Question

If $f(x) = \begin{cases} \frac{\sin 3x}{e^{2x} - 1} &; x \ne 0\\ k - 2 & ; x = 0 \end{cases}$ is
Continuous at x = 0, then k =

• A. $\frac{7}{2}$
• B. $\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $\frac{9}{5}$

Answer 1

Answer: (A)

$f(x) = \begin{cases} \frac{sin\:\:3x}{e^{2x}-1} ; & x\ne 0 \\ k-2 ; & \ x=0 \end{cases}$
Since f is continuous at x = 0
$\Rightarrow \, lim_{x\rightarrow0} f\left(x\right)=f\left(0\right)$
$lim _{x\rightarrow0} \frac{sin 3x}{e^{2x}-1}=k-2$
$\Rightarrow \,\frac{lim_{x\rightarrow0}\frac{sin 3x \, \times \, 3x}{3x}}{lim_{x\rightarrow0}\frac{e^{2x}-1 \, \times \, 2x}{2x}}=k-2$
$\Rightarrow \, \frac{3}{2}=k-2$
$\Rightarrow k=\frac{3}{2}+2=\frac{7}{2}$
$\therefore \, k=\frac{7}{2}$