1. Translate the problem requirements: Understand that [x] means the ceiling function (smallest integer ≥ x), and we need [x/2] = 0, meaning the smallest integer greater than or equal to x/2 must equal 0
2. Determine the range for x/2: Figure out what values x/2 can take for the ceiling function to equal 0
3. Convert back to find valid x values: Multiply the x/2 range by 2 to get the actual range for x
4. Test answer choices against our range: Check which given options fall within our determined valid range for x

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is asking in plain English. The symbol [x] represents something called the "ceiling function" - this means we take any number x and round it UP to the next whole number (integer). For example:

• If x = 2.3, then [x] = 3 (we round up to 3)
• If x = 5, then [x] = 5 (it's already a whole number)
• If x = -1.7, then [x] = -1 (we round up to -1, not down to -2)

Now our condition is [x/2] = 0. This means when we take x, divide it by 2, and then apply the ceiling function (round up to the next integer), we get 0.

Process Skill: TRANSLATE - Converting the ceiling function notation into understandable language

2. Determine the range for x/2

For the ceiling of x/2 to equal 0, we need to think: what values can x/2 take so that when rounded up, we get 0?

Let's think about this step by step:

• If x/2 is positive (like 0.1 or 0.9), then rounding up gives us 1, not 0
• If x/2 equals 0 exactly, then [0] = 0 ✓
• If x/2 is slightly negative (like -0.1 or -0.9), then rounding up brings us to 0 ✓
• If x/2 is -1 or less negative than -1 (like -1.5), then rounding up gives us -1 or lower, not 0

So for [x/2] = 0, we need: -1 < x/2 ≤ 0

Process Skill: INFER - Drawing the non-obvious conclusion about the range needed for the ceiling function

3. Convert back to find valid x values

Now we multiply our range for x/2 by 2 to find the valid range for x:

If -1 < x/2 ≤ 0, then multiplying all parts by 2:
-2 < x ≤ 0

This means x can be any value greater than -2 (but not equal to -2) and less than or equal to 0.

4. Test answer choices against our range

Let's check each answer choice against our range -2 < x ≤ 0:

• A. x = -2: This is NOT in our range because we need x > -2 (strictly greater than -2)
• B. x = -3/2 = -1.5: This IS in our range because -2 < -1.5 ≤ 0 ✓
• C. x = 1/2 = 0.5: This is NOT in our range because 0.5 > 0
• D. x = 1: This is NOT in our range because 1 > 0
• E. x = 2: This is NOT in our range because 2 > 0

Let's verify option B: If x = -3/2, then x/2 = (-3/2)/2 = -3/4 = -0.75
The ceiling of -0.75 is 0 (we round up from -0.75 to 0) ✓

Final Answer

The answer is B. -3/2
Only option B falls within our required range of -2 < x ≤ 0, and we verified that when x = -3/2, we get [x/2] = [-3/4] = 0.

Common Faltering Points

Errors while devising the approach

• Ceiling Function Confusion: Students often confuse the ceiling function [x] (round up to next integer) with the floor function (round down) or regular rounding. They might think [-1.7] = -2 instead of the correct [-1.7] = -1, leading to completely wrong inequality setup.
• Inequality Direction Misunderstanding: When determining what values make [x/2] = 0, students may incorrectly conclude that x/2 must be positive or may include -1 in their range, not realizing that [-1] = -1, not 0.
• Boundary Condition Errors: Students often struggle with whether endpoints should be included or excluded in inequalities. They might write -1 ≤ x/2 < 0 instead of the correct -1 < x/2 ≤ 0, missing that [0] = 0 but [-1] = -1.

Errors while executing the approach

• Multiplying Inequality Signs: When converting from -1 < x/2 ≤ 0 to the range for x, students might incorrectly multiply by 2 and flip inequality signs or forget to multiply all parts of the compound inequality consistently.
• Fraction Arithmetic Errors: When testing x = -3/2, students may incorrectly calculate x/2 = -3/4, making errors like getting -3/4 = -0.25 instead of -0.75, or confusing the sign during division.

Errors while selecting the answer

• Boundary Value Inclusion Errors: Students might incorrectly select choice A (x = -2) thinking it satisfies the range, forgetting that the range requires x > -2 (strictly greater than), not x ≥ -2.
• Verification Skip: Students may correctly identify the range -2 < x ≤ 0 but fail to verify their chosen answer by substituting back into the original condition [x/2] = 0, missing calculation errors or conceptual mistakes.