JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1

\lim \limits_{{x \rightarrow 0}}\left\{1+x \ln \left(1+b^2\right)\right\}^{1 / x}=2 b \sin ^2 \theta

b>0

and

\theta \in(-\pi, \pi]

, then the value of

\theta

, is

View answer and explanation

Correct option: D

\pm \frac{\pi}{2}

Option D (

\pm \frac{\pi}{2}

) is correct because by simplifying the given limit equation, we find that

\sin^2 \theta = 1

, which leads to

\theta = \pm \frac{\pi}{2}

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