If for nonzero x, af(x)+bf1x=1x−5. where a≠b, then f(2) =
3(2b+3a)2a2−b2
3(2b−3a)2a2−b2
3(3a−2b)2a2−b2
6a+b
afx+bf1x=1x-5...........1 put x=1/x⇒af1x+bfx=x-5..........2 a1-b2⇒a2-b2fx=ax-bx-5a-b
put x=2