Sums of Moduli | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Add the two distances, not their insides

$|x-1|+|x-5|$ adds the distances from 1 and 5. Between those points, moving towards one moves equally far from the other, so the total is always 4.

The minimum total distance is the gap between the two fixed points.

Outside that interval, both distances grow. Split at the two sign changes, $x=1$ and $x=5$ :

\begin{gathered}|x-1|+|x-5|\\=\begin{cases}6-2x&x<1,\\4&1\le x\le5,\\2x-6&x>5.\end{cases}\end{gathered}

For example, on the left both insides are negative, so the sum is $(1-x)+(5-x)$ . In the middle it is $(x-1)+(5-x)$ .

Common mistake

|u|+|v|

is not generally

|u+v|

. At

x=3

, the two distances sum to 4, while

|(3-1)+(3-5)|=0

Use each rule only on its own interval

Worked example

Solve |x − 1| + |x − 5| = 8.

Left: $6-2x=8$ gives $x=-1$ , which is below 1. Middle: $4=8$ is impossible. Right: $2x-6=8$ gives $x=7$ , which is above 5.

Answer

x=-1\quad\text{or}\quad x=7

For a total at most 8, use the portion of the graph at or below that height: $-1\le x\le7$ . For a total equal to 4, every $x$ between 1 and 5 works; for a total below 4, none do.

Your turnPractice

Solve $|x+2|+|x-4|\le6$ . Then solve the same expression equal to 10 and equal to 5.

Show worked answer

The centres are −2 and 4, six units apart. The minimum total is 6, reached throughout $[-2,4]$ .

Answer

-2\le x\le4

For total 10, the outer branches are $2-2x=10$ and $2x-2=10$ , giving $x=-4$ or $x=6$ . A total of 5 is impossible because it is below the minimum of 6.

Add the two distances, not their insides

$|x-1|+|x-5|$ adds the distances from 1 and 5. Between those points, moving towards one moves equally far from the other, so the total is always 4.

The minimum total distance is the gap between the two fixed points.

Outside that interval, both distances grow. Split at the two sign changes, $x=1$ and $x=5$ :

\begin{gathered}|x-1|+|x-5|\\=\begin{cases}6-2x&x<1,\\4&1\le x\le5,\\2x-6&x>5.\end{cases}\end{gathered}

For example, on the left both insides are negative, so the sum is $(1-x)+(5-x)$ . In the middle it is $(x-1)+(5-x)$ .

Common mistake

|u|+|v|

is not generally

|u+v|

. At

x=3

, the two distances sum to 4, while

|(3-1)+(3-5)|=0

Use each rule only on its own interval

Worked example

Solve |x − 1| + |x − 5| = 8.

Left: $6-2x=8$ gives $x=-1$ , which is below 1. Middle: $4=8$ is impossible. Right: $2x-6=8$ gives $x=7$ , which is above 5.

Answer

x=-1\quad\text{or}\quad x=7

For a total at most 8, use the portion of the graph at or below that height: $-1\le x\le7$ . For a total equal to 4, every $x$ between 1 and 5 works; for a total below 4, none do.

Your turnPractice

Solve $|x+2|+|x-4|\le6$ . Then solve the same expression equal to 10 and equal to 5.

Show worked answer

The centres are −2 and 4, six units apart. The minimum total is 6, reached throughout $[-2,4]$ .

Answer

-2\le x\le4

For total 10, the outer branches are $2-2x=10$ and $2x-2=10$ , giving $x=-4$ or $x=6$ . A total of 5 is impossible because it is below the minimum of 6.