Split at the absolute-value boundary · source reference

Complete question

Solve $x|x-2|\ge1$ exactly. [5] Hence solve $(x+1)|x-1|\ge1$. [3]

[modified 2008/ACJC/Prelim/I/5]

Printed working

Method 1 — graphical

For $x|x-2|\ge1$, the graph intersects $y=1$ at points A and B.

At A, consider $x(x-2)=1$:

$$x^2-2x-1=0\quad\Longrightarrow\quad x=\frac{2\pm\sqrt{4+4}}2=1\pm\sqrt2.$$

Since $x>2$, $x=1+\sqrt2$.

At B, consider $x[-(x-2)]=1$, giving $x^2-2x+1=0$. The source then prints $(x-1)^2=1\Rightarrow x=1$; the “=1” is a printed intermediate-line typo: this should be $(x-1)^2=0$.

The two curves intersect at $x=1+\sqrt2$ and $x=1$. From the graph, the solution is $x\ge1+\sqrt2$ or $x=1$.

Method 2 — algebraic

Case 1: For $x-2\ge0\Rightarrow x\ge2$:

$$\begin{aligned} x(x-2)&\ge1\\ x^2-2x-1&\ge0\\ (x-1)^2-2&\ge0\\ (x-1-\sqrt2)(x-1+\sqrt2)&\ge0\\ x&\le1-\sqrt2\quad\text{or}\quad x\ge1+\sqrt2. \end{aligned}$$

Intersecting this union with $x\ge2$ gives $x\ge1+\sqrt2$.

Case 2: For $x-2<0\Rightarrow x<2$:

$$x(2-x)\ge1\Rightarrow x^2-2x+1\le0\Rightarrow(x-1)^2\le0\Rightarrow x=1.$$

Intersecting $x<2$ and $x=1$ gives $x=1$.

Taking the union, the solution to $x|x-2|\ge1$ is $x\ge1+\sqrt2$ or $x=1$.

Replace $x$ with $x+1$. The solution for $(x+1)|x-1|\ge1$ is

$$x+1\ge1+\sqrt2\quad\text{or}\quad x+1=1\quad\Longrightarrow\quad x\ge\sqrt2\quad\text{or}\quad x=0.$$

Teacher margin notes

At A, $x>2$, so $|x-2|=x-2$.

At B, $x<2$, so $|x-2|=-(x-2)$.

The blue curve is above or intersects the red curve.

Source diagrams

Provenance

Question page references: 1. Solution page references: 6. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.