All real values of x which satisfy x2−3x+2>0 and x2−3x−4≤0 lie in the interval
A
[−1,1)∪(2,4]
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B
[3,6]
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C
(−5,0)
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D
[1,6]
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Solution
The correct option is A[−1,1)∪(2,4] x2−3x+2>0 and x2−3x−4≤0 x2−3x+2>0⇒(x−1)(x−2)>0⇒x∈(−∞,1)∪(2,∞)⋯(1) x2−3x−4≤0⇒(x+1)(x−4)≤0⇒x∈[−1,4]⋯(2) From (1) and (2), x∈[−1,1)∪(2,4]