Q105 / 160 · AP EAMCET 2021 (25 Aug Shift 2) · Mathematics

If $\lim _{x \rightarrow 0}\left\{1+x \log \left(1+a^2\right)\right\}^{1 / x}=2 a \sin ^2 \theta, a>0$ and $\theta \in R$, then

  • Correct answer
    A
    $\theta=n \pi \pm \frac{\pi}{2},(n \in Z)$
  • B
    $\theta=2 n \pi \pm \frac{\pi}{2},(n \in Z)$
  • C
    $\theta=n \pi+\frac{\pi}{2},(n \in Z)$
  • D
    $\theta=n \pi \pm \frac{\pi}{4},(n \in Z)$

Solution

$\lim _{x \rightarrow 0}\left\{1+x \log \left(1+a^2\right)\right\}^{1 / x}$ $$ =2 a \sin ^2 \theta, a>0 \text { and } \theta \in R $$ LHS $\lim _{x \rightarrow 0}\left\{1+x \log \left(1+a^2\right)\right\}^{1 / x}$ is of the form $\infty 1^{\infty}$ $$ \begin{aligned} & \Rightarrow e^{\lim _{x \rightarrow 0} \frac{1}{x}\left\{1+x \log \left(1+a^2\right)-1\right\}} \Rightarrow e^{\lim _{x \rightarrow 0} \log \left(1+a^2\right)} \\ & \Rightarrow 1+a^2=2 a \sin ^2 \theta \Rightarrow a^2-2 a \sin ^2 \theta+1=0 \end{aligned} $$ $\begin{aligned} \Rightarrow \quad a & =\frac{2 \sin ^2 \theta \pm \sqrt{4 \sin ^4 \theta-4}}{2} \\ a & =\frac{2 \sin ^2 \theta \pm \sqrt{4\left(\sin ^4 \theta-1\right)}}{2} \\ a & =\sin ^2 \theta \pm \sqrt{\sin ^4 \theta-1}\end{aligned}$ $$ \begin{aligned} & \because \quad a>0 \\ & \Rightarrow \sin ^4 \theta=1 \\ & \sin ^2 \theta=1=\sin ^2 \frac{\pi}{2} \\ & \Rightarrow \quad \theta=n \pi \pm \frac{\pi}{2} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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