Question
If the function f(x) defined by $\text{f(x)}=\begin{cases}\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, then k =
- 1
- 5
- -1
- None of these.
Solution:
Given, $\text{f(x)}=\begin{cases}\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$
If f(x) is continuous at x = 0, then $\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f(0)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}}\Big)=\text{k}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3\log(1+3\text{x})}{3\text{x}}-\frac{2\log(1-2\text{x})}{2\text{x}}\Big)=\text{k}$
$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})}{3\text{x}}\Big)-2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1-2\text{x})}{2\text{x}}\Big)=\text{k}$
$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})}{3\text{x}}\Big)+2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1-2\text{x})}{-2\text{x}}\Big)=\text{k}$
$\Rightarrow3\times1+2\times1=\text{k}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\Big]$
$\Rightarrow\text{k}=3+2$
$\Rightarrow\text{k}=5$
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