The function $f\left( x \right)=\dfrac{\ln \left( 1+ax \right)-\ln \left( 1-bx \right)}{x}$ not defined at x = 0. The value which should be assigned to f, at x = 0 so that it is continuous at x = 0, is
 a. a – b
b. a + b
c. ln a + ln b
d. None of these

Hint: In order to solve this question, we should know about the concept of limits on logarithmic function, like $\underset{f\left( x \right)\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+f\left( x \right) \right)}{f\left( x \right)}=1$. So, we will try to form the given function in this manner to apply the limit. Now, the question says that the function should be continuous at x = 0, it means, if $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ exists, $f\left( x \right)$ will be continuous at x = 0. By using this concept, we will get the answer.

Complete step-by-step answer:

In this question, we have been asked to find the value of $f\left( x \right)=\dfrac{\ln \left( 1+ax \right)-\ln \left( 1-bx \right)}{x}$ at x = 0 such that the function becomes continuous at x = 0. To solve this question, we will find the limits of $f\left( x \right)$ at x = 0 to get the value of $f\left( x \right)$ such that it will become continuous at x = 0. So, we can say, for continuous function, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ should exist. Now, let us calculate $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$. So, we can write it as,
$\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+ax \right)-\ln \left( 1-bx \right)}{x}$
We can further write it as,
$\begin{align}
  & \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{\ln \left( 1+ax \right)}{x}-\dfrac{\ln \left( 1-bx \right)}{x} \right] \\
 & \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+ax \right)}{x}-\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1-bx \right)}{x} \\
\end{align}$
Now, we will multiply and divide $\dfrac{\ln \left( 1+ax \right)}{x}$ by (a) and $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1-bx \right)}{x}$ by (-b). So, we get, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{a\ln \left( 1+ax \right)}{ax}-\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( -b \right)\ln \left( 1-bx \right)}{\left( -b \right)x}$
Now, we know that $\underset{f\left( x \right)\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+f\left( x \right) \right)}{f\left( x \right)}=1$. So, for $f\left( x \right)=ax$, we get $\underset{ax\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+ax \right)}{ax}=1$.
And for $f\left( x \right)=-bx$, we get $\underset{-bx\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+\left( -b \right)x \right)}{\left( -b \right)x}=1$.
Therefore, we can write, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ as,
$\begin{align}
  & \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=a\times 1-\left( -b \right) \\
 & \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=a+b \\
\end{align}$
Hence, we can say that $f\left( x \right)$ should be assigned as (a + b) at x= 0 so that it is continuous at x = 0.
Therefore, option (b) is the correct answer.

Note: While solving this question, there are possibilities that we might get stuck at $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1-bx \right)}{x}$. So, we can write this as $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+\left( -b \right)x \right)}{x}$ and then multiply its numerator and denominator by (-b) and then applying the property, $\underset{f\left( x \right)\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+f\left( x \right) \right)}{f\left( x \right)}=1$ and simplifying to get the answer. Also, there are chances that we might make calculation mistakes. So, we have to be very careful while solving this question.