Question

The number of solutions that are possible for the inequality \[\left[ x-5 \right] +\left| x -1\right|<2\] is

Answer 1

@satellite73

Answer 2

did this one yesterday
there is no solution

Answer 3

where....

Answer 4

The answer is 0 solution u r correct but how????

Answer 5

there are three intervals to consider \(x<1\) and \(1<x<5\) and \(x>5\)

if \(1<x<5\) then \(|x-1|=x-1\) and \(|x-5|=5-x\)

so
\(|x-5|+|x-1|=5-x+x-1=4\) in that interval and clearly 4>2

Answer 6

so...

Answer 7

if \(x>5\) then \(|x-1|=x-1, |x-5|=x-5\) and so \(|x-5|+|x-1|=x-5+x-1=2x-6\) in that interval
now if \(x>5\) then \(2x-6>4\) so it is larger than 4 in that interval as well, and if it is larger than 4 it is not less than 2

Answer 8

ok

Answer 9

the last interval to consider is if \(x<1\) in which case \(|x-1|=1-x\) and \(|x-5|=5-x\) and therefore ....
you can finish and conclude that if \(x<1\) your expression is still larger than 4, so it is never less than two

Answer 10

then for each case condradict

Answer 11

yes, i left the last case up to you, but it is pretty much like the second one

Answer 12

here is a nice visual confirmation that this expression is always \(\geq 4\)
http://www.wolframalpha.com/input/?i=y%3D |x-5|%2B|x-1|

Answer 13

you'll have to cut and past, absolute values reek havoc with links

Answer 14

so .... how do solution become 0

Answer 15

but the answer should be 0 dude...............

Answer 16

@glgan1 help lzz

Answer 17

@experimentX can u explain this more clearly....

Answer 18

how it condradicts can u show that

Answer 19

sorry ignore my last post ..

Answer 20

ok

Answer 21

it's more clearly explained here
http://openstudy.com/study#/updates/4fbf6570e4b0dd370c772b87

the lowest value of LHS is 4 ...
4 cannot be greater than 2... which is a contradiction

Answer 22

if i solutions condradicts....then others.....

Answer 23

what exactly are you expecting to be the solution??

Answer 24

i want to know why the solution become 0

Answer 25

the solution is no zero ...
simply there is not any solution

Answer 26

@experimentX anyway a many thanzz to u .... becozz u r the only person who reponded to my message quickly and help me thanzzzz a billion!!

Answer 27

sorry helped me!!

Answer 28

np ...

Answer 29

@heena

Answer 30

@open_study1 wat happen ur qn is already answered i guess

Answer 31

i did nt understand............