Question

If a function f satisfy $f\left(\frac{x+y}{3}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}$ for real x and y $f^{\prime }\left(2\right)=3$ then $f\left(x\right)$ is equal to

• A. $-\frac{1}{12}{x}^3+x^2$
• B. $24\log (3x+2)$
• C. $(3x+2)$
• D. $\frac{3}{4}{x}^2+2$

Answer 1

The correct option is C $(3x+2)$

$f\left(\frac{x+y}{3}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}$
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(\frac{3x+3h}{3}\right)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{\frac{2+f\left(3x\right)+f\left(3h\right)}{3}-\frac{2+f\left(3x\right)+f\left(0\right)}{3}}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(3h\right)-f\left(0\right)}{3h}=3\therefore f\left(x\right)=3x+c$