Question
If the function $\mathrm{f}$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x)=\left\{\begin{array}{ll}{\frac{1}{x} \log _{e}\left(\frac{1+3 x}{1-2 x}\right)} & {, \text { when } x \neq 0} \\ {k} & {, \text { when } x=0}\end{array}\right.$ is continuous, then $\mathrm{k}$ is equal to

Answer

b
$\mathrm{k}=\lim _{\mathrm{x} \rightarrow 0}\left(\frac{\ln (1+3 \mathrm{x})}{\mathrm{x}}-\frac{\ln (1-2 \mathrm{x})}{\mathrm{x}}\right)$

$\mathrm{k}=3+2=5$

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