The solution set of the linear inequation \(|4x-3|\le\displaystyle\frac{3}{4}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\left[\displaystyle\frac{9}{16},\frac{15}{16}\right]\)
For \(a \gt 0\), \(|t| \le a\) means \(-a \le t \le a\) (modulus inequality):
\( \displaystyle -\frac{3}{4} \le 4x - 3 \le \frac{3}{4} \)
Adding 3 throughout:
\( \displaystyle \frac{9}{4} \le 4x \le \frac{15}{4} \)
Dividing by 4:
\( \displaystyle \frac{9}{16} \le x \le \frac{15}{16} \)
The sign is \(\le\), so both endpoints belong to the set: a closed interval.
Why the other options are incorrect
Option (1) \(\left(\displaystyle\frac{9}{16},\frac{15}{16}\right)\) — Incorrect. The endpoints give \(|4x - 3| = \frac{3}{4}\), which satisfies \(\le\), so they must be included.
Option (2) \(\left(-\displaystyle\frac{15}{16},\frac{9}{16}\right)\) — Incorrect. It contains x = 0, but \(|0 - 3| = 3 \gt \frac{3}{4}\).
Option (4) \(\left[-\displaystyle\frac{15}{16},\frac{9}{16}\right]\) — Incorrect. It contains x = 0, but \(|0 - 3| = 3 \gt \frac{3}{4}\).