Let A = {x : |x-1|< 3} and B = {x : x2 – 2ax + a2 – 4 ≥ 0} be two sets. If A∩B={x:−2<x≤1}, then the smallest positive integral value of ‘a’ is
Let f(x)=x2−2ax+a2−4∵ A∩B={x:−2<x≤1} ⋯(1)∴ vertex≥1f(1)≥0}∩ ⋯ (2)(1)⇒a≥1(2)⇒a2−2a−3≥0⇒(a−3)(a+1)≥0⇒a≤−1 and a≥3
So smallest positive integral value of a is 3.