If `lim_(x->0)[1+x1n(1+b^2)]^(1/x)=2bsin^2theta,b >0`,where `theta in (-pi,pi],` then the value of `theta` is (a)`+-pi/4` (b) `+-pi/3` (c) `+-pi/6` (d) `+-pi/2`

Explanation

To solve the limit problem, we need to evaluate the limit expression given and find the value of \(\theta\).

### Step-by-Step Solution:

1. **Understanding the Limit Expression**:
We have the limit:
\[
\lim_{x \to 0} \left[1 + x \ln(1 + b^2)\right]^{\frac{1}{x}} = 2b \sin^2 \theta
...