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Q.
If the function $f$ defined as $f(x) = \frac{1}{x} - \frac{k - 1}{e^{2x} - 1} , x \neq 0$, is continuous at $x = 0$ , then the ordered pair $(k, f (0))$ is equal to :
JEE MainJEE Main 2018Continuity and Differentiability
Solution:
Given: $f (x)=\frac{1}{x}-\frac{k-1}{e^{2 x}-1} ; x \neq 0$
$f(x)$ is continuous at $x=0$. Therefore,
$f(0)=\displaystyle\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{k-1}{e^{2 x}-1}\right)$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\left(1+(2 x)+\frac{1}{2 !}(2 x)^{2}+\cdots(-1-x(k-1))\right.}{2 x^{2}\left(\frac{e^{2 x}-1}{2 x}\right)}$
Therefore, clearly $k=3$ for $f (0)=1$