Let A={x∈Z:3(x+1)(x2−7x+12)=1}and B={x∈Z:−5<2x−1≤7}, where Z is the set of integers. If the number of relations from A to B is 2k then the value of k is
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Solution
3(x+1)(x2−7x+12)=1=30 ⇒(x+1)(x−3)(x−4)=0⇒x=−1,3,4 ∴A={−1,3,4} −5<2x−1≤7⇒−4<2x≤8⇒−2<x≤4 ∴x=−1,0,1,2,3,4 ∴B={−1,0,1,2,3,4} No. of relations from A to B=2n(A)⋅n(B)=218 Hence k=18