The number of integers satisfying log1x2(x−2)(x+1)(x−5)≥1 is
0
1
2
3
Case I: 1/x>1 or 0<x<1∴ log1x2(x−2)(x+1)(x−5)≥1⇒ 2(x−2)(x+1)(x−5)≥1x⇒ 2(x−2)(x+1)(x−5)−1x≥0⇒ 2x(x−2)−(x+1)(x−5)x(x+1)(x−5)≥0⇒ x2+5x(x+1)(x−5)≥0 ⇒xx+1x-5>0 since x2+5 >0 for all x∈R⇒ x∈(−1,0)∪(5,∞)
Hence, no solution in this case.
Case II: 0<1x<1 or x>1 →1 ∴ 0<x<5 →2 Also 2(x−2)(x+1)(x−5)>0 ⇒x∈ -1,2∪5,∞→ 3 from 1,2,3 we can take x∈(1,2)