\(\frac{1}{{\text{x}}-1} \le 2, x \in R\)
Subtracting 2 from both the sides in the above equation
Signs of 3 – 2x:
3 – 2x = 0 → x = \(\frac{3}{2}\)
(Subtracting by 3 on both the sides, then multiplying by -1 on both the sides and then dividing both the sides by 2)
3 – 2x < 0 → x > \(\frac{3}{2}\)
(Subtracting by 3 on both the sides, then multiplying by -1 on both the sides and then dividing both the sides by 2)
3 – 2x > 0 → x < \(\frac{3}{2}\)
(Subtracting by 3 on both the sides, then multiplying by -1 on both the sides and then dividing both the sides by 2)
Signs of x – 1:
x – 1 = 0 → x = 1 (Adding 1 on both the sides)
x – 1 < 0 → x < 1 (Adding 1 on both the sides)
x – 1 > 0 → x > 1 (Adding 1 on both the sides)
Zeroes of denominator:
x – 1 = 0 → x = 1
At x = 1, \(\frac{3-2x}{x-1}\) is not defined
Intervals satisfying the condition: ≤ 0
x < 1 and x ≥ \(\frac{3}{2}\)
Therefore,
x ∈(-∞,1) U \(\bigg[ \frac{3}{2}, \infty\bigg)\)