Modulus Inequalities | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Less than is inside; greater than is outside

For a positive number $k$ , $|x-a|<k$ means less than $k$ units from $a$ . Both bounds must hold. By contrast, $|x-a|>k$ allows either of the two outside regions.

Open endpoints exclude equality. Use filled endpoints for ≤ or ≥.

\begin{gathered}|u|<k\ \Longleftrightarrow\ -k<u<k\\[6pt]|u|>k\\\Updownarrow\\u<-k\ \text{or}\ u>k\end{gathered}

Replace strict signs by $\le$ or $\ge$ when equality is allowed. An open circle excludes an endpoint; a filled circle includes it. “And” means overlap; “or” means either region.

Worked example

Solve |2x − 5| < 3.

\begin{aligned}-3<2x-5<3\\2<2x<8\\1<x<4\end{aligned}

Add 5 to all three parts, then divide all three by 2. For $|2x-5|\ge3$ , the answer is instead $x\le1$ or $x\ge4$ .

If you multiply or divide an inequality by a negative number, reverse its sign. For example, $-2x<6$ gives $x>-3$ .

Your turnPractice

Solve $|1-2x|\le5$ and $|1-2x|>5$ .

Show worked answer

\begin{gathered}-5\le1-2x\le5\\-6\le-2x\le4\end{gathered}

Dividing by −2 reverses both signs: $3\ge x\ge-2$ .

Answer

\begin{gathered}-2\le x\le3;\\x<-2\ \text{or}\ x>3\end{gathered}

Check zero or a negative bound directly

Do not apply the positive-distance picture to a negative bound. Use $|u|\ge0$ instead.

\begin{aligned}|u|<0&:\ \text{no solutions},\\|u|\le0&:\ u=0,\\|u|>0&:\ u\ne0,\\|u|\ge0&:\ \text{all allowed inputs}.\end{aligned}

For a negative right side, $|u|<k$ and $|u|\le k$ have no solutions; $|u|>k$ and $|u|\ge k$ hold for all inputs where $u$ is defined.

Your turnPractice

Solve $|x-2|>0$ and $|x-2|\le-1$ .

Show worked answer

Answer

x\ne2;\qquad\text{no solutions}

Less than is inside; greater than is outside

For a positive number $k$ , $|x-a|<k$ means less than $k$ units from $a$ . Both bounds must hold. By contrast, $|x-a|>k$ allows either of the two outside regions.

Open endpoints exclude equality. Use filled endpoints for ≤ or ≥.

\begin{gathered}|u|<k\ \Longleftrightarrow\ -k<u<k\\[6pt]|u|>k\\\Updownarrow\\u<-k\ \text{or}\ u>k\end{gathered}

Replace strict signs by $\le$ or $\ge$ when equality is allowed. An open circle excludes an endpoint; a filled circle includes it. “And” means overlap; “or” means either region.

Worked example

Solve |2x − 5| < 3.

\begin{aligned}-3<2x-5<3\\2<2x<8\\1<x<4\end{aligned}

Add 5 to all three parts, then divide all three by 2. For $|2x-5|\ge3$ , the answer is instead $x\le1$ or $x\ge4$ .

If you multiply or divide an inequality by a negative number, reverse its sign. For example, $-2x<6$ gives $x>-3$ .

Your turnPractice

Solve $|1-2x|\le5$ and $|1-2x|>5$ .

Show worked answer

\begin{gathered}-5\le1-2x\le5\\-6\le-2x\le4\end{gathered}

Dividing by −2 reverses both signs: $3\ge x\ge-2$ .

Answer

\begin{gathered}-2\le x\le3;\\x<-2\ \text{or}\ x>3\end{gathered}

Check zero or a negative bound directly

Do not apply the positive-distance picture to a negative bound. Use $|u|\ge0$ instead.

\begin{aligned}|u|<0&:\ \text{no solutions},\\|u|\le0&:\ u=0,\\|u|>0&:\ u\ne0,\\|u|\ge0&:\ \text{all allowed inputs}.\end{aligned}

For a negative right side, $|u|<k$ and $|u|\le k$ have no solutions; $|u|>k$ and $|u|\ge k$ hold for all inputs where $u$ is defined.

Your turnPractice

Solve $|x-2|>0$ and $|x-2|\le-1$ .

Show worked answer

Answer

x\ne2;\qquad\text{no solutions}