If lceil x rceil denotes the least integer greater than or equal to x and lceil x/2 rceil = 0,...

1. Translate the problem requirements: Understand that \(\lceil \mathrm{x} \rceil\) means the ceiling function (smallest integer ≥ x), and we need \(\lceil \mathrm{x}/2 \rceil = 0\), meaning the smallest integer greater than or equal to \(\mathrm{x}/2\) must equal 0
2. Determine the range for x/2: Figure out what values \(\mathrm{x}/2\) can take for the ceiling function to equal 0
3. Convert back to find valid x values: Multiply the \(\mathrm{x}/2\) 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 \(\lceil \mathrm{x} \rceil\) 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 \(\mathrm{x} = 2.3\), then \(\lceil \mathrm{x} \rceil = 3\) (we round up to 3)
• If \(\mathrm{x} = 5\), then \(\lceil \mathrm{x} \rceil = 5\) (it's already a whole number)
• If \(\mathrm{x} = -1.7\), then \(\lceil \mathrm{x} \rceil = -1\) (we round up to -1, not down to -2)

Now our condition is \(\lceil \mathrm{x}/2 \rceil = 0\). 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 \(\mathrm{x}/2\) to equal 0, we need to think: what values can \(\mathrm{x}/2\) take so that when rounded up, we get 0?

Let's think about this step by step:

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

So for \(\lceil \mathrm{x}/2 \rceil = 0\), we need: \(-1 < \mathrm{x}/2 \leq 0\)

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 \(\mathrm{x}/2\) by 2 to find the valid range for x:

If \(-1 < \mathrm{x}/2 \leq 0\), then multiplying all parts by 2:
\(-2 < \mathrm{x} \leq 0\)

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 \(-2 < \mathrm{x} \leq 0\):

1. \(\mathrm{x} = -2\): This is NOT in our range because we need \(\mathrm{x} > -2\) (strictly greater than -2)
2. \(\mathrm{x} = -\frac{3}{2} = -1.5\): This IS in our range because \(-2 < -1.5 \leq 0\) ✓
3. \(\mathrm{x} = \frac{1}{2} = 0.5\): This is NOT in our range because \(0.5 > 0\)
4. \(\mathrm{x} = 1\): This is NOT in our range because \(1 > 0\)
5. \(\mathrm{x} = 2\): This is NOT in our range because \(2 > 0\)

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

Final Answer

The answer is B. \(-\frac{3}{2}\)

Only option B falls within our required range of \(-2 < \mathrm{x} \leq 0\), and we verified that when \(\mathrm{x} = -\frac{3}{2}\), we get \(\lceil \mathrm{x}/2 \rceil = \lceil -\frac{3}{4} \rceil = 0\).

Common Faltering Points

Errors while devising the approach

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

Errors while executing the approach

• Multiplying Inequality Signs: When converting from \(-1 < \mathrm{x}/2 \leq 0\) 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 \(\mathrm{x} = -\frac{3}{2}\), students may incorrectly calculate \(\mathrm{x}/2 = -\frac{3}{4}\), making errors like getting \(-\frac{3}{4} = -0.25\) 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 (\(\mathrm{x} = -2\)) thinking it satisfies the range, forgetting that the range requires \(\mathrm{x} > -2\) (strictly greater than), not \(\mathrm{x} \geq -2\).
• Verification Skip: Students may correctly identify the range \(-2 < \mathrm{x} \leq 0\) but fail to verify their chosen answer by substituting back into the original condition \(\lceil \mathrm{x}/2 \rceil = 0\), missing calculation errors or conceptual mistakes.