Question

The number of solutions of the inequation $\left| {x - 2} \right| + \left| {x + 2} \right| < 4$ is
A) $1$
B) $2$
C) $4$
D) $0$
E) $3$

Answer 1

Hint: $\left| {ax + b} \right| < k$ is a modulus or absolute inequality. These kind of inequalities can be solved by keeping in mind that if $\left| x \right| < a$ then $ - a < x < a$ .We know that graph of a modulus function is V-shaped and the point at which the graph changes its nature from decreasing to increasing is $ - a$ if the modulus function is $\left| {x + a} \right|$ . This point is found by putting the value of modulus function equal to zero. So, we will consider the points 2, -2 while simplifying the given inequality.

Complete step by step solution:
The given modulus inequality is $\left| {x - 2} \right| + \left| {x + 2} \right| < 4$ . This expression contains two different absolute values so we will take different cases for different values of $x$ :
Case 1:
When $x < - 2$
Then
$x - 2 < 0$ and $x + 2 < 0$
So, $\left| {x - 2} \right| = 2 - x$ and $\left| {x + 2} \right| = - x - 2$
Thus,
$
  \left| {x - 2} \right| + \left| {x + 2} \right| < 4 \\
   \Rightarrow 2 - x - x - 2 < 4 \\
   \Rightarrow - 2x < 4 \\
   \Rightarrow x > - 2 \\
 $
Which is contradicting, so $x$ is not smaller than $ - 2$
Case 2:
When $x > 2$
$x - 2 > 0$ and $x + 2 > 0$
So, $\left| {x - 2} \right| = x - 2$ and $\left| {x + 2} \right| = x + 2$
Thus,
$
  \left| {x - 2} \right| + \left| {x + 2} \right| < 4 \\
   \Rightarrow x - 2 + x + 2 < 4 \\
   \Rightarrow 2x < 4 \\
   \Rightarrow x < 2 \\
 $
Which is contradicting, so $x$ is not greater than \[2\]
Case 3:
When $ - 2 < x < 2$
Then $x - 2 < 0$ and $x + 2 > 0$
So, $\left| {x - 2} \right| = 2 - x$ and $\left| {x + 2} \right| = x + 2$
Thus,
$
  \left| {x - 2} \right| + \left| {x + 2} \right| < 4 \\
   \Rightarrow 2 - x + x + 2 < 4 \\
   \Rightarrow 4 < 4 \\
 $
No value of $x$ satisfies the given modulus inequality, so there is no solution to the given inequality.

The correct option is (D).

Note:
When observing different cases of different values of $x$ , note the value of $\left| {x - 2} \right|$ and $\left| {x + 2} \right|$ carefully. The given inequality contains two absolute terms that’s why we solve it using different values of $x$ .