JEE Main 30 June 2022 Shift 1 Solved Paper

Section: Mathematics

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Question : 1 of 90

Marks: +1, -0

Let S1={x∈R−{1,2}:‌

(x+2)(x2+3x+5)

−2+3x−x2

≥0} and S2={x∈R:32x−3x+1−3x+2+27≤0}. Then, S1∪S2 is equal to :

[30-Jun-2022-Shift-1]

(−∞,−2]∪(1,2)
(−∞,−2]∪[1,2]
(−2,1]∪[2,∞)
(−∞,2]

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