Question

Let $f:R\to R$ be a function such that $f\left(\frac{x+y}{3}\right)=\frac{f\left(x\right)+f\left(y\right)}{3}$, $f\left(0\right)=0$ and $f^{\prime }\left(0\right)=3$ then

• A. $f\left(x\right)$ is a quadratic equation
• B. $f\left(x\right)$ is continuous but not differentiable
• C. $f\left(x\right)$ is differentiable in $R$
• D. $f\left(x\right)$ is bounded in $R$

Answer 1

Answer: (C)

$f\left(x\right)$ is differentiable in $R$

$f\left(\frac{x+y}{3}\right)=\frac{f\left(x\right)+f\left(y\right)}{3}\to 1$

Partial differentiation w.r.t. x

$=f^{\prime }\left(\frac{x+y}{3}\right)\times \left(\frac{1}{3}\right)=\frac{f^{\prime }\left(x\right)}{3}$

Put $x=0$

$f^{\prime }\left(\frac{y}{3}\right)=f^{\prime }\left(0\right)=k$

$\Rightarrow f\left(\frac{y}{3}\right)=ky+c$

$\Rightarrow f\left(y\right)=3ky+c$

$f\left(0\right)=0$

$\Rightarrow 0=3k\left(0\right)+c\Rightarrow c=0$

& $f^{\prime }\left(y\right)=3k$

given $f^{\prime }\left(0\right)=3$

$3=3k\Rightarrow k=1$

$\Rightarrow f\left(y\right)=3y\Rightarrow f\left(x\right)=3x$

$\Rightarrow f\left(x\right)$ is a linear equation

$\Rightarrow f\left(x\right)$ is continuous on R

$\Rightarrow f\left(x\right)$ is differentiable on R

Option C but given is Option A.