Question

Let $f\left(x\right)=x^n,n$ being a positive integer. The value of $n$ for which the equality $f\text{'}(a+b)=f\text{'}\left(a\right)+f\text{'}\left(b\right).f$ is valid for all $a,b>0$ is

• A. $0,2$
• B. $1,3$
• C. $3,4$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for correct option.

Finding the value of $n$

Given: $f\left(x\right)=x^n,$ equality $f\text{'}(a+b)=f\text{'}\left(a\right)+f\text{'}\left(b\right)$

Differentiating$f\left(x\right)=x^n$ with respect to $x$
$f\prime \left(x\right)=n{x}^{n-1}$

Therefore,
$$\begin{gathered} f\text{'}(a+b)=n{(a+b)}^{n-1} \\ f\text{'}\left(a\right)=n{\left(a\right)}^{n-1} \\ f\prime \left(b\right)=n{\left(b\right)}^{n-1} \end{gathered}$$

Using the given condition,
$$\begin{gathered} n{(a+b)}^{n-1}=n{\left(a\right)}^{n-1}+n{\left(b\right)}^{n-1} \\ {(a+b)}^{n-1}={\left(a\right)}^{n-1}+{\left(b\right)}^{n-1} \end{gathered}$$
The above equation will only be true when $n-1=1$

Putting $n-1=1$ in the above equation,

${(a+b)}^1={\left(a\right)}^1+{\left(b\right)}^1$
So, $n=2$

Hence the correct answer is Option (D).