Let A={x:xisrealsatisfyingx4<5x−23−(7x−3)5} and B={x:xisrealsatisfying2x2x2+5x+2>1x+1}, then
A
A∩B=ϕ
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B
A∪B=Z,Z is an integer
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C
A∩B=A
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D
A∩B=B
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Solution
The correct option is AA∩B=ϕ x4<5x−23−(7x−3)5⇒x>4⇒xϵ(4,∞)∴A=(4,∞)2x2x2+5x+2>1x+1⇒−(3x−2)(2x+1)(x+2)(x+1)>0⇒3x+2(2x+1)(x+2)(x+1)<0⇒xϵ(−2,−1)∪(−23,−12)∴B=(−2,−1)∪(−23,−12)⇒A∩B=ϕ