Question

If $ lim_{ x \to 0 } [ 1 + x \, log \, (1 + b^2) ]^{\frac{1}{x}} = 2 b sin^2 \, \theta, b > 0 \, and \, \theta \in ( - \pi , \pi), $ then the value of $ \theta $ is

• A. $ \pm \frac{\pi}{4}$
• B. $ \pm \frac{\pi}{3}$
• C. $ \pm \frac{\pi}{6}$
• D. $ \pm \frac{\pi}{2}$

Answer 1

Answer: (D)

Here, $ lim_{ x \to 0 } \{ 1 + x \, log (1 + b^2) \}^{1/x}$
= $ lim_{ x \to 0 } \{ x \, log (1 + b^2) \} . \frac{1}{x} $
= $ e^{log (1 + b^2)} = (1 + b^2) $
Given, $ lim_{ x \to 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 $ \ge $ GM, $ \frac{ b + \frac{1}{b}}{ 2} \ge \bigg(b. \frac{1}{b}\bigg)^{1/2}$
$\Rightarrow \frac{b^2 + 1}{2b} \ge 1 $
From Eqs. (ii) and (iii),
$ $ $ sin^2 \theta = 1 \Rightarrow = \pm \frac{\pi}{2}, \, as \, \theta \, \in \, (- \pi, \pi] $