Question

If $\displaystyle \lim_{x \to 0}{{\left[ 1+x\ln \left( 1+{{b}^{2}} \right) \right]}^{\dfrac{1}{x}}}=2b{{\sin }^{2}}\theta $ , $b>0$ and $\theta \in (-\pi ,\pi ]$ ,then the value of $\theta $ is
(A) $\pm \dfrac{\pi }{4}$
(B) $\pm \dfrac{\pi }{3}$
(C) $\pm \dfrac{\pi }{6}$
(D) $\pm \dfrac{\pi }{2}$

Answer 1

Hint: For answering the question we need to find the value of $\theta $ when $\displaystyle \lim_{x \to 0}{{\left[ 1+x\ln \left( 1+{{b}^{2}} \right) \right]}^{\dfrac{1}{x}}}=2b{{\sin }^{2}}\theta $ , $b>0$ and $\theta \in (-\pi ,\pi ]$. For doing that we will use the limit formula given as $\displaystyle \lim_{x \to 0}{{\left( 1+ax \right)}^{\dfrac{1}{x}}}={{e}^{a}}$ and the range specified as ${{\sin }^{2}}\theta \in \left[ 0,1 \right]$ and $b+\dfrac{1}{b}\in [2,\infty )\forall b>0$.

Complete step by step solution:
Now considering from the question we have been asked to find the value of $\theta $ when $\displaystyle \lim_{x \to 0}{{\left[ 1+x\ln \left( 1+{{b}^{2}} \right) \right]}^{\dfrac{1}{x}}}=2b{{\sin }^{2}}\theta $ , $b>0$ and $\theta \in (-\pi ,\pi ]$ .
For doing that we will use the limit formula which we have learnt in basics of limits given as $\displaystyle \lim_{x \to 0}{{\left( 1+ax \right)}^{\dfrac{1}{x}}}={{e}^{a}}$.
From the basic concepts of trigonometry we know that the range of sine square function is specified as ${{\sin }^{2}}\theta \in \left[ 0,1 \right]$ and similarly from basic algebra we know that the range is $b+\dfrac{1}{b}\in [2,\infty )\forall b>0$ .
Now we will simplify the given expression. After simplifying we will have
$\begin{align}
  & \displaystyle \lim_{x \to 0}{{\left[ 1+x\ln \left( 1+{{b}^{2}} \right) \right]}^{\dfrac{1}{x}}}=2b{{\sin }^{2}}\theta \\
 & \Rightarrow {{e}^{\ln \left( 1+{{b}^{2}} \right)}}=2b{{\sin }^{2}}\theta \\
 & \Rightarrow 1+{{b}^{2}}=2b{{\sin }^{2}}\theta \\
 & \Rightarrow \dfrac{1+{{b}^{2}}}{b}=2{{\sin }^{2}}\theta \\
 & \Rightarrow b+\dfrac{1}{b}=2{{\sin }^{2}}\theta \\
\end{align}$
By verifying the common range for the expressions on both sides of the simplified expression we will have only 2 in the common range.
Hence we can say that ${{\sin }^{2}}\theta =1$ .
The solutions of $\sin \theta =\pm 1$ are given as $n\pi \pm \dfrac{\pi }{2}$ .
As here we have been asked within the given limits $\theta \in (-\pi ,\pi ]$ the answer is $\pm \dfrac{\pi }{2}$ .
Hence we will mark the option “D” as correct.

Note: While answering questions of this type we should be sure with our concepts that we are going to apply and the calculations that we are going to perform. Similarly the solutions of $\sin \theta =\pm \dfrac{1}{\sqrt{2}}$ are given as $n\pi \pm \dfrac{\pi }{4}$ and many more.