Let S=\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\} and T=\{x \in \mathbb{Z}: x^2-7|x|+9 \leq 0\} Then the number of elements in {\S} \cap {\T} is : [28-Jul-2022-Shift-2]

JEE Mains 28-July-2022 Shift 2 Solved Paper

Section: Mathematics

Question : 1 of 90

Marks: +1, -0

Let S={x∈[−6,3]−{−2,2}:‌

|x+3|−1

|x|−2

≥0} and T={x∈ℤ:x2−7|x|+9≤0}
Then the number of elements in S∩T is :

[28-Jul-2022-Shift-2]

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