If the function f defined as f(x) = (1)/(x) - k - 1 e ^ 2x - 1 , x...

Given $f\left(x\right) = \dfrac{1}{x} - \dfrac{k - 1}{{e}^{2x} - 1}, x \ne 0$

and $f\left(x\right)$ is continuous at $x = 0$.

$⇒ f\left(0\right) = \lim_{x \to 0}\left(\dfrac{1}{x} - \dfrac{k - 1}{{e}^{2x} - 1}\right)$

$= \lim_{x \to 0}\dfrac{\left(1 + \left(2x\right) + \dfrac{1}{2 !}{\left(2x\right)}^{2} + \ldots \left(- 1 - x\left(k - 1\right)\right)\right)}{2{x}^{2}\left(\dfrac{{e}^{2x} - 1}{2x}\right)}$

$= \lim_{x \to 0}\dfrac{\left(1 + \left(2x\right) + \dfrac{1}{2 !}{\left(2x\right)}^{2} + \ldots \left(- 1 - x\left(k - 1\right)\right)\right)}{2{x}^{2}}$

Clearly, $k = 3$ and $f\left(0\right) = 1$. $($ $∵$Equating coefficient of $x = 0$$)$