Q117 / 160 · AP EAMCET 2010 · Mathematics

If $\tanh ^{-1} x=a \log \left(\frac{1+x}{1-x}\right),|x| < 1$ then $a$ is equal to

  • A
    1
  • B
    2
  • Correct answer
    C
    $\frac{1}{2}$
  • D
    $\frac{1}{4}$

Solution

$\tanh ^{-1} x=a \log \left(\frac{1+x}{1-x}\right),|x| < 1$ ...(i) But $\tanh ^{-1} x=\frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$ ...(ii) From Eqs. (i) and (ii), we get $a=\frac{1}{2}$

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

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