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 If a=xi+(x1)j+k and b=(x+1)i+j+ak always make an acute angle for all xR ,then the least integral value of a is

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3
b
4
c
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d
6

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detailed solution

Correct option is D

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


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