Question

If \[x\] satisfies \[\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6\] then prove that all the numbers of \[x\] which satisfy the above relation are given by \[x \leqslant 0\] or \[x \geqslant 4\].

Answer 1

Hint: To solve this problem we need to identify the change points of the equation.
From these points we will consider certain cases to find the range of \[x\]
The cases that fall into the range as well as satisfy the given condition
We will get the required answer.

Complete step-by-step solution:
Here the change points will be \[x = 1,2,3\].
\[(i)x \leqslant 1\] \[(ii)1 < x \leqslant 2\] \[(iii)2 < x \leqslant 3\] \[(iv)x > 3\]
We will find for each case if the value of x is possible such that it satisfies \[\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6\]
\[(i)x \leqslant 1 \to (1)\]
Here the given condition is\[\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6\] can be written as \[1 - x + 2 - x + 3 - x \geqslant 6\] since modulus of any number gives its absolute value.
\[ \Rightarrow 1 - x + 2 - x + 3 - x \geqslant 6\]
We add the constant terms together and \[x\] terms together.
\[ \Rightarrow 6 - 3x \geqslant 6\]
Taking \[6\] to right hand side as \[ - 6\]
\[ \Rightarrow - 3x \geqslant 6 - 6\]
Subtracting \[6\] from \[6\]results in \[0\]
\[ \Rightarrow - 3x \geqslant 0\]
Since there is a \[-\] sign on the left-hand side, we will multiply both sides by \[-\] to get the value of\[x\].
In doing so, the inequality will change and result in
\[ \Rightarrow x \leqslant 0\]
As we have considered in equation\[\left( 1 \right)\], \[x \leqslant 1\] means that the values of \[x\] can be any value less than or equal to \[1\] but not greater than\[1\].
Hence \[x \leqslant 0\] is possible when \[x \leqslant 1\], this case is true.
\[(ii)1 < x \leqslant 2 \to (2)\]
Here \[\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6\] can be written as \[x - 1 + 2 - x + 3 - x \geqslant 6\]
Since modulus of any number gives its absolute value.
\[ \Rightarrow x - 1 + 2 - x + 3 - x \geqslant 6\]
We add the constant terms together and \[x\] terms together.
\[ \Rightarrow 4 - x \geqslant 6\]
Taking \[4\] from left hand side to right hand side
\[ \Rightarrow - x \geqslant 6 - 4\]
On subtracting the term and we get,
\[ \Rightarrow - x \geqslant 2\]
Since there is a \[-\] sign on the left-hand side, we will multiply both sides by \[-\] to get the value of \[x\].
In doing so, the inequality will change and result in
\[ \Rightarrow x \leqslant - 2\]
As we have considered in equation \[\left( 2 \right)\], \[1 < x \leqslant 2\] means that the values of x can be any value greater than but not including \[1\] up to\[2\], and not greater than \[2\].
Since \[x \leqslant - 2\] cannot be possible when\[1 < x \leqslant 2\], this case is not possible.
\[(iii)2 < x \leqslant 3 \to (3)\]
Here the given condition is \[\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6\] can be written as \[x - 1 + x - 2 + 3 - x \geqslant 6\] since modulus of any number gives its absolute value.
\[ \Rightarrow x - 1 + x - 2 + 3 - x \geqslant 6\]
We add the constant terms together and \[x\] terms together.
\[ \Rightarrow 2x - x \geqslant 6\]
Let us subtract the term and we get,
\[ \Rightarrow x \geqslant 6\]
As we have considered in equation \[\left( 3 \right)\], \[2 < x \leqslant 3\] means that the values of x can be any value greater than but not including \[2\] up to \[3\], and not greater than \[3\].
Since \[x \geqslant 6\] cannot be possible when \[2 < x \leqslant 3\], this case is not possible.
\[(iv)x > 3 \to (4)\]
Here the given condition is \[\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6\] can be written as \[x - 1 + x - 2 + x - 3 \geqslant 6\]
Since modulus of any number gives its absolute value.
\[ \Rightarrow x - 1 + x - 2 + x - 3 \geqslant 6\]
We add the constant terms together and \[x\] terms together.
\[ \Rightarrow 3x - 6 \geqslant 6\]
We take 6 from left hand side to right hand side
\[ \Rightarrow 3x \geqslant 6 + 6\]
On adding we get,
\[ \Rightarrow 3x \geqslant 12\]
Let us divide \[3\] on both side we get,
\[ \Rightarrow x \geqslant 4\]
As we have considered in equation \[\;\left( 4 \right)\], \[x > 3\] means that the values of x can be any value greater than \[3\] but not \[3\] or any value less than\[3\].
Since \[x \geqslant 4\] is possible when\[x > 3\], this case is true.
Now, from the results of the above \[4\] cases, it is evident that when \[x\] satisfies \[\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6\] the value of \[x\] will be given only when \[x \leqslant 0\] or \[x \geqslant 4\]
Hence proved.

Note: In this type of question we need to be aware that the modulus of any terms will give its absolute value. That is the negative sign can be neglected.
For example, $\left| 5 \right| = \left| 5 \right|$ and $\left| { - 25} \right| = \left| {25} \right|$
Here some of the properties of the modulus value as follow:
$\left| x \right| + \left| y \right| \geqslant \left| {x + y} \right|$
$\left| 0 \right| = 0$
$\left| x \right| \geqslant 0$
The next thing we need to be careful when we deal with inequalities; If we have to multiply or divide both sides of an inequality by a negative number, we will reverse the direction of the given inequality sign.