For what values of $x$, the following identity is valid and holds? tan $\mathrm{h}^{-1}(x)=\frac{1}{2} \log…

For what values of $x$, the following identity is valid and holds? tan $\mathrm{h}^{-1}(x)=\frac{1}{2} \log _e\left(\frac{1+x}{1-x}\right)$.
  1. $(-\infty, \infty)$
  2. $(1, \infty)$
  3. $(-\infty, 1)$
  4. $(-1,1)$

Solution

$ \text \tanh ^{-1} x-\frac{1}{2} \log _e\left(\frac{1+x}{1-x}\right) $
When $x=1$ identity does not hold as $\log _e\left(\frac{2}{0}\right)$ will form. At $x=-1$ also identity does not hold as $\log _e\left(\frac{0}{-2}\right)$ will form and $\log _e 0$ does not exist. When $x < -1$, then $\frac{1+x}{1-x}$ becomes negative, then identity fails to exist. When $x>1$, then $\frac{1+x}{1-x}$ becomes negative, then identity fails to exist. $\therefore$ Identity valid only for $x \in(-1,1)$

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

Practice more Functions questions on Aicharya