The modulus functionEdexcel A-Level Maths: Revision notes

The modulus $|a|$ of a number is its non-negative size: $|5|=5$ and $|-5|=5$. For a function, $|f(x)|$ keeps the value when $f(x)\ge0$ and changes the sign when $f(x)<0$, so it is never negative. The graph of $y=|ax+b|$ is V-shaped. Sketch $y=ax+b$, then reflect the part below the $x$-axis upwards. The vertex lies on the $x$-axis where $ax+b=0$, at $x=-\frac ba$. The arms have gradients $\pm a$. Example: $y=|2x-1|$ has vertex $\left(\frac12,0\right)$, meets the $y$-axis at $(0,1)$ and has $f(-2)=|-5|=5$.

If $|ax+b|=c$ with $c>0$, then there are two cases: $ax+b=c$ or $ax+b=-c$. If $c<0$ there is no solution, and if $c=0$ there is one. Example: $|2x-1|=3$ gives $2x-1=3$ so $x=2$, or $2x-1=-3$ so $x=-1$. Both lie on the line $y=3$, which meets the V-shape twice.

Again there are two cases: $ax+b=cx+d$ or $ax+b=-(cx+d)$. Solve both, then check each answer in the original equation, since a solution is invalid if it makes the right-hand side negative. Example: $|2x-1|=x$. Case 1: $2x-1=x$ gives $x=1$. Case 2: $2x-1=-x$ gives $x=\frac13$. Check: $|2(1)-1|=1$ and $\left|\frac23-1\right|=\frac13$, so both are valid. Example of a rejected answer: $|x|=2x-3$. Case 1: $x=2x-3$ gives $x=3$, valid. Case 2: $-x=2x-3$ gives $x=1$, but then $|1|=1\ne-1$, so it is rejected.

For $|ax+b|<c$ (with $c>0$): $-c<ax+b<c$, a single interval. For $|ax+b|>c$: $ax+b>c$ or $ax+b<-c$, two outer regions. Example: $|3x+2|<4$ gives $-6<3x<2$, so $-2<x<\frac23$. $|3x+2|>4$ gives $x<-2$ or $x>\frac23$. To compare with a line, find the intersections, then read the graph. For $|2x-1|>x$ the intersections are $x=\frac13$ and $x=1$ and the V is above the line outside them: $x<\frac13$ or $x>1$. For $|3x-6|<x+2$ the intersections are $x=1$ and $x=4$ and the V is below the line between them: $1<x<4$. Alternatively square both sides, which is safe because both sides are non-negative: $(2x-1)^2>x^2\Rightarrow(x-1)(3x-1)>0$.

The solutions of $|3x-6|=x+k$ are the intersections of a V-shape with vertex $(2,0)$ and the line $y=x+k$. Since the line's gradient $1$ is smaller than the arms' gradients $\pm3$, the line meets the V twice when the vertex is below the line, once when the vertex is on the line, and not at all when the vertex is above it. Here the vertex is below the line when $0<2+k$, i.e. $k>-2$; exactly one solution when $k=-2$ (at $x=2$); none when $k<-2$. The two case solutions are $x=\frac{k+6}{2}$ and $x=\frac{6-k}{4}$.