Question
The function $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{x}\ne0\\\frac{\text{k}}{2},&\text{x}=0\end{cases}$ is continuous at x = 0, then k =
- 3
- 6
- 9
- 12
Solution:
Given, $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{x}\ne0\\\frac{\text{k}}{2},&\text{x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{\text{x}}=\text{f}(0)$
$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}=\frac{\text{k}}{2}$
$\Rightarrow3\times1=\frac{\text{k}}{2}$
$\Rightarrow\frac{\text{k}}{2}=3$
$\Rightarrow\text{k}=6$
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