Question

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

• A. $k=3$
• B. $k=1$
• C. $f\left(0\right)=1$
• D. $f\left(0\right)=3$

Answer 1

Answer: (A), (C)

$f\left(0\right)=1$

$f\left(x\right)=\frac{1}{x}-\frac{k-1}{e^{2x}-1},x\ne 0$
$f\left(x\right)\text{is continuous at}x=0$
$\Rightarrow f\left(0\right)=\underset{x\to 0}{lim}(\frac{1}{x}-\frac{k-1}{e^{2x}-1})$
$\Rightarrow f\left(0\right)=\underset{x\to 0}{lim}\frac{1+\left(2x\right)+\frac{1}{2!}(2x{)}^2+\cdots +(-1-x(k-1))}{2x^2(\frac{e^{2x}-1}{2x})}$
For limit to exist the coefficient of $x$ should be $0.$
$\therefore k=3,f\left(0\right)=1$