If the function f (x) defined by


If the function f (x) defined by

$f(x)=\left\{\begin{array}{cl}\frac{\log (1+3 x)-\log (1-2 x)}{x}, & x \neq 0 \\ k & , x=0\end{array}\right.$ is continuous at $x=0$, then $k=$

(a) 1

(b) 5

(c) $-1$


(d) none of these


(b) 5

Given: $f(x)=\left\{\begin{array}{l}\frac{\log (1+3 x)-\log (1-2 x)}{x}, x \neq 0 \\ k, x=0\end{array}\right.$

If $f(x)$ is continuous at $x=0$, then $\lim _{x \rightarrow 0} f(x)=f(0)$.

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

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

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

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

$\Rightarrow 3 \times 1+2 \times 1=k$           $\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$

$\Rightarrow k=3+2=5$

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