Question

plz solve this...modulus inequality
$| |x - 1| + 2 \leqslant 4$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given inequality is

$\rm :\longmapsto\: |x - 1| + 2 \leqslant 4$

On Subtracting 2 from both sides, we get

$\rm :\longmapsto\: |x - 1| + 2 - 2 \leqslant 4 - 2$

$\rm :\longmapsto\: |x - 1| \leqslant 2$

We know,

$\boxed{ \rm{ \: |x - a| \leqslant b \: \implies \: - b + a \leqslant x \leqslant b + a}}$

So, using this, we get

$\rm :\longmapsto\: - 2 + 1 \leqslant x \leqslant 2 + 1$

$\rm :\longmapsto\: - 1 \leqslant x \leqslant 3$

$\bf\implies \:x \: \in \: [ - 1, \: 3]$

Additional Information :-

Let solve one more problem of same type!!

Question :- Solve the inequality

$\rm :\longmapsto\: | |x - 3| - 4| \leqslant 2$

Solution :-

Given that

$\rm :\longmapsto\: | |x - 3| - 4| \leqslant 2$

We know that,

$\boxed{ \rm{ \: |x| \leqslant b \: \implies \: - b \leqslant x \leqslant b}}$

So, using this result, we get

$\rm :\longmapsto\: - 2 \leqslant |x - 3| - 4 \leqslant 2$

On adding 4 in each term, we get

$\rm :\longmapsto\: - 2 + 4 \leqslant |x - 3| - 4 + 4\leqslant 2 + 4$

$\rm :\longmapsto\: 2 \leqslant |x - 3| \leqslant 6$

Now, 2 cases arises

Case :- 1

$\rm :\longmapsto\: 2 \leqslant |x - 3|$

$\rm :\longmapsto\: |x - 3| \geqslant 2$

We know,

$\boxed{ \rm{ \: |x| \geqslant b \: \implies \: x \leqslant - b \: \: or \: \: x \geqslant b}}$

So, using this result, we have

$\rm :\longmapsto\:x - 3 \leqslant - 2 \: \: or \: \: x - 3 \geqslant 2$

$\rm :\longmapsto\:x \leqslant 1 \: \: or \: \: x \geqslant 5 - - - (1)$

Case :- 2

$\rm :\longmapsto\: |x - 3| \leqslant 6$

We know,

$\boxed{ \rm{ \: |x - a| \leqslant b \: \implies \: - b + a \leqslant x \leqslant b + a}}$

So, using this result, we get

$\rm :\longmapsto\: - 6 + 3 \leqslant x \leqslant 3 + 6$

$\rm :\longmapsto\: -3 \leqslant x \leqslant 9 - - (2)$

So, from equation (1) and (2), we get

$\bf\implies \:x \: \in \: [ - 3, \: 1] \: \cup \: [5, \: 9]$