Question

If $f\left(x\right)=\{\begin{array}{ccc}lo{g}_{(1-3x)}(1+3x),& for& x\ne 0\\ K,& for& x=0\end{array}$ is continuous at $x=0$, then k is equal to

• A. -2
• B. 2
• C. 1
• D. -1

Answer 1

The correct option is D -1

Given, $f\left(x\right)=\{\begin{array}{cc}lo{g}_{(1-3x)}(1+3x),& forx\ne 0\\ k,& forx=0\end{array}$
Now, $\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{log(1+3x)}{log(1-3x)}$
$=-\underset{x\to 0}{lim}\frac{log(1+3x)}{3x}\cdot \frac{-3x}{log(1-3x)}$
$=-1$
and $f\left(0\right)=k$
$\because f\left(x\right)$ is continuous at $x=0$
$\therefore k=-1$