Question

If the function $f$ defined as $f(x)=\frac{1}{x}-\frac{k-1}{e^{2x}-1},x\ne 0$, is continuous at $x=0$ , then the ordered pair $(k,f(0))$ is equal to :

• A. (3, 2)
• B. (3, 1)
• C. (2, 1)
• D. $\left(\frac{1}{3},2\right)$

Answer 1

Answer: (B)

Given: $f(x)=\frac{1}{x}-\frac{k-1}{e^{2x}-1};x\ne 0$
$f(x)$ is continuous at $x=0$. Therefore,
$f(0)=\underset{x\to 0}{\lim }\left(\frac{1}{x}-\frac{k-1}{e^{2x}-1}\right)$
$=\underset{x\to 0}{\lim }\frac{(1+(2x)+\frac{1}{2!}(2x{)}^2+\cdots (-1-x(k-1))}{2x^2\left(\frac{e^{2x}-1}{2x}\right)}$
Therefore, clearly $k=3$ for $f(0)=1$