JEE Main 2026 Jan 21 Shift 2, Mathematics Q13: Set

JEE Main2026Jan 21, Shift 2Mathematics

Let $A = {x : ∣ x^{2} - 10∣ \leq 6}$ and $B = {x : ∣ x - 2∣ > 1}$ . Then

Solution

Solve A: $∣ x^{2} - 10∣ \leq 6 \Leftrightarrow - 6 \leq x^{2} - 10 \leq 6 \Leftrightarrow 4 \leq x^{2} \leq 16$ . So $A = [- 4, - 2] \cup [2, 4]$ .

Solve B: $∣ x - 2∣ > 1 \Leftrightarrow x < 1$ or $x > 3$ , i.e. $B = (- \infty, 1) \cup (3, \infty)$ .

Compute each option carefully:

• $A \cup B = (- \infty, 1) \cup [2, \infty)$ — option (1) is off (uses $(- \infty, 1]$ and $(2, \infty)$ ).

• $A - B$ = elements of A not in B = $[2, 3]$ , not $[2, 3)$ , so option (2) is wrong.

• $A \cap B = [- 4, - 2] \cup (3, 4]$ , not $[- 4, - 2] \cup [3, 4]$ — option (3) is wrong.

• $B - A$ = elements of B not in A = $(- \infty, - 4) \cup (- 2, 1) \cup (4, \infty)$ . Option (4) matches.

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.