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 which of the following options(s) is/are correct:

• A. $\theta =\frac{\pi }{2}$
• B. $\theta =\frac{-\pi }{2}$
• C. $b=1$
• D. $b=2$

Answer 1

Answer: (A), (B), (C)

$b=1$

We have,
$\underset{x\to 0}{lim}[1+x\ln (1+b^2){]}^{\frac{1}{x}}=2b{\sin }^2\theta$
$L.H.S=\underset{x\to 0}{lim}[1+x\ln (1+b^2){]}^{\frac{1}{x}},[1^{\infty }\text{Form}]$
$=e^{x\ln (1+b^2)\times \frac{1}{x}}$
$=e^{\ln (1+b^2)}$

$\Rightarrow e^{\ln (1+b^2)}=2b{\sin }^2\theta$
$\Rightarrow \frac{1+b^2}{2b}={\sin }^2\theta$
$\Rightarrow \frac{1}{b}+b=2{\sin }^2\theta \cdots \left(1\right)$
We know $\frac{1}{b}+b\ge 2$ (using A.M. $\ge$ G.M.)
Also, $2{\sin }^2\theta \le 2$
So, equation $\left(1\right)$ is true if and only if $2{\sin }^2\theta =2$ and $\frac{1}{b}+b=2$
$\therefore \theta =\pm \frac{\pi }{2}$ and $b=1$