Number of positive integers satisfying (x+2)x2−2x+1−4+3x−x2≥ 0 is
0
1
2
3
(x+2)x2−2x+1−x2+3x−4≥0⇒ (x+2)(x−1)2x2−3x+4≤0 but x2−3x+4>0∀x∈R since ∆ =9-4·1·4=-7<0 ∴ (x+2)(x−1)2≤0⇒ x∈(−∞,−2]∪{1} therefore only 1 is the positive integral solution