A) \[\sqrt{2}\]
B) 1
C) \[\sqrt{5}\]
D) \[\sqrt{3}\]
Correct Answer: D
Solution :
Given, \[|a+b|=1,|a|=|b|=1\] \[\Rightarrow \]\[|a{{|}^{2}}+|{{b}^{2}}|+2|a||b|=1\] \[\Rightarrow \]\[2|a||b|=-1\] Now, \[|a-b{{|}^{2}}=|a{{|}^{2}}+|b{{|}^{2}}-2|a||b|\] \[={{1}^{2}}+{{1}^{2}}-(-1)=3\][From Eq. (i)] \[\Rightarrow \] \[|a-b|=\sqrt{3}\]
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