Set Theory – Sets – JEE Main 21 Jan 2026 Shift 2

Solution:

First, solve the inequality for set $A$:
$$|x^{2}-10|\le6$$

$$-6 \le x^{2}-10 \le 6$$

$$4 \le x^{2} \le 16$$

Taking the square root, we get two disjoint intervals:
$$x \in [-4, -2] \cup [2, 4]$$
So, $A = [-4, -2] \cup [2, 4]$.

Next, solve the inequality for set $B$:
$$|x-2| > 1$$

$$x-2 > 1 \quad \text{or} \quad x-2 < -1$$
$$x > 3 \quad \text{or} \quad x < 1$$
So, $B = (-\infty, 1) \cup (3, \infty)$.

Now, calculate $B – A$ (elements present in $B$ but not in $A$):
$B – A = ((-\infty, 1) \cup (3, \infty)) – ([-4, -2] \cup [2, 4])$

Subtracting the interval $[-4, -2]$ from $(-\infty, 1)$:
The remaining part is $(-\infty, -4) \cup (-2, 1)$.

Subtracting the interval $[2, 4]$ from $(3, \infty)$:
The intersection is $(3, 4]$. Removing this from $(3, \infty)$ leaves $(4, \infty)$.

Combining the remaining parts:
$$B – A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$$

Ans. (4)