Answer
Verified
480k+ views
Hint: Since we have to test the differentiability, we have to find the derivative of the function. We can find derivatives using first principles and using the functional relation given in the question.
It is given in the question $f\left( xy \right)=f\left( x \right)+f\left( y \right)\forall x,y\ne 0\text{ }\And f'\left( 1 \right)=3,$
To check differentiability, we have to find $f'\left( x \right)$. To find $f'\left( x \right)$ we have to follow certain no. of steps,
1. Use first principle to find $f'\left( x \right)$
We know by first principle,
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x\left( 1+\dfrac{h}{x} \right) \right)-f\left( x \right)}{h}.........\left( I \right) \\
\end{align}$
Since it is given $f\left( xy \right)=f\left( x \right)+f\left( y \right)$, we can substitute $f\left( x\left( 1+\dfrac{h}{x} \right) \right)=f\left( x \right)+f\left( 1+\dfrac{h}{x} \right)$ in $\left( I \right)$,
$\begin{align}
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)+f\left( 1+\dfrac{h}{x} \right)-f\left( x \right)}{h} \\
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)}{h} \\
\end{align}$
Now, we cannot proceed further in step 1. So, we proceed to step 2.
2. We will find some boundary values of $f\left( x \right)$.
Given $f\left( xy \right)=f\left( x \right)+f\left( y \right)$
Substituting $x=1,y=1$ in the above functional relation, we get 🡪
$\begin{align}
& f\left( 1 \right)=f\left( 1 \right)+f\left( 1 \right) \\
& \Rightarrow f\left( 1 \right)=2f\left( 1 \right) \\
& \Rightarrow f\left( 1 \right)=0......................\left( II \right) \\
\end{align}$
Now we will go to step 1. Adding/subtracting $0$to any term will not cause any change in value.
So, in the final expression of step1, we will do the following changes,
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-0}{h}$
Now, we will multiple and divide the denominator with $x$. Also, from $\left( II \right)$, we can substitute 0 as f1.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{x.\dfrac{h}{x}}$
Since limit is of $h$, we can take $\dfrac{1}{x}$ out of the limit.
$f'\left( x \right)=\dfrac{1}{x}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{\dfrac{h}{x}}$
Since it is given $x\ne 0,$ we can replace $\underset{h\to 0}{\mathop{\lim }}\,\text{ by }\underset{\dfrac{h}{x}\to 0}{\mathop{\lim }}\,$
$f'\left( x \right)=\dfrac{1}{x}\underset{\dfrac{h}{x}\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{\dfrac{h}{x}}...........\left( III \right)$
If we substitute $x=1$ in the first principle of derivative, we will get,
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\
& x=1\to \\
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}.........\left( IV \right) \\
\end{align}$
In $\left( IV \right)$, we can write $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}=f'\left( 1 \right)$
Similarly, in $\left( III \right)$, we can write
\[\underset{\dfrac{h}{x}\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{\dfrac{h}{x}}=f'\left( 1 \right)\]
So, substituting in $\left( III \right)$, we get,
\[f'\left( x \right)=\dfrac{1}{x}\times of'\left( 1 \right)\]
It is given \[f'\left( 1 \right)=3\],
$\Rightarrow f'\left( x \right)=\dfrac{3}{x}$
If we draw the graph of $f'\left( x \right)=\dfrac{3}{x}$, we get 🡪
This graph is not continuous at $x=0$ since $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f'\left( x \right)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f'\left( x \right)$.
So we can say that $f\left( x \right)$ is differentiable $\ for all x\in R-\left\{ 0 \right\}$.
Note: There is a possibility of mistake while finding the boundary value of the function in step 2. The boundary value is to be found by taking help of the information given in the question. For example, it was given in the question that $f'\left( 1 \right)=3$ that is why we found the value of $f\left( 1 \right)$ in step 2.
It is given in the question $f\left( xy \right)=f\left( x \right)+f\left( y \right)\forall x,y\ne 0\text{ }\And f'\left( 1 \right)=3,$
To check differentiability, we have to find $f'\left( x \right)$. To find $f'\left( x \right)$ we have to follow certain no. of steps,
1. Use first principle to find $f'\left( x \right)$
We know by first principle,
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x\left( 1+\dfrac{h}{x} \right) \right)-f\left( x \right)}{h}.........\left( I \right) \\
\end{align}$
Since it is given $f\left( xy \right)=f\left( x \right)+f\left( y \right)$, we can substitute $f\left( x\left( 1+\dfrac{h}{x} \right) \right)=f\left( x \right)+f\left( 1+\dfrac{h}{x} \right)$ in $\left( I \right)$,
$\begin{align}
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)+f\left( 1+\dfrac{h}{x} \right)-f\left( x \right)}{h} \\
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)}{h} \\
\end{align}$
Now, we cannot proceed further in step 1. So, we proceed to step 2.
2. We will find some boundary values of $f\left( x \right)$.
Given $f\left( xy \right)=f\left( x \right)+f\left( y \right)$
Substituting $x=1,y=1$ in the above functional relation, we get 🡪
$\begin{align}
& f\left( 1 \right)=f\left( 1 \right)+f\left( 1 \right) \\
& \Rightarrow f\left( 1 \right)=2f\left( 1 \right) \\
& \Rightarrow f\left( 1 \right)=0......................\left( II \right) \\
\end{align}$
Now we will go to step 1. Adding/subtracting $0$to any term will not cause any change in value.
So, in the final expression of step1, we will do the following changes,
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-0}{h}$
Now, we will multiple and divide the denominator with $x$. Also, from $\left( II \right)$, we can substitute 0 as f1.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{x.\dfrac{h}{x}}$
Since limit is of $h$, we can take $\dfrac{1}{x}$ out of the limit.
$f'\left( x \right)=\dfrac{1}{x}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{\dfrac{h}{x}}$
Since it is given $x\ne 0,$ we can replace $\underset{h\to 0}{\mathop{\lim }}\,\text{ by }\underset{\dfrac{h}{x}\to 0}{\mathop{\lim }}\,$
$f'\left( x \right)=\dfrac{1}{x}\underset{\dfrac{h}{x}\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{\dfrac{h}{x}}...........\left( III \right)$
If we substitute $x=1$ in the first principle of derivative, we will get,
$\begin{align}
& f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\
& x=1\to \\
& \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}.........\left( IV \right) \\
\end{align}$
In $\left( IV \right)$, we can write $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}=f'\left( 1 \right)$
Similarly, in $\left( III \right)$, we can write
\[\underset{\dfrac{h}{x}\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{\dfrac{h}{x}}=f'\left( 1 \right)\]
So, substituting in $\left( III \right)$, we get,
\[f'\left( x \right)=\dfrac{1}{x}\times of'\left( 1 \right)\]
It is given \[f'\left( 1 \right)=3\],
$\Rightarrow f'\left( x \right)=\dfrac{3}{x}$
If we draw the graph of $f'\left( x \right)=\dfrac{3}{x}$, we get 🡪
This graph is not continuous at $x=0$ since $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f'\left( x \right)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f'\left( x \right)$.
So we can say that $f\left( x \right)$ is differentiable $\ for all x\in R-\left\{ 0 \right\}$.
Note: There is a possibility of mistake while finding the boundary value of the function in step 2. The boundary value is to be found by taking help of the information given in the question. For example, it was given in the question that $f'\left( 1 \right)=3$ that is why we found the value of $f\left( 1 \right)$ in step 2.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Change the following sentences into negative and interrogative class 10 english CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Application to your principal for the character ce class 8 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE