Question

If the function $ƒ$ defined on $\left(-\frac{1}{3},\frac{1}{3}\right)$ by

$f\left(x\right)=\left\{\begin{array}{l}\left(\frac{1}{\mathrm{k}}\right){\log }_{\mathrm{e}}\left(\frac{1+3\mathrm{x}}{1-2\mathrm{x}}\right)\mathrm{when}\mathrm{x}\ne 0\\ \mathrm{k}\mathrm{when}\mathrm{x}=0\end{array}\right.$

is continuous, then $k$ is equal to:

Answer 1

Solve for value of $k$

Given that the function $ƒ$ defined on $\left(-\frac{1}{3},\frac{1}{3}\right)$ by $f\left(x\right)=\left\{\begin{array}{l}\left(\frac{1}{\mathrm{k}}\right){\log }_{\mathrm{e}}\left(\frac{1+3\mathrm{x}}{1-2\mathrm{x}}\right)\mathrm{when}\mathrm{x}\ne 0\\ \mathrm{k}\mathrm{when}\mathrm{x}=0\end{array}\right.$is continuous

We know that for any continuous function $\underset{x\to 0}{\lim }f\left(x\right)=f\left(x^{-}\right)=f\left(x^{+}\right)$

So, for the given function, $\underset{x\to 0}{\lim }\left(\frac{1}{x}\right){\log }_e\frac{1+3x}{1-2x}=k$

$$\begin{gathered} \Rightarrow \underset{x\to 0}{\lim }\left(\frac{1}{x}\right)\left[{\log }_e(1+3x)-{\log }_e(1-2x)\right]=k\left[\because \log \left(\frac{m}{n}\right)=\log \left(m\right)-\log \left(n\right)\right] \\ \Rightarrow \underset{x\to 0}{\lim }\left[\frac{{\log }_e(1+3x)}{x}-\frac{{\log }_e(1-2x)}{x}\right]=k \\ \Rightarrow \underset{x\to 0}{\lim }\left[\frac{3{\log }_e(1+3x)}{3x}+\frac{2{\log }_e(1-2x)}{-2x}\right]=k \\ \Rightarrow 3+2=k\left[\because \underset{x\to 0}{\lim }\frac{\log \left(1+x\right)}{x}=1\right] \\ \Rightarrow 5=k \end{gathered}$$

Therefore, $f\left(x\right)$ is continuous when $k=5$.