Answer

Verified

450.3k+ views

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 ðŸ¡ª

\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)+f\left( h \right)-f\left( x \right)}{h}\]

\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)}{h}\]

(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 doesnâ€™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 ðŸ¡ª

\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)+f\left( h \right)-f\left( x \right)}{h}\]

\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)}{h}\]

(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 doesnâ€™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$\left( ii \right)$.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

How many crores make 10 million class 7 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths