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Question

Let Z be the set of integers. If A={xZ:ex34x27x+10=1} and B={xZ:(1x2)(x22x8)0}, then n[(AB)×(AB)] is

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Solution

ex34x27x+10=1=e0
x34x27x+10=0
By trial and error method, we can find that x=1 is a solution.

Applying synthetic division
x=1147100131013100

x34x27x+10=(x1)(x23x10)=0
(x1)(x5)(x+2)=0
x=2,1,5
A={2,1,5}

B={xZ:(1x)(1+x)(x4)(x+2)0}
B={2,1,1,2,3,4}

AB={2,1,1,2,3,4,5}
AB={2,1}

n[(AB)×(AB)]
=n(AB)×n(AB)=7×2=14


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