The solution set of \(6 \le -3(2x-4) \lt 12,\ \ x \in R\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (0, 1]
Since \(-3(2x - 4) = -6x + 12\),
\(6 \le -6x + 12 \lt 12\)
Subtracting 12 throughout:
\(-6 \le -6x \lt 0\)
Dividing by \(-6\) reverses both inequality signs:
\(1 \ge x \gt 0\)
So \(0 \lt x \le 1\), that is, \(x \in (0, 1]\).
Why the other options are incorrect
Option (2) [1, 0) — Incorrect. Its left end is larger than its right end, so it is not a valid interval.
Option (3) (0, 1) — Incorrect. x = 1 gives \(-3(-2) = 6\), and \(6 \le 6\) holds, so 1 must be included.
Option (4) [0, 1] — Incorrect. x = 0 gives \(-3(-4) = 12\), and \(12 \lt 12\) fails, so 0 must be excluded.