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B
2√3
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C
−2
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D
4√3
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Solution
The correct option is C−2 Given:x=2+√3,xy=1Onsolvingforx2−x+y2−y⇒x2−x+xy2x−xy=x2−x+12x−1⇒2+√32−(2+√3)+12(2+√3)−1⇒(2+√3)−√3+12(2+√3)−1⇒(2+√3)−√3+13+2√3=−6−7√3−6+√3√3(3+2√3)...(Addingboththefractions)⇒−(12+6√3)√3(3+2√3)⇒−6(2+√3)3(2+√3)=−63=−2.