If limx→0 1+xlog1+b21/x=2bsin2θ , b>0 and θ∈[−π,π] then the value of θ is
±π4
±π3
±π6
±π2
limx→0 1+xlog1+b21/x=limx→0 1+xlog1+b21xlog1+b2×log1+b2=elog1+b2=1+b2
thus,
1+b2=2bsin2θ⇒ sin2θ=1+b22b≥1
sin2θ=1⇒sinθ=±1 or θ=±π/2