Solving a Modulus Function Equation

A modulus equation $|ax + b| = k$ (with $k \ge 0$) splits into two linear equations, because the inside can be either $+k$ or $-k$. Show both branches, solve each, and the two marks for the solutions are secure.

The function $f$ is defined by $f(x) = |2x - 6|$ for $x \in \mathbb{R}$. (i) Solve $f(x) = 4$. [3] (ii) State the range of $f$. [2]

The working

(i) $|2x - 6| = 4$ means the inside is $4$ or $-4$: $2x - 6 = 4 \qquad \text{or} \qquad 2x - 6 = -4 \quad \text{(M1 for both cases)}$

Solve each: $2x = 10 \Rightarrow x = 5 \qquad \text{or} \qquad 2x = 2 \Rightarrow x = 1 \quad \text{(A1, A1)}$

Check: $|2(5) - 6| = |4| = 4$ ✓ and $|2(1) - 6| = |-4| = 4$ ✓.

(ii) A modulus output is never negative, and $|2x - 6|$ reaches $0$ when $2x - 6 = 0$ (at $x = 3$): $\text{range: } f(x) \ge 0 \quad \text{(M1, A1)}$

Where the marks are won and lost

• The two cases are the method mark. A single equation $2x - 6 = 4$ gives only $x = 5$ and throws away $x = 1$.
• The range of any modulus function is $\ge 0$ (it can touch $0$). Writing $f(x) > 0$ misses that $f(3) = 0$ is attainable.
• Both answers should be checked back in the modulus, cheap insurance against a sign slip.

Common mistakes

• Solving only the $+4$ branch.
• Squaring both sides and mis-handling the resulting quadratic (valid but slower and error-prone here).
• Giving the range as $f(x) > 0$ or, worse, all reals.

Full method: Modulus Functions & Their Graphs notes. For inequalities of this type see Equations, Inequalities and Graphs. Topic home: Functions pillar.