solve: `|x-2|+|x-4| ge 8`

Options

A. x`in (-infty, -1]cup [7, infty)`
B. x`in (-infty, 3]cup [7, infty)`
C. x`in (-7, -1]cup [7, infty)`
D. x`in [7, infty)`

Answer

A

Explanation

If `x-2 =0 and x-4 = 0, then x =2 and x=4`
Now, divide the number line in three intervals
`(-infty, 2) [2,4) amd [4,infty )`
If `x lt2, then`
`|x-2|=-(x-2) and |x-4|=-(x-4)`
`:. |x-2|+|x-4| ge` 8
`rArr -(x-2)-(x-4)ge`8
`rArr -2x ge`2
`rarr x le` -1
`rArr xin (-infty` , -1] ...(1)
If 2`le` x lt 4, then
`|x-2|= x-2`
`|x-4| lt -(x-4)`
`:. |x-2| + |x-4| ge` 8
` rArr (x-2)-(x-4)ge` 8
`rarr 2ge` 8
which is meaningless.
If 3`le x lt infty` , then
`|x-2| = x-2`
`|x-4|=x-4`
`:. |x-2|+|x-4| ge` 8
`rarr x-2+x-4ge` 8
`rArr 2xge`14
`rArr x ge` 7
` :. x in [7,infty` ) ...(2)
Now from eqs. (1) and (2)
x`in (-infty, -1]cup [7, infty` ).