IGCSE · Practice questions · The modulus of a function

Reflect, label, count.

Six original Cambridge-style questions on the graph of \(y = |f(x)|\): intercepts and sharp corners, the reflected maximum, cubic and trigonometric curves, the range of \(|f(x)|\) and the number of solutions of \(|f(x)| = k\).

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

Sketch \(f\), reflect what is negative, then read off.

Roots first: solve \(f(x) = 0\); these \(x\)-intercepts do not move and are the sharp corners.
Signs: decide where \(f(x) < 0\) (factors, or the shape of the trig curve) and reflect only those pieces.
Key points: \(y\)-intercept \((0, |f(0)|)\); a minimum \((p, -q)\) below the axis becomes a maximum \((p, q)\).
Range: if \(f\) changes sign the least value of \(|f(x)|\) is \(0\); the greatest is the larger of the two extreme sizes.
Count: solutions of \(|f(x)| = k\) are intersections with \(y = k\); watch corners, maxima and end points.

Recall

[2 marks]

Sketch the graph of \(y = |2x + 3|\), stating the coordinates of the points where the graph meets the axes.

\(2x + 3 = 0\) at \(x = -\dfrac{3}{2}\): the V shape has its sharp corner at \(\left(-\dfrac{3}{2}, 0\right)\). ✓
\(|2(0) + 3| = 3\), so it meets the \(y\)-axis at \((0, 3)\). For \(x < -\dfrac{3}{2}\) the line \(y = 2x + 3\) is reflected to \(y = -2x - 3\). ✓

Calculation

[4 marks]

(a) Express \(x^{2} + 2x - 8\) in the form \((x + a)^{2} + b\).

(b) Hence sketch \(y = |x^{2} + 2x - 8|\), stating the coordinates of the points where it meets the axes and of its local maximum.

(a) \(x^{2} + 2x - 8 = (x + 1)^{2} - 9\). ✓
(b) \(x^{2} + 2x - 8 = (x + 4)(x - 2)\), so the graph meets the \(x\)-axis at \((-4, 0)\) and \((2, 0)\), its sharp corners. ✓
\(f(0) = -8\), so the \(y\)-intercept of \(y = |f(x)|\) is \((0, 8)\). ✓
The minimum \((-1, -9)\) of \(y = f(x)\) is reflected to the local maximum \((-1, 9)\). ✓

Calculation

[4 marks]

\(f(x) = (x + 2)(x - 1)(x - 4)\).

(a) Find the values of \(x\) for which \(f(x) < 0\).

(b) Hence describe how the graph of \(y = |f(x)|\) is obtained from the graph of \(y = f(x)\), and state the coordinates of the points where \(y = |f(x)|\) meets the axes.

(a) The roots are \(-2\), \(1\), \(4\) and the \(x^{3}\) coefficient is positive, so the signs are \(-,\ +,\ -,\ +\) from left to right. ✓
\(f(x) < 0\) for \(x < -2\) and for \(1 < x < 4\). ✓
(b) The parts of \(y = f(x)\) for \(x < -2\) and \(1 < x < 4\) are reflected in the \(x\)-axis; the rest is unchanged. ✓
\(f(0) = (2)(-1)(-4) = 8\), so the graph meets the axes at \((-2, 0)\), \((1, 0)\), \((4, 0)\) and \((0, 8)\). ✓

Analysis

[4 marks]

Find the set of values of \(k\) for which the equation \(|x^{2} - 6x + 5| = k\) has

(a) exactly 4 solutions, (b) exactly 2 solutions.

\(x^{2} - 6x + 5 = (x - 3)^{2} - 4 = (x - 1)(x - 5)\): the graph of \(y = |x^{2} - 6x + 5|\) has corners at \((1, 0)\), \((5, 0)\) and a local maximum \((3, 4)\). ✓
(a) The line \(y = k\) cuts both arms and the reflected arch twice when \(0 < k < 4\). ✓
(b) Exactly 2 solutions when the line meets only the two arms: \(k > 4\), or when \(k = 0\) (the corners \(x = 1\) and \(x = 5\)). ✓
(\(k = 4\) gives 3 solutions and \(k < 0\) gives none.) Answers: (a) \(0 < k < 4\); (b) \(k = 0\) or \(k > 4\). ✓

Analysis

[5 marks]

(a) Sketch the graph of \(y = |3\sin x - 1|\) for \(0^{\circ} \le x \le 360^{\circ}\), and state the range of \(y\).

(b) Use your graph to find the number of solutions of (i) \(|3\sin x - 1| = 1\), (ii) \(|3\sin x - 1| = 3\), for \(0^{\circ} \le x \le 360^{\circ}\).

(a) \(y = 3\sin x - 1\) starts at \(-1\), rises to \(2\) at \(90^{\circ}\), falls to \(-4\) at \(270^{\circ}\) and ends at \(-1\). Reflect the negative pieces: \(|3\sin x - 1|\) starts at \(1\), falls to \(0\), rises to \(2\) at \(90^{\circ}\), falls to \(0\), rises to \(4\) at \(270^{\circ}\) and falls to \(1\) at \(360^{\circ}\). ✓
It takes every value from \(0\) to \(4\): range \(0 \le y \le 4\). ✓
(b)(i) The line \(y = 1\) passes through both end points (\(0^{\circ}\) and \(360^{\circ}\)) and cuts the middle three pieces once each: 5 solutions. ✓
Check: \(\sin x = \dfrac{2}{3}\) gives 2 solutions; \(\sin x = 0\) gives \(0^{\circ}, 180^{\circ}, 360^{\circ}\); total 5. ✓
(ii) \(y = 3\) is above the local maximum \(2\), so it meets only the peak at \(270^{\circ}\), twice: 2 solutions. ✓

Analysis

[5 marks]

(a) \(f(x) = x^{2} - 4x - 1\) for \(0 \le x \le 5\). Find the range of \(f\) and the range of \(|f|\).

(b) Sketch \(y = |\tan x - 1|\) for \(0^{\circ} \le x \le 180^{\circ}\), \(x \ne 90^{\circ}\), and state the number of solutions of \(|\tan x - 1| = 1\) in this interval.

(a) \(f(x) = (x - 2)^{2} - 5\): minimum \(-5\) at \(x = 2\); \(f(0) = -1\) and \(f(5) = 4\), so \(-5 \le f(x) \le 4\). ✓
\(f\) changes sign in the domain (at \(x = 2 + \sqrt{5}\)), so \(0 \le |f(x)| \le 5\). ✓
(b) \(\tan x - 1 = 0\) at \(x = 45^{\circ}\). For \(0^{\circ} \le x < 45^{\circ}\) and \(90^{\circ} < x \le 180^{\circ}\), \(\tan x - 1 < 0\), so those pieces are reflected: the graph starts at \(1\), falls to a corner at \(45^{\circ}\), rises towards the asymptote, then comes down from above to \(1\) at \(180^{\circ}\). ✓
The line \(y = 1\) meets it at \(x = 0^{\circ}\), once between \(45^{\circ}\) and \(90^{\circ}\) (where \(\tan x = 2\)), and at \(x = 180^{\circ}\): 3 solutions. ✓

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.