Answer:
Option D
Explanation:
Here, $\lim_{x \rightarrow0}[1+x \log (1+b^{2})]^{1/x}$
$[1^{\infty}$ from ]
$\Rightarrow$ $e^{\lim_{x \rightarrow 0} ( x \log [1+b^{2}))}.\frac{1}{x}$
$\Rightarrow$ $ e^{\log(1+b^{2})}=(1+b)^{2}$
Given,
$\lim_{x \rightarrow 0}{(1+x \log (1+b^{2}))}^{1/x}=2b \sin^{2} \theta$
$\Rightarrow$ $(1+b^{2})=2b \sin^{2} \theta$
$\therefore$ $\sin^{2} \theta = \frac{1+b^{2}}{2b}$
By AM $\geq$ GM,
$\frac{b+\frac{1}{b}}{2}\geq \left( b.\frac{1}{b}\right)^{1/2} \Rightarrow \frac{b^{2}+1}{2b}\geq 1$...(iii)
From Eqs.(ii) and (iii) , we get $\sin^{2} \theta=1$
$\Rightarrow$ $\theta = \pm \frac{\pi}{2} $ as $\theta \epsilon (- \pi, \pi )$
