Question

The number of solutions of the inequation \( |x - 2| + |x + 2| < 4 \) is:

• A. \( 1 \)
• B. \( 2 \)
• C. \( 4 \)
• D. \( 0 \)
• E. infinite

Hint:

Geometrically, \( |x-a| + |x-b| \ge |a-b| \). The sum of distances to two points can never be less than the distance between the points themselves (Triangle Inequality).

Answer 1

The Correct Option is D

Concept: The expression \( |x - a| + |x - b| \) represents the sum of distances from a point \( x \) to the points \( a \) and \( b \) on the number line. The minimum value of this sum is the distance between \( a \) and \( b \), which occurs for any \( x \) in the interval \( [a, b] \).

Step 1: Analyze the distance property.
We have the expression \( f(x) = |x - 2| + |x - (-2)| \). This is the sum of the distance from \( x \) to \( 2 \) and the distance from \( x \) to \( -2 \). On the number line, the distance between \( -2 \) and \( 2 \) is: \[ |2 - (-2)| = 4 \]

Step 2: Determine the minimum value.
1. If \( x > 2 \): Both terms are positive. \( (x-2) + (x+2) = 2x \). Since \( x > 2 \), \( 2x > 4 \). 2. If \( -2 \le x \le 2 \): \( |x-2| = 2-x \) and \( |x+2| = x+2 \). Sum = \( (2-x) + (x+2) = 4 \). 3. If \( x < -2 \): Both terms are negative. \( -(x-2) - (x+2) = -2x \). Since \( x < -2 \), \( -2x > 4 \). Conclusion: The value of \( |x - 2| + |x + 2| \) is always \(\ge 4\).

Step 3: Conclusion.
The inequation asks for values where the sum is strictly less than 4. Since the minimum value is 4, there are no real values of \( x \) that satisfy \( f(x) < 4 \).