Question

The function $\mathrm{f}\left(\mathrm{x}\right)=\frac{\ln \left(1+\mathrm{ax}\right)-\ln (1-\mathrm{bx})}{\mathrm{x}}$ is not defined at $\mathrm{x}=0$. The value which should be assigned to $\mathrm{f}$ at $\mathrm{x}=0$, so that it is continuous at $\mathrm{x}=0$, is

• A. $\mathrm{a}-\mathrm{b}$
• B. $\mathrm{a}+\mathrm{b}$
• C. $\ln \mathrm{a}-\ln \mathrm{b}$
• D. none of these

Answer 1

The correct option is B

$\mathrm{a}+\mathrm{b}$

Explanation for correct option:

We will define the value of the function at $x=0$ whose function becomes continuous at that point.

L-’Hospital’s rule is a method of evaluating indeterminate forms such as $\frac{0}{0}$ or $\frac{\infty }{\infty }$.

The given function is $\mathrm{f}\left(\mathrm{x}\right)=\frac{\ln \left(1+\mathrm{ax}\right)-\ln (1-\mathrm{bx})}{\mathrm{x}}$

For $\mathrm{f}\left(\mathrm{x}\right)$ to be continuous,

$\mathrm{f}\left(0\right)=\underset{\mathrm{x}\to 0}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

$\therefore$ $\underset{\mathrm{x}\to 0}{\lim }\frac{\log \left(1+\mathrm{ax}\right)-\log \left(1-\mathrm{bx}\right)}{\mathrm{x}}$.

$=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{a}·\log \left(1+\mathrm{ax}\right)}{\mathrm{ax}}+\frac{\mathrm{b}·\log \left(1-\mathrm{bx}\right)}{-\mathrm{bx}}$

. $=\mathrm{a}\times 1+\mathrm{b}\times 1$ $\left[\because \underset{\mathrm{x}\to 0}{\lim }\frac{\log \left(1+\mathrm{x}\right)}{\mathrm{x}}=1\right]$

$\therefore \mathrm{f}\left(0\right)=\mathrm{a}+\mathrm{b}$

Hence, option B is correct answer.