Question

If the function f (x) defined by
$f\left(x\right)=\left\{\begin{array}{ll}\frac{\log \left(1+3x\right)-\log \left(1-2x\right)}{x},& x\ne 0\\ k,& x=0\end{array}\right.$is continuous at x = 0, then k =
(a) 1
(b) 5
(c) −1
(d) none of these

Answer 1

$\left(b\right)5$

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

If f(x) is continuous at x = 0, then​$\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$.

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

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