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 =
  1. 3
  2. 6
  3. 9
  4. 12

Answer

  1. 6

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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