(-3x = -3\), \(x = 1\); check \(|-2| = 2\). ✓
From the sketch the line is above the V between the crossings (test \(x = 2\): \(0 < 3\)). ✓
\(1 < x < 5\). ✓

Analysis

[5 marks]

Solve the inequality \(|x^{2} - x - 4| \le 2\).

\(-2 \le x^{2} - x - 4 \le 2\). Split into two inequalities, both of which must hold. ✓
\(x^{2} - x - 4 \le 2\): \(x^{2} - x - 6 \le 0\), \((x - 3)(x + 2) \le 0\), so \(-2 \le x \le 3\). ✓
\(x^{2} - x - 4 \ge -2\): \(x^{2} - x - 2 \ge 0\), \((x - 2)(x + 1) \ge 0\), so \(x \le -1\) or \(x \ge 2\). ✓
Combine: inside \([-2, 3]\) but outside \((-1, 2)\). ✓
\(-2 \le x \le -1\) or \(2 \le x \le 3\). ✓

Proof

[6 marks]

(a) Show that the line \(y = 3\) meets the curve \(y = |x^{2} - 4x|\) at exactly four points, and find their \(x\)-coordinates in exact form.

(b) Hence solve the inequality \(|x^{2} - 4x| > 3\).

(a) \(|x^{2} - 4x| = 3\) means \(x^{2} - 4x = 3\) or \(x^{2} - 4x = -3\). ✓
\(x^{2} - 4x - 3 = 0\): \(x = \dfrac{4 \pm \sqrt{16 + 12}}{2} = \dfrac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}\). ✓
\(x^{2} - 4x + 3 = 0\): \((x - 1)(x - 3) = 0\), so \(x = 1\) or \(x = 3\). ✓
These four values are different, so there are exactly four points: \(x = 2 - \sqrt{7},\ 1,\ 3,\ 2 + \sqrt{7}\). ✓
(b) The curve is \(y = x^{2} - 4x\) with the part between \(x = 0\) and \(x = 4\) reflected up, giving a hump of height 4 at \(x = 2\). It is above \(y = 3\) on the two outer arms and at the top of the hump. ✓
\(x < 2 - \sqrt{7}\) or \(1 < x < 3\) or \(x > 2 + \sqrt{7}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.