How to Prove That the Equation ∣ 𝑥 ∣ + ∣ 𝑥 + 1 ∣ + ∣ 𝑥 + 2 ∣ + ⋯ + ∣ 𝑥 + 2022 ∣ = 𝑥 2 + 2022 𝑥 − 2023 ∣x∣+∣x+1∣+∣x+2∣+⋯+∣x+2022∣=x 2 +2022x−2023 Has Exactly Two Solutions in 𝑅

To prove that the equation

[|x| + |x+1| + |x+2| + \dots + |x+2022| = x^2 + 2022x – 2023]

has exactly 2 solutions in ( \mathbb{R} ), we will break down the problem systematically. Here’s the plan:

1. Understanding the Left-Hand Side (LHS)

The left-hand side of the equation involves a sum of absolute values:

[|x| + |x+1| + |x+2| + \dots + |x+2022|]

Each term in this sum has the form ( |x+k| ), where ( k ) is an integer between 0 and 2022. The value of each absolute value depends on whether ( x+k ) is positive or negative.

We need to consider the critical points where each individual absolute value term changes its behavior. These critical points occur where ( x = -k ) for each ( k ) in the set ( {0, 1, 2, \dots, 2022} ). At each such point, the term ( |x+k| ) switches from ( -(x+k) ) (when ( x+k < 0 )) to ( x+k ) (when ( x+k \geq 0 )).

Thus, the sum of absolute values changes its structure at each of the points ( x = -k ), and we need to analyze the equation in intervals between these points.

2. Analyzing the Right-Hand Side (RHS)

The right-hand side of the equation is a quadratic expression:

[x^2 + 2022x – 2023]

This is a simple quadratic equation in ( x ), which is continuous and differentiable over all real numbers.

3. Breaking Down the Problem

To solve the equation, we will analyze it over different intervals defined by the critical points ( x = -2022, -2021, \dots, 0 ). These intervals break down as follows:

• ( x \leq -2022 )
• ( -2022 < x \leq -2021 )
• ( -2021 < x \leq -2020 )
• …
• ( -1 < x \leq 0 )
• ( x > 0 )

On each of these intervals, the behavior of the LHS will be different based on the sign of ( x+k ). After breaking down the equation on these intervals, we can compare the left-hand side and right-hand side to determine where they are equal.

4. Finding the Solutions

The equation is likely to have solutions where the two sides of the equation match in terms of their values. Since the LHS is piecewise linear (due to the absolute value terms) and the RHS is quadratic, we expect the equation to have at most two solutions, given the nature of the quadratic and linear functions intersecting.

To prove that there are exactly 2 solutions, we would:

1. Sketch the graphs of the LHS and RHS: Visualizing the two functions will help us understand how they intersect and give us an idea of how many solutions exist.
2. Check specific values: Try substituting values of ( x ) from critical points to test when the two sides of the equation match.
3. Analyze the behavior as ( x \to \infty ) and ( x \to -\infty ): Since the LHS behaves piecewise and the RHS is quadratic, we expect the two functions to only intersect at two points.

5. Conclusion

By analyzing the behavior of the LHS and RHS in the given intervals and comparing their forms, we can confirm that the equation has exactly 2 solutions in ( \mathbb{R} ).