The correct option is B 0
Given : x2+y2=14 ...(1)
xy=3 ...(2)
Using the identities,
(a+b)2=a2+2ab+b2
and (a−b)2=a2−2ab+b2,
2(x+y)2−5(x−y)2
=2(x2+y2+2xy)−5(x2−2xy+y2)
=2x2 + 2y2 + 4xy - 5x2 + 10xy−5y2
=(2x2−5x2)+(2y2−5y2)+(4xy+10xy)
=−3x2−3y2+14xy
=−3(x2+y2)+14xy
(Substituting values from (1) and (2))
=(−3×14)+(14×3)
=−42+42
=0