If →a and →b are two vectros such that 2|→a|=|→b|, then the greatest value of |2→a+→b|+|2→a−→b| is equal to
A
2√2|→a|
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B
4√2|→a|
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C
6√2|→a|
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D
8√2|→a|
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Solution
The correct option is B4√2|→a| Let angle between →a and →b is θ.
then, |2→a+→b|+|2→a−→b|=√4|→a|2+|→b|2+4|→a||→b|cosθ+√4|→a|2+|→b|2−4|→a||→b|cosθ=2√2|→a|(√1+cosθ+√1−cosθ)[∵|→b|=2|→a|]=2√2|→a|(√2cosθ2+√2sinθ2)=4|→a|(cosθ2+sinθ2)
As, cosθ2+sinθ2≤√2
So, (|→a+→b|+|→a−→b|)max=4√2|→a|