The function f defined on (-1 3 , 1 3 ) by f(x)= cc 1 x (1+3 x 1-… - Vidyadip

MCQ

The function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x)=\left\{\begin{array}{cc}\frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}\right) & , x \neq 0 \\ k & , \quad x=0\end{array}\right.$ is continuous at $x=0$, then $k$ is

A
$6$
B
$1$
✓
$5$
D
$-5$

✓

Answer

Correct option: C.

$5$

So, $\lim _{x \rightarrow 0} f(x)=f(0)=k$
$\Rightarrow k=\lim _{x \rightarrow 0} \frac{\log \frac{(1+3 x)}{(1-2 x)}}{x}$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\frac{(1+3 x)}{(1-2 x)}-1\right)}{\left(\frac{3 x+1}{1-2 x}-1\right) x} \times\left[\left(\frac{3 x+1}{1-2 x}\right)-1\right]$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\frac{5 x}{1-2 x}\right)}{\frac{5 x}{1-2 x}} \times \frac{5}{1-2 x}$
$[$As we know that $\lim _{f(x) \rightarrow 0} \frac{\log (1+f(x))}{f(x)}=1]$
$\Rightarrow \lim _{x \rightarrow 0} \frac{5}{1-2 x}=5$
$\Rightarrow k=5$

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