Question

If $3\cdot {\tan }^{-1}x={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$, then $x$ belongs to

• A. $[-1,1]$
• B. $(-1,1)$
• C. $\left[-\frac{1}{3},\frac{1}{\sqrt{3}}\right]$
• D. $(-\infty ,\infty )$

Answer 1

Answer: (D)

$(-\infty ,\infty )$

Consider the given function

$3{\tan }^{-1}x={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$

Let put $x=\tan \theta then\theta ={\tan }^{-1}x$ we get,

$3{\tan }^{-1}\tan \theta ={\tan }^{-1}\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)$

$\Rightarrow 3\theta ={\tan }^{-1}\tan 3\theta$

$\Rightarrow 3\theta =3\theta$

$\Rightarrow \theta =\theta$

$\theta ={\tan }^{-1}x$

$x=\tan \theta$

We know that,

$\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$ then

$x\in (-\infty ,\infty )$

Hence,the answer is option ${}^{\prime }{D}^{\prime }$