Question

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

Answer 1

$\left(b\right)a+b$

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

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

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

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