We have,
\(\frac{(x-1)(x-2)}{(x-3)(x-4)}\) ≥ 0
Equating (x − 3)(x − 4) and (x – 1)(x − 2) to zeto we obtain x = 3,4, 1, 2 as critical points. Plot these points on the real line as shown below. The real line is divided into 6 regions.
When x > 4 ∴ \(\frac{(x-1)(x-2)}{(x-3)(x-4)}\) ≥ 0
When 3 < x < 4 ∴ \(\frac{(x-1)(x-2)}{(x-3)(x-4)}\) ≤ 0
When 2 ≤ x < 3 ∴ \(\frac{(x-1)(x-2)}{(x-3)(x-4)}\) ≥ 0
When 1 ≤ ≤ 2 ∴ \(\frac{(x-1)(x-2)}{(x-3)(x-4)}\) ≤ 0
When 1 < < 0 ∴ \(\frac{(x-1)(x-2)}{(x-3)(x-4)}\) ≥ 0
When < 0 ∴ \(\frac{(x-1)(x-2)}{(x-3)(x-4)}\)≥ 0
Hence Solution Set,
{−∞1] ∪ [2,3)(4, ∞)
Hence, solution set:
≥ 4 ∈ [4, ∞)
