If limx→0[1+xln(1+b2)]1/x = 2bsin2θ, b>0 and θ∈(-π, π], then the value of θ is ±π/3, ±π/4, ±π/6, ±π/2

Given limx→0[1+xln(1+b2)]1/x = 2bsin2θ
We know that limx→0(1+ax)1/x = ea
So, applying limit in the LHS of given equation, we get
ln(1+b2) = 2bsin2θ
or, 1+b2 = e2bsin2θ
b2 - 2bsin2θ + 1 = 0
This is a quadratic equation in b. For b to be real, the discriminant must be non-negative:
(2sin2θ)2 - 4(1)(1) ≥ 0
4sin4θ - 4 ≥ 0
sin4θ ≥ 1
This implies |sin2θ| ≥ 1, which is only possible if sin2θ = 1.
Therefore, sinθ = ±1, which gives θ = ±π/2.
Alternatively, solving the quadratic equation for b using the quadratic formula:
b = [2sin2θ ± √(4sin4θ - 4)] / 2 = sin2θ ± √(sin4θ - 1)
Since b must be real and positive, we require sin4θ ≥ 1. This means sin2θ = 1, so sinθ = ±1. This gives θ = ±π/2.