Question

If $x$ satisfies $|x - 1| + |x - 2| + |x - 3| \geqslant 6{\text{ }}$ then prove that all number $x$ which satisfy the
above relation are given by $x \leqslant 0$ or $x \geqslant 4$

Answer 1

In the given question, we have given one equation for some value and the large of variable to also given. And we have to prove that all the numbers $x$ which satisfy the given equation is given by $2$ value of variable.

Complete step-by-step answer:
In the question, we are given that any variable $x$ satisfiers
$|x - 1| + |x - 2| + |x - 3| \geqslant 6.........(1)$
then from here we have to prove that all numbers $x$ which satisfies the above relation are given by $x \leqslant 0{\text{ or }}x \geqslant 4$
Firstly we substitute $x - 1,x - 2{\text{ and }}x - 3$ equal to new and hence we get there point which are

$$x - 1 = 0{\text{ }}x - 2 = 0{\text{ }}x - 3 = 0{\text{ }} \\\ {\text{ }}x = 1{\text{ }}x = 2{\text{ }}x = 3 \\\ (1,2,3){\text{ }} \\\$$

Therefore change points are $(1,2,3)$
Hence the following cases are possible
$(i){\text{ }}x < 1 \\\ (ii){\text{ 1}} < x < 2 \\\ (iii){\text{ 2 < }}x < 3 \\\ (iv){\text{ }}x > 3 \\\$
We will solve each case one by one and verify that all numbers$x$is given by $x \leqslant 0{\text{ or }}x \geqslant 4$

In case $(i){\text{ }}x < 1$
For this $1$becomes
$- (x - 1) - (x - 2) - (x - 3) \geqslant 6$
Which on solving given
$- x + 1 - x + 2 - x + 3 \geqslant 6$
$\Rightarrow - 3x + 6 \geqslant 6$
Which means $- 3x \geqslant 6 - 6$
Or we get $- 3x \geqslant 0$
$x \leqslant 0$
Which satisfies the condition

In case $\left( {ii} \right)$, $1 < x < 2$
Equation $(1)$ becomes
$(x + 1) - (x - 2) - (x - 3) \geqslant 6$
$\Rightarrow x - 1 - x - 2 - x - 3 \geqslant 6$
Which on solving given
$- x + 4 \geqslant 6$
$\Rightarrow - x \geqslant 2$
This implies $x \leqslant 2$
Which does not satisfy given condition
Hence no solution

In case $(iii){\text{ }}2 < x < 3$
Equation $(1)$becomes
$(x - 1) - (x - 2) - (x - 3) \geqslant 6$
$\Rightarrow x - 1 - x + 2 - x + 3 \geqslant 6$
$- x + y \geqslant 6$
Which on solving gives $- x \geqslant 2$
Which implies $x \leqslant 2$
Which does not given condition
Hence, No solution

In case $(iv){\text{ }}x > 3$
Equation$(1)$ becomes
$p( - 1) - (n - 2) - (x - 3) \geqslant 6$
$\Rightarrow x - 1 - x + 2 - n + 3 \geqslant 6$
Which on solving given
$x \geqslant 4$
There from loses $(i)$ and $(iv)$, all numbers $x$ which satisfy the given relations is given by $x \leqslant 0$and $x \geqslant 4$.

Note: In the question modulus functions is given which means the value of is position $|x|$ when $x > 0$ and the value of $|x|$ is negative $x$ when$x < 0$
Which means |x| = \left\\{ \begin{gathered} \+ x{\text{ for }}x > 0 \\\ \- x{\text{ for }}x < 0 \\\ \end{gathered} \right.
Their value of $x$ (either positive or negative defends or varies as the range of $x$ changes for greater and lesser.