If $\tanh ^{-1} x = a \log \left(\frac{1+x}{1-x}\right)$, $|x| < 1$, then $a$ is equal to
- A$1$
- B$2$
- C$\frac{1}{2}$
- D$\frac{1}{4}$
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Similar Questions
If $0 < y < 2^{1/3}$ and $x(y^3 - 1) = 1$,then $\frac{2}{x} + \frac{2}{3x^3} + \frac{2}{5x^5} + \dots$ is equal to:
If $|a| < 1$ and $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$,then $a$ is equal to
DifficultTS EAMCET 2005
View SolutionIf $\alpha, \beta$ are the roots of the equation $x^2 - px + q = 0$,then $\log_e(1 + px + qx^2) = $
Medium
View SolutionIf $|x| < 1$ and $y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$,then $x$ is equal to :
The value of the series $x \log _e a + \frac{x^3}{3!} (\log _e a)^3 + \frac{x^5}{5!} (\log _e a)^5 + \dots$ is
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