Answer: B
\(2+bā3=\cfrac{1}{2+ā3}\)
or, \(b=-1 \)
\(2+bā3=\cfrac{1}{2+ā3}\)
\(=\cfrac{(2-ā3)}{(2+ā3)(2-ā3)}\)
\( =\cfrac{2-ā3}{4-3}\)
\( =2-ā3\)
\(ā“bā3=-ā3\)or, \(b=-1 \)
\(=\cfrac{(2-ā3)}{(2+ā3)(2-ā3)}\)
\( =\cfrac{2-ā3}{4-3}\)
\( =2-ā3\)
\(ā“bā3=-ā3\)