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 the ordered pair $(k,f\left(0\right))$ is equal to:

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

Answer 1

The correct option is B $(3,1)$

$f\left(x\right)=\frac{1}{x}-\frac{k-1}{e^{2x}-1},x\ne 0,$
$f\left(x\right)\text{is continuous at}x=0$
$\Rightarrow f\left(0\right)=\underset{x\to 0}{lim}(\frac{1}{x}-\frac{k-1}{e^{2x}-1})$
$\Rightarrow f\left(0\right)=\underset{x\to 0}{lim}\frac{1+\left(2x\right)+\frac{1}{2!}(2x{)}^2+.......+(-1-x(k-1))}{2x^2(\frac{e^{2x}-1}{2x})}$
For limit to exist the numerator must be zero
$\Rightarrow$
$\text{clearly}k=3\text{and}f\left(0\right)=1$