Question

Solve the following: $1 \leqslant |x - 2| \leqslant 3$

Answer 1

Hint: In the question we are given an equation which is a linear inequality. First, we will split our equation in two parts, then each part will give us two conditions, then using those conditions we will find the solution of each part, then we will take the intersection of solutions to both parts to get our required answer.

Complete step by step answer:
Our given equation is $1 \leqslant |x - 2| \leqslant 3$
Let us split this equation in two parts
So, first part is
$1 \leqslant |x - 2|$
Or we can write it as
$|x - 2| \geqslant 1$
From the property of modulus function
If $|x| \leqslant a$ then $ - a \leqslant x \leqslant a$ where $a$ is any real number.
From this we can also generalize that
If $|x| \geqslant a$ then $ - a \geqslant x \geqslant a$ where $a$ is any real number.
Using this in our given inequality.
$|x - 2| \geqslant 1$
This will give us two conditions
$ - 1 \geqslant x - 2$ Or $x - 2 \geqslant 1$
Simplifying first condition
$ - 1 \geqslant x - 2$
Or we can write it as
$x - 2 \leqslant - 1$
Now we will simplify this linear inequality
Transposing $ - 2$ to right hand side
$
 x - 2 \leqslant - 1 \\
 \Rightarrow x \leqslant - 1 + 2 \\
 \Rightarrow x \leqslant 1 \\
 $
According to this condition $x$ should be equal to less than $1$ which means $x \in ( - \infty ,1]$
We can take $x = 1$ so we have taken a closed bracket at the end.
Now we will simplify our second condition
$x - 2 \geqslant 1$
Now we will simplify this linear inequality
Transposing $ - 2$ to right hand side
$
 x - 2 \geqslant 1 \\
 \Rightarrow x \geqslant 1 + 2 \\
 \Rightarrow x \geqslant 3 \\
 $
According to this condition we can take $x$ greater than or equal to $3$ which means $x \in [3,\infty )$
We can take $x = 3$ so we have taken a closed bracket in the starting.
Now we will join both the conditions
$|x - 2| \geqslant 1$
According to first condition $x \in ( - \infty ,1]$
According to second condition $x \in [3,\infty )$
Joining both we will get
$x \in ( - \infty ,1] \cup [3,\infty )$
Now we will move to second part of our given equation
$|x - 2| \leqslant 3$
This will give us two conditions
$ - 3 \leqslant x - 2$ Or $x - 2 \leqslant 3$
Simplifying first condition
$ - 3 \leqslant x - 2$
Or we can write it as
$x - 2 \geqslant - 3$
Now we will simplify this linear inequality
Transposing $ - 2$ to right hand side
$
 x - 2 \geqslant - 3 \\
 \Rightarrow x \geqslant - 3 + 2 \\
 \Rightarrow x \geqslant - 1 \\
 $
According to this condition $x$ should be equal to greater than $ - 1$ which means $x \in [ - 1,\infty )$
We can take $x = - 1$ so we have taken a closed bracket in the starting.
Now we will simplify our second condition
$x - 2 \leqslant 3$
Now we will simplify this linear inequality
Transposing $ - 2$ to right hand side
$
 x - 2 \leqslant 3 \\
 \Rightarrow x \leqslant 3 + 2 \\
 \Rightarrow x \leqslant 5 \\
 $
According to this condition we can take $x$ less than or equal to $5$ which means $x \in ( - \infty .5]$
We can take $x = 5$ so we have taken a closed bracket at the end.
Now we will join both the conditions
$|x - 2| \leqslant 3$
According to first condition $x \in [ - 1,\infty )$
According to second condition $x \in ( - \infty .5]$
Joining both we will get
$x \in [ - 1,5]$
Now we will join solutions of both parts
$x \in ( - \infty ,1] \cup [3,\infty )$And $x \in [ - 1,5]$
Taking intersection $( \cap )$
We will get
$x \in [ - 1,1] \cup [3,5]$
$\therefore x \in [ - 1,1] \cup [3,5]$ is our required solution.

Note:
When we want to join any numbers of sets then we take union which is represented as $ \cup $ and when we want to take out the common part in any number of sets then we take intersection which is represented as $ \cap $. Let $A$ and $B$ are two sets then union is represented as $A \cup B$ and intersection is represented as $A \cap B$ .