If a→=xi→+(x−1)j→+k→ and b→=(x+1)i→+j→+ak→ always make an acute angle for all x∈R ,then the least integral value of a is
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a→.b→=[xi→+(x-1)j→+k→]⋅[(x+1)i→+j→+ak→]=x(x+1)+x−1+a=x2+2x+a−1>0,∀x∈R,⇒a>2