Mathematics · 2026
JEE Main · 21 January 2026, Shift 2 · Q8
Let A ={x:|x^2-10| ≤ 6} and B ={x:|x-2|>1}. Then
Let $\displaystyle \mathrm{A}=\left\{x:\left|x^2-10\right| \leq 6\right\}$ and $\displaystyle \mathrm{B}=\{x:|x-2|>1\}$. Then
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \mathrm{B}-\mathrm{A}=(-\infty,-4) \cup(-2,1) \cup(4, \infty)$
JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.