Question involving absolute function

Question

I saw this interesting problem in a math puzzle forum:-

Find all integral values of $t$ such that the equation $|s-1| - 3|s+1| + |s+2| = t $ has no solutions.

How does one approach these kind of problems?

Answer 1

Divide into regions like so:

Case 1: Assume $s\ge 1$

Your equation reduces to:

$s-1-3(s+1)+(s+2)=t$

Case 2: Assume $-1 \le s \le 1$

Your equation reduces to:

$-(s-1)-3(s+1)+(s+2)=t$

and so on.

Answer 2

First you have to know the properties of absolute value to confront these kind of problems.When x is negative, $ |x| =-x $ and when x is positive $ |x|=(+x) $. Then you check if you are given $|x-a|$ then for $x>a$ ,$|x-a|$=positive or for $x<a$ ,$|x-a|$ is negative.

Now look to your problem:$|s-1|-3|s+1|+|s+2|=t$,Consider the critical points of the curve $y=|x-1| - 3|x+1| + |x+2|$ are the points where $x=(-2),x=(-1),x=1$.So there are some distinctive parts to be considered.$$ case 1. $$When $s<-2$;$|s-1|=(-ve),|s+1|=(-ve),|s+2|=(-ve)$.Then the $|s-1|-3|s+1|+|s+2|$ becomes $=-(s-1)-3*{-(s+1)}+{-(s+2)} $=$(1-s+3s+3-s-2)$=$(s+2)$...So “t” can take value within the range (-infinity,0) for all$ s<-2$. $$ case 2.$$When $-2<s<-1$;$|s-1|=(-ve),|s+1|=(-ve),|s+2|=(+ve)$.Then the”t” becomes=$(|s-1|-3|s+1|+|s+2|)$=$(1-s+3s+3+s+2)$=$(3s+6)$ .Therefore “t” can take value within the range (0,3) for all $-2<s<-1$..$$case 3.$$when $-1<s<1$;$|s-1|=(-ve),|s+1|=(+ve),|s+2|=(+ve)$.Then the “t” becomes=$(1-s-3(s+1)+s+2)=1-s-3s-3+s+2=-3s $.So “t” can take value within the range (-3,3) for all$ -1<s<1$...$$case 4.$$ When $x>1$;$|s-1|=(+ve),|s+1|=(+ve),|s+2|=(+ve)$.Then the “t” becomes=$(s-1-3s-3+s+2)=(-s-2)$.So, “t” can take value within range (-3,-infinity)….$$CONCLUSION…~~$$SO.now you can see that as “s” varies over real values $“t” can get its maximum value as [+3](including)$ and $”t” can have minimum value as (-infinity)$...Then to find all such integral values of “t” such that the equation given has no solution “t” must lie within the range (+3,+infinity]...I think I could help you with my knowledge...Thanks in advance if you point out any of my fault to be rectified...:)