Question

A function $f:R\to R$, hwre $R$ is the set of real numbers satisfies the equation $f\left(\frac{x+y}{3}\right)=\frac{f\left(x\right)+f\left(y\right)+f\left(0\right)}{3}$ for all $x,y$ in $R$. If the function $f$ is differentiable at $x=0$, then $f$ is

• A. linear
• B. quadratic
• C. cubic
• D. biquadratic

Answer 1

The correct option is A linear

Since $f\left(x\right)$ is differentiable at $x=0$

$\Rightarrow \underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=a$ (say) ...(1)

Now, $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(\frac{3x+3h}{3}\right)-f\left(\frac{3x+3.0}{3}\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(3x\right)+f\left(3h\right)+f\left(0\right)-f\left(3x\right)-f\left(0\right)-f\left(0\right)}{3h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(3h\right)-f\left(0\right)}{3h}$

$\Rightarrow f^{\prime }\left(x\right)=f^{\prime }\left(0\right)$

$\Rightarrow f^{\prime }\left(x\right)=a$ [from (1)] (say)

$\therefore f\left(x\right)=ax+b,$ which is linear.