Questions & Answers
Question
Answers
Related Questions
If $f\left( x+y \right)=f\left( x \right)+f\left( y \right)\forall x,y\ And f'\left( 0 \right)=2,$ then test the differentiability of $f\left( x \right)$.
Answer
![Verified]()
Verified
Verified
Hint: Since we have to test the differentiability, we have to find the derivative of the function. We can find the derivative of any function using the first principle of derivative and using the functional relation given in the question.
It is given in the question $f\left( x+y \right)=f\left( x \right)+f\left( y \right)\And f'\left( 0 \right)=2,$
In such types of questions, it is necessary to find the function first in order to check its differentiability.
To find the function, we will follow some steps:
(i) Use first principle to find $f'\left( x \right)$
We know by first principle;
\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}.............\left( I \right)\]
Since it is given in the question that \[f\left( x+y \right)=f\left( x \right)+f\left( y \right)\], we can write\[f\left( x+h \right)=f\left( x \right)+f\left( h \right)\]. Now, substituting \[f\left( x+h \right)=f\left( x \right)+f\left( h \right)\] in $\left( I \right)$, we get 🡪
\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)+f\left( h \right)-f\left( x \right)}{h}\]
\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)}{h}\]
(ii)Now, we cannot proceed further in step $\left( i \right)$, we proceed to step $\left( ii \right)$ where we will find some boundary value of \[f\left( x \right)\]
$f\left( x+y \right)=f\left( x \right)+f\left( y \right)$
Let us substitute $\text{ }x=y=0$
$\begin{align}
& \Rightarrow f\left( 0+0 \right)=f\left( 0 \right)+f\left( 0 \right) \\
& \Rightarrow f\left( 0 \right)=2f\left( 0 \right) \\
& \Rightarrow f\left( 0 \right)=0 \\
\end{align}$
Now, we will do some changes in step $\left( i \right)$ . From step $\left( ii \right)$, we have;
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)}{h}$
Adding or subtracting $0$ to any term doesn’t affect the value. Hence, we can write;
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0+h \right)-0}{h}$
Since $f\left( 0 \right)=0$, we can replace $0$ by $f\left( 0 \right)$ in the above equation;
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0+h \right)-f\left( 0 \right)}{h}.........\left( II \right)$
From the first principle, we know;
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( h \right)}{h}$
If we substitute $x=0$, we get;
$f'\left( 0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-f\left( 0 \right)}{h}.........\left( III \right)$
From $\left( III \right)$, substituting $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-\left( 0 \right)}{h}=f'\left( 0 \right)$in $\left( II \right)$, we get;
$f'\left( x \right)=f'\left( 0 \right)$
It is given in the question $f'\left( 0 \right)=2$
$f'\left( x \right)=2.......\left( IV \right)$
From$\left( IV \right)$, we can find $f'\left( x \right)$ is constant and continuous $\forall x\in R$. This shows that $f\left( x \right)$ is differentiable for $ \ all x\in R$.
To find the function, we will integrate $\left( IV \right)$ with respect to $x$.
\[\begin{align}
& \Rightarrow \int{f'\left( x \right)dx=\int{2dx}} \\
& \Rightarrow f\left( x \right)=2x+C \\
\end{align}\]
Where $C$ is constant of integration.
Note: There is a possibility of mistake while finding the boundary value of the function in step$\left( ii \right)$. The boundary value is to be found by taking help of the information given in the question. For example, since it was given in the question that $f'\left( 0 \right)=2$, that is why we found the value of $f\left( 0 \right)$ in step$\left( ii \right)$.
It is given in the question $f\left( x+y \right)=f\left( x \right)+f\left( y \right)\And f'\left( 0 \right)=2,$
In such types of questions, it is necessary to find the function first in order to check its differentiability.
