Questions & Answers

Question

Answers

a. a – b

b. a + b

c. ln a + ln b

d. None of these

Answer
Verified

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.

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.

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