Question

he solution set of the inequality ||x – 2| – 4| ≥ 1 is

Answer 1

Solution :-

$\footnotesize \sf \implies | |x - 2| - 4 | \geqslant 1$

$\footnotesize \sf \implies |x - 2| - 4 \geqslant \pm1$

$\footnotesize \sf \implies |x - 2| - 4 \geqslant 1; |x - 2| - 4 \geqslant - 1$

$\footnotesize \sf \implies |x - 2| \geqslant 1 + 4; |x - 2| \geqslant - 1 + 4$

$\footnotesize \sf \implies |x - 2| \geqslant 5; |x - 2| \geqslant 3$

$\footnotesize \sf \implies x - 2 \geqslant \pm \: 5; x - 2 \geqslant \pm 3$

$\footnotesize \sf \implies x - 2 \geqslant \: 5;x - 2 \geqslant - 5 ;x - 2 \geqslant 3;x - 2 \geqslant - 3$

$\footnotesize \sf \implies x \geqslant \: 5 + 2;x \geqslant - 5 + 2 ;x \geqslant 3 + 2;x \geqslant - 3 + 2$

$\footnotesize \sf\implies \purple{\boxed{ \sf x \geqslant \: 7}; \boxed{ \sf x \geqslant - 3} ; \boxed{ \sf x \geqslant 5}; \boxed{ \sf{x \geqslant - 1}}}$

Plot these all points on a number line and substitute values between the intervals and check whether the interval is a solution set or not.

For interval ( 7, ∞ ) :-

We can substitute any value greater than 7 in the inequity.

Substitute x = 8

$\footnotesize\leadsto\sf ||x-2|-4| \geqslant 1$

$\footnotesize\leadsto\sf ||8 - 2|-4| \geqslant 1$

$\footnotesize\leadsto\sf ||6|-4| \geqslant 1$

$\footnotesize\leadsto\sf |6-4| \geqslant 1$

$\footnotesize\leadsto\sf |2| \geqslant 1$

$\footnotesize\leadsto\sf 2 \geqslant 1$

It is true

For interval ( 5 , 7 ) :-

We can substitute any value greater than 5 and smaller than 7.

Substitute x = 6

$\footnotesize\leadsto\sf ||x-2|-4| \geqslant 1$

$\footnotesize\leadsto\sf ||6 - 2|-4| \geqslant 1$

$\footnotesize\leadsto\sf ||4|-4| \geqslant 1$

$\footnotesize\leadsto\sf |4-4| \geqslant 1$

$\footnotesize\leadsto\sf |0| \geqslant 1$

$\footnotesize\leadsto\sf 0 \geqslant 1$

It is not true so this is not a solution.

For interval ( -1, 5 ) :-

We can substitute any value greater than -1 and smaller than 5.

Substitute x = 0

$\footnotesize\leadsto\sf ||x-2|-4| \geqslant 1$

$\footnotesize\leadsto\sf || 0 - 2|-4| \geqslant 1$

$\footnotesize\leadsto\sf || - 2|-4| \geqslant 1$

$\footnotesize\leadsto\sf |2-4| \geqslant 1$

$\footnotesize\leadsto\sf | - 2| \geqslant 1$

$\footnotesize\leadsto\sf 2 \geqslant 1$

It is true

For interval ( -3 , -1 ) :-

We can substitute any value greater than -3 and smaller than -1.

Substitute x = -2

$\footnotesize\leadsto\sf ||x-2|-4| \geqslant 1$

$\footnotesize\leadsto\sf || - 2-2|-4| \geqslant 1$

$\footnotesize\leadsto\sf || -4|-4| \geqslant 1$

$\footnotesize\leadsto\sf |4-4| \geqslant 1$

$\footnotesize\leadsto\sf |0| \geqslant 1$

$\footnotesize\leadsto\sf 0 \geqslant 1$

It is false so not a solution.

For interval (-∞ , -3 ) :-

We can substitute any value smaller than -3.

Substitute x = -4

$\footnotesize\leadsto\sf ||x-2|-4| \geqslant 1$

$\footnotesize\leadsto\sf || - 4-2|-4| \geqslant 1$

$\footnotesize\leadsto\sf || - 6|-4| \geqslant 1$

$\footnotesize\leadsto\sf |6-4| \geqslant 1$

$\footnotesize\leadsto\sf |2| \geqslant 1$

$\footnotesize\leadsto\sf 2 \geqslant 1$

It is true

$\rule{300}{1}$

We have checked for all the solution intervals but the critical points have not been checked such as -3 , -1 , 5 and 7. So let's check it out.

For -3 :-

$\footnotesize\to ||x-2|-4| \geqslant1$

$\footnotesize\to || - 3 - 2|-4| \geqslant1$

$\footnotesize\to || - 5|-4| \geqslant1$

$\footnotesize\to |5-4| \geqslant1$

$\footnotesize\to | 1| \geqslant1$

$\footnotesize\to 1 \geqslant1$

It is true

For -1 :-

$\footnotesize\to ||x-2|-4| \geqslant1$

$\footnotesize\to || - 1-2|-4| \geqslant1$

$\footnotesize\to || - 3|-4| \geqslant1$

$\footnotesize\to |3-4| \geqslant1$

$\footnotesize\to | - 1| \geqslant1$

$\footnotesize\to 1 \geqslant1$

It is true

For 5 :-

$\footnotesize\to ||x-2|-4| \geqslant1$

$\footnotesize\to ||5-2|-4| \geqslant1$

$\footnotesize\to ||3|-4| \geqslant1$

$\footnotesize\to |3-4| \geqslant1$

$\footnotesize\to | - 1| \geqslant1$

$\footnotesize\to 1 \geqslant1$

It is true

For 7 :-

$\footnotesize\to ||x-2|-4| \geqslant1$

$\footnotesize\to ||7-2|-4| \geqslant1$

$\footnotesize\to ||5|-4| \geqslant1$

$\footnotesize\to |5-4| \geqslant1$

$\footnotesize\to |1 | \geqslant1$

$\footnotesize\to 1 \geqslant1$

It is true

By combining our above results, the final solution :

$\sf \longrightarrow \footnotesize x \in ( - \infty,-3] \cup[-1,5]\cup[7,\infty)$