If √3−1√3+1=a+b√3, then the value of 'a' and 'b' is
A
a=2, b=−1
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B
a=2, b=1
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C
a=−2, b=1
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D
a=−2, b=−1
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Solution
The correct option is Aa=2, b=−1 Given √3−1√3+1=a+b√3 LHS =(√3−1)23−1(Multiplyingby(√3−1)withnumeratoranddenominator) =3+1−2√32 =4−2√32 =2−√3 =2−√3=a+b√3 a=2,b=−1