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Answer 1

Given,

$\longrightarrow\sf{f(x)=\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)}$

Here we can find $\sf{f(x)}$ is not defined for $\sf{x=0,}$ but has limit.

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}\left(\lim_{x\to0}\dfrac{\sqrt{1+x^2}-1}{x}\right)}$

By L'hospital's Rule,

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}\left(\lim_{x\to0}\dfrac{x}{\sqrt{1+x^2}}\right)}$

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=0}$

$\displaystyle\longrightarrow\sf{\lim_{x\to0}f(x)=0}$

We can see $\sf{f(x)}$ is an odd function.

$\longrightarrow\sf{f(-x)=\tan^{-1}\left(\dfrac{\sqrt{1+(-x)^2}-1}{-x}\right)}$

$\longrightarrow\sf{f(-x)=\tan^{-1}\left(-\dfrac{\sqrt{1+x^2}-1}{x}\right)}$

$\longrightarrow\sf{f(-x)=-\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)}$

$\longrightarrow\sf{f(-x)=-f(x)}$

Let's check about the nature of $\sf{f(x).}$

$\longrightarrow\sf{f'(x)=\dfrac{d}{dx}\,\left[\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)\right]}$

$\longrightarrow\sf{f'(x)=\dfrac{1}{\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)^2+1}\cdot\dfrac{\dfrac{x^2}{\sqrt{1+x^2}}-\sqrt{1+x^2}+1}{x^2}}$

$\longrightarrow\sf{f'(x)=\dfrac{\dfrac{x^2}{\sqrt{1+x^2}}-\sqrt{1+x^2}+1}{x^2\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)^2+x^2}}$

$\longrightarrow\sf{f'(x)=\dfrac{1-\dfrac{1}{\sqrt{1+x^2}}}{\left(\sqrt{1+x^2}-1\right)^2+x^2}}$

$\longrightarrow\sf{f'(x)=\dfrac{\sqrt{1+x^2}-1}{\left(2(1+x^2)-2\sqrt{1+x^2}\right)\sqrt{1+x^2}}}$

$\longrightarrow\sf{f'(x)=\dfrac{\sqrt{1+x^2}-1}{2\sqrt{1+x^2}\left(\sqrt{1+x^2}-1\right)\sqrt{1+x^2}}}$

$\longrightarrow\sf{f'(x)=\dfrac{1}{2(1+x^2)}}$

Here we get $\sf{f'(x),}$ having domain $\mathbb{R,}$ is always positive for every values of $\sf{x.}$

Therefore, $\sf{f(x)}$ is a strictly increasing function.

Finding limit of $\sf{f(x)}$ as $\sf{x}$ tends to infinity,

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}\left(\lim_{x\to\infty}\dfrac{\sqrt{1+x^2}-1}{x}\right)}$

By L'hospital's Rule,

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}\left(\lim_{x\to\infty}\dfrac{x}{\sqrt{1+x^2}}\right)}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}\left(\lim_{x\to\infty}\sqrt{\dfrac{x^2}{1+x^2}}\right)}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}\sqrt{\lim_{x\to\infty}\dfrac{x^2}{1+x^2}}}$

Again by L'hospitals rule,

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}\sqrt{\lim_{x\to\infty}\dfrac{2x}{2x}}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}\sqrt{\lim_{x\to\infty}1}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}\left(1\right)}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\,\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\dfrac{\pi}{4}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}f(x)=\dfrac{\pi}{4}}$

Hence, since the function is strictly increasing,

$\longrightarrow\sf{For\quad x\in(0,\ \infty),\quad f(x)\in\left(0,\ \dfrac{\pi}{4}\right)}$

And since the function is odd,

$\longrightarrow\sf{For\quad x\in(-\infty,\ 0),\quad f(x)\in\left(-\dfrac{\pi}{4},\ 0\right)}$

Hence the range of $\sf{f(x)}$ is,

$\displaystyle\longrightarrow\sf{f(x)\in\left(-\dfrac{\pi}{4},\ 0\right)\cup\left(0,\ \dfrac{\pi}{4}\right)}$

$\displaystyle\longrightarrow\sf{\underline{\underline{f(x)\in\left(-\dfrac{\pi}{4},\ \dfrac{\pi}{4}\right)-\{0\}}}}$

Hence (b) is the answer.