Question

Let $f\left(\frac{x_1+x_2+...+x_n}{n}\right)=\frac{f\left(x_1\right)+f\left(x_2\right)+...+f\left(x_n\right)}{n}$ where all $x_i\in R$ are independent to each other and $n\in N$. if $f\left(x\right)$ is differentiable and $f^{\prime }\left(0\right)=a,f\left(0\right)=b$ and $f^{\prime }\left(x\right)$ is equal to

• A. $a$
• B. $0$
• C. $b$
• D. None of these

Answer 1

The correct option is A $a$

Differentiating the given equation w.r.t. $x_1,$ we get

$\frac{1}{n}{f}^{\prime }\left(\frac{x_1+x_2+...+x_n}{n}\right)=\frac{f^{\prime }\left(x_1\right)}{n}$

$[$Since all $x_i^{\prime }s$ are independent to each other, $\therefore \frac{d{x}_i}{{dx}_j}=0$ if $i\ne j$ and $\frac{{dx}_i}{{dx}_j}=1$ if $(i=j)]$

On putting $x_1=x_2=...=x_{n-1}=0$ and $x_n=x,$ we get $f^{\prime }\left(\frac{x}{n}\right)=f^{\prime }\left(0\right)=a.$

On integrating, we get $n{f}^{\prime }\left(\frac{x}{n}\right)=ax+c$

Since $f\left(0\right)=b,$ we have $c=nb$

$\therefore n{f}^{\prime }\left(\frac{x}{n}\right)=ax+nb\Rightarrow n{f}^{\prime }\left(x\right)=nax+nb\Rightarrow f\left(x\right)=ax+b.$

$\therefore f^{\prime }\left(x\right)=a,\forall x\in R$