Q65 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

If for $|x|>1, \tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$, then $f(-5)=$

  • A
    $\frac{3}{2}$
  • B
    $\frac{-2}{3}$
  • Correct answer
    C
    $\frac{2}{3}$
  • D
    $\frac{1}{3}$

Solution

Given that, $\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1} x=\log _e(f(x))$ for $|x|>1$ $\because \quad \tanh ^{-1}\left(\frac{1}{x}\right)=\frac{1}{2} \log _e\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right)$ $\left(\because \tanh ^{-1} x=\frac{1}{2} \log _e\left(\frac{1+x}{1-x}\right)\right)$ $\tanh ^{-1}\left(\frac{1}{x}\right)=\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right)$ $\ldots(i)$ and $\operatorname{coth}^{-1} x=\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right)$ $\ldots(ii)$ Adding the both Eqs. (i) and (ii), we get $\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1} x=\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right)$ $+\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right)$ $=\log _e\left(\frac{x+1}{x-1}\right)$ $\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1} x=\log _e(f(x))$ On comparing, we get $f(x)=\frac{x+1}{x-1}$ At $x=-5, f(-5)=\frac{-5+1}{-5-1}=\frac{-4}{-6} \Rightarrow f(-5)=\frac{2}{3}$

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

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