Answer:
Option D
Explanation:
We have function,
$f(x)=\frac{(e^{kx}-1)\tan kx}{4x^{2}},x\neq0$
=16 x=0
is continuous at x=0
$\therefore$ $\lim_{x \rightarrow 0} f(x)= f(0)$
$\Rightarrow$ $ \lim_{x \rightarrow 0} \frac{(e^{kx}-1)\tan kx}{4x^{2}}=16\left(\frac{0}{0}form\right)$
$\Rightarrow$ $ \lim_{x \rightarrow 0} \frac{(e^{kx}-1)}{kx}\left(\frac{\tan kx}{kx}\right)\times\frac{k^{2}}{4}=16$
$\Rightarrow$ $ \frac{k^{2}}{4}=16\Rightarrow k^{2}=64\Rightarrow k\pm 8$
