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The function $\mathrm{f}$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by
$\mathrm{f}(x)=\left\{\begin{array}{cc}
\frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}\right), & x \neq 0 \\
\mathrm{k} & , \quad x=0
\end{array}\right.$
is continuous at $x=0$, then $\mathrm{k}$ is
Options:
$\mathrm{f}(x)=\left\{\begin{array}{cc}
\frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}\right), & x \neq 0 \\
\mathrm{k} & , \quad x=0
\end{array}\right.$
is continuous at $x=0$, then $\mathrm{k}$ is
Solution:
2133 Upvotes
Verified Answer
The correct answer is:
5
$\mathrm{f}$ is continuous at $x=0$.
$\begin{aligned}
\therefore \quad \mathrm{f}(0) & =\lim _{x \rightarrow 0} \mathrm{f}(x) \\
\therefore \quad \mathrm{k} & =\lim _{x \rightarrow 0}\left(\frac{1}{x} \log (1+3 x)-\frac{1}{x} \log (1-2 x)\right) \\
& =\lim _{x \rightarrow 0}\left(\frac{3 \log (1+3 x)}{3 x}+\frac{2 \log (1-2 x)}{-2 x}\right) \\
& =3+2=5
\end{aligned}$
$\begin{aligned}
\therefore \quad \mathrm{f}(0) & =\lim _{x \rightarrow 0} \mathrm{f}(x) \\
\therefore \quad \mathrm{k} & =\lim _{x \rightarrow 0}\left(\frac{1}{x} \log (1+3 x)-\frac{1}{x} \log (1-2 x)\right) \\
& =\lim _{x \rightarrow 0}\left(\frac{3 \log (1+3 x)}{3 x}+\frac{2 \log (1-2 x)}{-2 x}\right) \\
& =3+2=5
\end{aligned}$
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