AS & A Level · Practice questions · The modulus function

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Papers 2 and 3. Six original Cambridge-style questions on the modulus function: sketches, equations, inequalities and checking for false roots.

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

The method

Choose the method from the shape of the question.

Sketch: vertex where \(ax + b = 0\), \(y\)-intercept \(|b|\), arms of gradient \(\pm a\).
\(|f| = |g|\): square both sides, or solve \(f = g\) and \(f = -g\).
\(|f| = \) line: split into cases and CHECK each root in the original equation.
\(|x - a| < b\): \(a - b < x < a + b\); for \(>\) take the two outside pieces.
\(|f| < |g|\): square, then solve the quadratic inequality.
Modulus against a line: sketch, find valid crossings, read off the region.

Recall

[2 marks]

Sketch the graph of \(y = |3x + 6|\), stating the coordinates of the points where the graph meets the axes.

Vertex: \(3x + 6 = 0\) gives \(x = -2\), so \((-2, 0)\). ✓
\(y\)-intercept: \(y = |6| = 6\), so \((0, 6)\). ✓
A V shape with arms of gradient \(-3\) (left) and \(3\) (right). ✓

Calculation

[3 marks]

Solve the equation \(|x - 4| = |2x + 1|\).

Square: \(x^2 - 8x + 16 = 4x^2 + 4x + 1\), so \(3x^2 + 12x - 15 = 0\). ✓
\(x^2 + 4x - 5 = 0\), so \((x + 5)(x - 1) = 0\). ✓
\(x = -5\) or \(x = 1\). (Check: \(|-9| = |-9|\), \(|-3| = |3|\).) ✓

Calculation

[2 marks]

Solve the inequality \(|2x - 3| \ge 5\).

\(2x - 3 \ge 5\) gives \(x \ge 4\). ✓
\(2x - 3 \le -5\) gives \(x \le -1\). ✓
So \(x \le -1\) or \(x \ge 4\). ✓

Analysis

[4 marks]

Solve the inequality \(|x + 2| > |2x - 5|\).

Both sides are non-negative, so square: \(x^2 + 4x + 4 > 4x^2 - 20x + 25\). ✓
\(0 > 3x^2 - 24x + 21\), so \(x^2 - 8x + 7 < 0\). ✓
\((x - 1)(x - 7) < 0\), so the quadratic is negative between its roots. ✓
\(1 < x < 7\). (Check \(x = 3\): \(5 > 1\).) ✓

Analysis

[4 marks]

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

For \(x \ge -5\): \(3x - 1 = x + 5\) gives \(x = 3\), which satisfies \(x \ge -5\). Valid crossing. ✓
For \(x < -5\): \(3x - 1 = -x - 5\) gives \(x = -1\), which does not satisfy \(x < -5\). Reject. ✓
The line \(y = 3x - 1\) is steeper than the right arm, so it is above the V to the right of \(x = 3\). ✓
\(x > 3\). (Check \(x = 0\): \(-1 > 5\) is false, as expected.) ✓

Analysis

[6 marks]

(a) On the same axes, sketch the graphs of \(y = |2x - 5|\) and \(y = x + 1\).

(b) Solve the equation \(|2x - 5| = x + 1\).

(a) V with vertex \((2.5, 0)\) and \(y\)-intercept \((0, 5)\); straight line through \((-1, 0)\) and \((0, 1)\), crossing both arms of the V. ✓
(b) Right arm: \(2x - 5 = x + 1\) gives \(x = 6\). Check: \(|7| = 7\). Valid. ✓
Left arm: \(5 - 2x = x + 1\) gives \(3x = 4\), \(x = \dfrac{4}{3}\). Check: \(\left|-\dfrac{7}{3}\right| = \dfrac{7}{3}\). Valid. ✓
So \(x = \dfrac{4}{3}\) or \(x = 6\). ✓
(c) The V is below the line between the crossings: \(\dfrac{4}{3} < x < 6\). ✓

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.