Question

If $\underset{x\to 0}{lim}[1+x\ln (1+b^2){]}^{\frac{1}{x}}=2b{\sin }^2\theta ,b>0$ 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

The correct option is D $\pm \frac{\pi }{2}$

Given ${lim}_{x\to 0}{[1+x\ln (1+b^2)]}^{\frac{1}{x}}=2b{\sin }^2\theta$

We know that

${lim}_{x\to 0}{(1+ax)}^{\frac{1}{x}}=e^a$

So, applying limit in the LHS of given equation, we get

$e^{\ln (1+b^2)}=2b{\sin }^2\theta$

or, $1+b^2=2b{\sin }^2\theta$

$b^2-2b{\sin }^2\theta +1=0$

$b=\frac{2{\sin }^2\theta \pm \sqrt{4{\sin }^4\theta -4}}{2}$

$b={\sin }^2\theta \pm \sqrt{{\sin }^2\theta {\sin }^2\theta -1}$

$b={\sin }^2\theta \pm \sqrt{{\sin }^2\theta (1-{\cos }^2\theta )-1}$

$b={\sin }^2\theta \pm \sqrt{-{\cos }^2\theta (1+{\sin }^2\theta )}$

Now for b to be real , $\cos \theta =0$

implies, $\theta =\pm \frac{\pi }{2}$