If[x+ [2x ]] <3 where [] denotes the greatest integer function, then

We are given the inequality \( x + \left\lfloor 2x \right\rfloor < 3 \), where \( \left\lfloor y \right\rfloor \) denotes the greatest integer function, meaning the largest integer less than or equal to \( y \).

### Step-by-step solution:

1. **Understanding the greatest integer function:**

   The function \( \left\lfloor 2x \right\rfloor \) represents the greatest integer less than or equal to \( 2x \).

2. **Breaking the problem down:**
   
   We need to find the values of \( x \) that satisfy the inequality:
   \[
   x + \left\lfloor 2x \right\rfloor < 3.
   \]
   
   To approach this, consider different intervals based on the value of \( x \), as the floor function \( \left\lfloor 2x \right\rfloor \) changes at integer values of \( 2x \).

3. **Case analysis:**

   Let's look at different intervals for \( x \):

   - **When \( 0 \leq x < 0.5 \):**
     - In this range, \( 0 \leq 2x < 1 \), so \( \left\lfloor 2x \right\rfloor = 0 \).
     - The inequality becomes:
       \[
       x + 0 < 3 \quad \text{or} \quad x < 3.
       \]
     - Since \( x \) is already constrained by \( 0 \leq x < 0.5 \), the inequality holds for this range.

   - **When \( 0.5 \leq x < 1 \):**
     - In this range, \( 1 \leq 2x < 2 \), so \( \left\lfloor 2x \right\rfloor = 1 \).
     - The inequality becomes:
       \[
       x + 1 < 3 \quad \text{or} \quad x < 2.
       \]
     - Since \( x \) is already constrained by \( 0.5 \leq x < 1 \), the inequality holds for this range.

   - **When \( 1 \leq x < 1.5 \):**
     - In this range, \( 2 \leq 2x < 3 \), so \( \left\lfloor 2x \right\rfloor = 2 \).
     - The inequality becomes:
       \[
       x + 2 < 3 \quad \text{or} \quad x < 1.
       \]
     - This condition contradicts \( 1 \leq x < 1.5 \), so there are **no solutions** in this range.

   - **When \( 1.5 \leq x < 2 \):**
     - In this range, \( 3 \leq 2x < 4 \), so \( \left\lfloor 2x \right\rfloor = 3 \).
     - The inequality becomes:
       \[
       x + 3 < 3 \quad \text{or} \quad x < 0.
       \]
     - This condition contradicts \( 1.5 \leq x < 2 \), so there are **no solutions** in this range.

4. **Conclusion:**

   The only values of \( x \) that satisfy the inequality are those in the range \( 0 \leq x < 1 \).

   Therefore, the solution is:
   \[
   0 \leq x < 1.
   \]

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Would you like more details on how the greatest integer function affects inequalities, or any other clarification?

Here are some related questions you may find useful:

1. How does the greatest integer function impact the solution of quadratic inequalities?
2. Can the greatest integer function be used in solving rational inequalities?
3. What are the key properties of the greatest integer function that are useful in algebraic manipulation?
4. How can we extend this method to inequalities involving higher powers of \( x \)?
5. What happens if we replace the greatest integer function with a ceiling function?

**Tip:** To solve inequalities involving the greatest integer function, always break down the problem into cases depending on the integer boundaries for \( 2x \). This allows you to simplify and solve the inequality step by step.

Learn how to solve the inequality x + ⌊2x⌋ < 3 involving the greatest integer (floor) function. This step-by-step solution explores different intervals for x and explains how to apply the greatest integer function in solving inequalities. Suitable for high school students.

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