Modulus Inequalities with a Variable Side | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Split where the inside changes sign

When only one side has modulus bars, the other side may be negative. Squaring can reverse which side is larger: $2>-3$ , but $4<9$ . Instead, replace the modulus by its correct branch and keep that branch’s input restriction.

Worked example

Solve 2x + 5 < |x + 1|.

The inside changes sign at $x=-1$ .

For $x<-1$ , use $|x+1|=-x-1$ . Then $2x+5<-x-1$ gives $x<-2$ , which lies in this branch.
For $x\ge-1$ , use $|x+1|=x+1$ . Then $x<-4$ , which has no overlap with $x\ge-1$ . This branch supplies no solutions.

Answer

x<-2

Check $x=-3$ : $-1<2$ is true. At −2 the sides are equal, so the strict inequality excludes it.

Your turnPractice

Solve $|2x-3|\le x+1$ . Keep each branch condition.

Show worked answer

For $x<3/2$ , $3-2x\le x+1$ gives $2/3\le x<3/2$ . For $x\ge3/2$ , $2x-3\le x+1$ gives $3/2\le x\le4$ .

Answer

2/3\le x\le4

The intervals meet at $3/2$ ; there is no gap. Both endpoints satisfy equality.

Keep an unknown constant as a letter

A parameter is a fixed but unspecified number. If a question says $a>0$ , use that information to compare expressions such as $5a/3<2a$ ; do not choose your own value of $a$ .

Keep parameter labels: choosing a numerical value would show only one member of the family.

Your turn9709/32/M/J/24 Q1(a–b)

For $a>0$ , sketch $y=|x-2a|$ , then solve $2x-3a<|x-2a|$ .

Show worked answer

The V has vertex $(2a,0)$ , y-intercept $(0,2a)$ and gradients −1 and 1. Extend it into negative $x$ .

For $x<2a$ , solve $2x-3a<2a-x$ , giving $x<5a/3$ ; this is within the branch since $a>0$ .

For $x\ge2a$ , solve $2x-3a<x-2a$ , giving $x<a$ . It has no overlap with the branch.

Answer

x<5a/3

Split where the inside changes sign

Worked example

Solve 2x + 5 < |x + 1|.

The inside changes sign at $x=-1$ .

For $x<-1$ , use $|x+1|=-x-1$ . Then $2x+5<-x-1$ gives $x<-2$ , which lies in this branch.
For $x\ge-1$ , use $|x+1|=x+1$ . Then $x<-4$ , which has no overlap with $x\ge-1$ . This branch supplies no solutions.

Answer

x<-2

Check $x=-3$ : $-1<2$ is true. At −2 the sides are equal, so the strict inequality excludes it.

Your turnPractice

Solve $|2x-3|\le x+1$ . Keep each branch condition.

Show worked answer

For $x<3/2$ , $3-2x\le x+1$ gives $2/3\le x<3/2$ . For $x\ge3/2$ , $2x-3\le x+1$ gives $3/2\le x\le4$ .

Answer

2/3\le x\le4

The intervals meet at $3/2$ ; there is no gap. Both endpoints satisfy equality.

Keep an unknown constant as a letter

A parameter is a fixed but unspecified number. If a question says $a>0$ , use that information to compare expressions such as $5a/3<2a$ ; do not choose your own value of $a$ .

Keep parameter labels: choosing a numerical value would show only one member of the family.

Your turn9709/32/M/J/24 Q1(a–b)

For $a>0$ , sketch $y=|x-2a|$ , then solve $2x-3a<|x-2a|$ .

Show worked answer

The V has vertex $(2a,0)$ , y-intercept $(0,2a)$ and gradients −1 and 1. Extend it into negative $x$ .

For $x<2a$ , solve $2x-3a<2a-x$ , giving $x<5a/3$ ; this is within the branch since $a>0$ .

For $x\ge2a$ , solve $2x-3a<x-2a$ , giving $x<a$ . It has no overlap with the branch.

Answer

x<5a/3