Answer
Verified
39.9k+ views
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.
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.
Recently Updated Pages
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
Let gx 1 + x x and fx left beginarray20c 1x 0 0x 0 class 12 maths JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
Other Pages
The percentage error in quantities P Q R and S are class 11 physics JEE_Main
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
How many grams of concentrated nitric acid solution class 11 chemistry JEE_Main