To find the function, we will follow some steps:
(i) Use first principle to find $f'\left( x \right)$
We know by first principle;
\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}.............\left( I \right)\]
Since it is given in the question that \[f\left( x+y \right)=f\left( x \right)+f\left( y \right)\], we can write\[f\left( x+h \right)=f\left( x \right)+f\left( h \right)\]. Now, substituting \[f\left( x+h \right)=f\left( x \right)+f\left( h \right)\] in $\left( I \right)$, we get 🡪
\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)+f\left( h \right)-f\left( x \right)}{h}\]
\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)}{h}\]
(ii)Now, we cannot proceed further in step $\left( i \right)$, we proceed to step $\left( ii \right)$ where we will find some boundary value of \[f\left( x \right)\]
$f\left( x+y \right)=f\left( x \right)+f\left( y \right)$
Let us substitute $\text{ }x=y=0$
$\begin{align}
& \Rightarrow f\left( 0+0 \right)=f\left( 0 \right)+f\left( 0 \right) \\
& \Rightarrow f\left( 0 \right)=2f\left( 0 \right) \\
& \Rightarrow f\left( 0 \right)=0 \\
\end{align}$
Now, we will do some changes in step $\left( i \right)$ . From step $\left( ii \right)$, we have;
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)}{h}$
Adding or subtracting $0$ to any term doesn’t affect the value. Hence, we can write;
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0+h \right)-0}{h}$
Since $f\left( 0 \right)=0$, we can replace $0$ by $f\left( 0 \right)$ in the above equation;
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0+h \right)-f\left( 0 \right)}{h}.........\left( II \right)$
From the first principle, we know;
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( h \right)}{h}$
If we substitute $x=0$, we get;
$f'\left( 0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-f\left( 0 \right)}{h}.........\left( III \right)$
From $\left( III \right)$, substituting $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-\left( 0 \right)}{h}=f'\left( 0 \right)$in $\left( II \right)$, we get;
$f'\left( x \right)=f'\left( 0 \right)$
It is given in the question $f'\left( 0 \right)=2$
$f'\left( x \right)=2.......\left( IV \right)$
From$\left( IV \right)$, we can find $f'\left( x \right)$ is constant and continuous $\forall x\in R$. This shows that $f\left( x \right)$ is differentiable for $ \ all x\in R$.
To find the function, we will integrate $\left( IV \right)$ with respect to $x$.
\[\begin{align}
& \Rightarrow \int{f'\left( x \right)dx=\int{2dx}} \\
& \Rightarrow f\left( x \right)=2x+C \\
\end{align}\]
Where $C$ is constant of integration.
Note: There is a possibility of mistake while finding the boundary value of the function in step$\left( ii \right)$. The boundary value is to be found by taking help of the information given in the question. For example, since it was given in the question that $f'\left( 0 \right)=2$, that is why we found the value of $f\left( 0 \right)$ in step$\left( ii \right)$.
×
Sorry!, This page is not available for now to bookmark.
Like this
Liked
.png)
-
LIVECourses
-
Crash Courses 2021
-
Vedantu Pro
-
Vedantu Assist
-
Class 12
-
-
JEE
-
JEE Main & Advanced 2021
-
JEE Main & Advanced 2022
-
JEE Main & Advanced 2023
-
-
NEET
-
NEET 2021
-
NEET 2022
-
NEET 2023
-
-
NDA
-
CBSE
-
Commerce
-
Class 11
-
Class 12
-
-
ICSE
-
Maharashtra Board
-
Micro Courses
-
-
Reading & Spoken English -
FREEMaster Classes
-
PREMIUM1-on-1 Classes
-
Study Material
-
NCERT Solutions
-
Previous Year Papers
-
Mock Tests
-
Sample Papers
-
Reference Book Solutions
-
ICSE Solutions
-
School Syllabus
-
Revision Notes
-
Important Questions
-
Math Formula Sheets
-
-
More
Book your FREE Online counselling session
Share your contact information
Your FREE Online counselling session has been booked!
Vedantu academic counsellor will be calling you shortly for your Online Counselling session.
DONE
Please try again later
OK
NCERT Book Solutions
Reference Book Solutions
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
(Mon-Sun: 9am - 11pm IST)
Questions?
Chat with us
