Question

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

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

Answer 1

The correct option is A $(3,2)$

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{+}}f\left(x\right)=f\left(0\right)$

${lim}_{x\to 0}\frac{1}{x}-\frac{k-1}{e^{2x}-1}\Rightarrow {lim}_{x\to 0}\frac{e^{2x}-1-(k-1)x}{x(e^{2x}-1)}$

As we know, $e^{2x}=1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}\dots .$

$\begin{array}{cc}& \Rightarrow \underset{x\to 0}{lim}\frac{(1+2x+\frac{4x^2}{2!}+\dots )-1-(k-1)x}{x[1+2x+\frac{4x^3}{2!}\cdots -1]}\\ \Rightarrow & \underset{x\to 0}{lim}\frac{(2+\frac{4x}{2!}+\dots )-(k-1{)}^{-}}{x[2+\frac{4x}{2!}+\frac{8x^2}{3!}\dots ]}\end{array}$

$\begin{array}{cc}\text{for the limit to exist}& \\ 2-(k-1)=& 0\\ \Rightarrow 2=k-1\Rightarrow k=3& \end{array}$

Hence, ${lim}_{x\to 0}\frac{\frac{4}{2!}+\frac{8x}{3!}+\dots }{2+\frac{4x}{21}+\cdots }={lim}_{x\to 0}\frac{2}{2}=1$

Hence, $(k,f(0\left)\right)=(3,1)$

option (B).