Question

If $f\left(x\right)=x^n,n$ being a non-negative integer, then the values of $n$ for which $f^{\prime }(\alpha +\beta )=f^{\prime }\left(\alpha \right)+f^{\prime }\left(\beta \right)$ for all $\alpha ,\beta >0$ is

• A. $1$
• B. $2$
• C. $0$
• D. $5$

Answer 1

Answer: (B)

$2$

$f\left(x\right)=x^n$
Therefore $f^{\prime }\left(x\right)=n{x}^{n-1}$
$f^{\prime }(\alpha +\beta )=n(\alpha +\beta {)}^{n-1}$
$f^{{}^{\prime }}\left(\alpha \right)=n(\alpha {)}^{n-1}$
$f^{{}^{\prime }}\left(\beta \right)=n(\beta {)}^{n-1}$
Using the given condition
$n(\alpha +\beta {)}^{n-1}=n(\alpha {)}^{n-1}+n(\beta {)}^{n-1}$
$(\alpha +\beta {)}^{n-1}=(\alpha {)}^{n-1}+(\beta {)}^{n-1}$
The above equation will only be true when $n-1=1$
That is $n=2$