Question

Let A = {x : |x-1|< 3} and B = {x : $x^2$ – 2ax + $a^2$ – 4 $\ge$ 0} be two sets. If $A\cap B=\{x:-2<x\le 1\}$, then the smallest positive integral value of ‘a’ is ___

Answer 1

$Letf\left(x\right)=x^2-2ax+a^2-4\because A\cap B=\{x:-2<x\le 1\}\cdots \left(1\right)\therefore \begin{array}{ccc}vertex& \ge & 1\\ f\left(1\right)& \ge & 0\end{array}\}\cap \cdots \left(2\right)\left(1\right)\Rightarrow a\ge 1\left(2\right)\Rightarrow a^2-2a-3\ge 0\Rightarrow (a-3)(a+1)\ge 0\Rightarrow a\le -1anda\ge 3$
So smallest positive integral value of a is 3.