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Question

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

$\mathrm{a}-\mathrm{b}$

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B

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

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C

$\mathrm{ln}\mathrm{a}-\mathrm{ln}\mathrm{b}$

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D

none of these

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Solution

## 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 $\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ln}\left(1+\mathrm{ax}\right)-\mathrm{ln}\left(1-\mathrm{bx}\right)}{\mathrm{x}}$For $\mathrm{f}\left(\mathrm{x}\right)$ to be continuous, $\mathrm{f}\left(0\right)=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)$$\therefore$ $\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left(1+\mathrm{ax}\right)-\mathrm{log}\left(1-\mathrm{bx}\right)}{\mathrm{x}}$. $=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{a}·\mathrm{log}\left(1+\mathrm{ax}\right)}{\mathrm{ax}}+\frac{\mathrm{b}·\mathrm{log}\left(1-\mathrm{bx}\right)}{-\mathrm{bx}}$. $=\mathrm{a}×1+\mathrm{b}×1$ $\left[\because \underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left(1+\mathrm{x}\right)}{\mathrm{x}}=1\right]$$\therefore \mathrm{f}\left(0\right)=\mathrm{a}+\mathrm{b}$Hence, option B is correct answer.

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