Q70 / 160 · TS EAMCET 2005 · Mathematics

$2 \tan \mathrm{h}^{-1} \frac{1}{2}$ is equal to

  • A
    0
  • B
    $\log 2$
  • Correct answer
    C
    $\log 3$
  • D
    $\log 4$

Solution

$\begin{aligned} 2 \tanh ^{-1} & \left(\frac{1}{2}\right) \\ & =\tanh ^{-1} \frac{2\left(\frac{1}{2}\right)}{1+\left(\frac{1}{2}\right)^2} \\ & {\left[\because 2 \tanh ^{-1} x=\tanh ^{-1} \frac{2 x}{1+x^2}\right] }\end{aligned}$ $\begin{aligned} & =\tanh ^{-1} \frac{4}{5}=\frac{1}{2} \log \left(\frac{1+\frac{4}{5}}{1-\frac{4}{5}}\right) \\ & \qquad \quad\left[\because \tanh ^{-1} x=\frac{1}{2} \log \left(\frac{1+x}{1-x}\right)\right] \\ & =\frac{1}{2} \log \left(\frac{\frac{9}{5}}{\frac{1}{5}}\right)=\frac{1}{2} \log 3^2=\log 3\end{aligned}$

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

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