The solution set of the given inequation \( \displaystyle\frac{2x+3}{x-1}\lt 4\ (x\ne 1) \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( (-\infty,1)\cup\left(\displaystyle\frac{7}{2},\infty\right) \)
The sign of x − 1 is not known, so the inequality cannot be cross-multiplied. Subtracting 4 from both sides:
\( \displaystyle \frac{2x+3}{x-1}-4\lt 0 \)
\( \displaystyle \frac{2x+3-4x+4}{x-1}\lt 0 \)
\( \displaystyle \frac{7-2x}{x-1}\lt 0 \)
Multiplying by −1 (the inequality reverses):
\( \displaystyle \frac{2x-7}{x-1}\gt 0 \)
The critical points are x = 1 and \( x=\frac{7}{2} \). Sign of the fraction:
\( \displaystyle x\lt 1:\ \frac{(-)}{(-)}\gt 0 \)
\( \displaystyle 1\lt x\lt\frac{7}{2}:\ \frac{(-)}{(+)}\lt 0 \)
\( \displaystyle x\gt\frac{7}{2}:\ \frac{(+)}{(+)}\gt 0 \)
The inequality is strict and x ≠ 1, so both end points are excluded:
\( \displaystyle x\in(-\infty,1)\cup\left(\frac{7}{2},\infty\right) \)
Why the other options are incorrect
Option (1) \( (-\infty,-1]\cup\left[\displaystyle\frac{3}{2},\infty\right) \) — Incorrect. This set contains x = 2, where the left side is \( \frac{7}{1}=7 \), not less than 4; the critical points are 1 and \( \frac{7}{2} \), not −1 and \( \frac{3}{2} \).
Option (2) \( \left(1,\displaystyle\frac{7}{2}\right) \) — Incorrect. \( \left(1,\frac{7}{2}\right) \) is where the fraction is greater than 4; for example x = 2 gives 7.
Option (4) \( (-\infty,1]\cup\left[\displaystyle\frac{7}{2},\infty\right) \) — Incorrect. The square brackets include x = 1, where the expression is not defined, and \( x=\frac{7}{2} \), where it equals 4, not less than 4.