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

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

Answer 1

Answer: (B)

$(3,1)$

$f\left(x\right)=\frac{e^{2x}-1-x(k-1)}{x(e^{2x}-1)}$

$f\left(x\right)=\frac{e^{2x}-(k-1)x-1}{x(e^{2x}-1)}=\frac{0}{0}$

Apply L Hospital rule

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

$k-1=2$

$k=3$

Apply L Hospital rule again

$\underset{x\to 0}{lim}\frac{4e^{2x}}{4x{e}^{2x}+2e^{2x}+2e^{2x}}$

$=\frac{4}{2+2}=1$

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

$B$ is correct