Question

If $f:R\to R$ is a function satisfying $f(x+y)=f\left(xy\right)$ for all $x,y\in R$ and $f\left(\frac{3}{4}\right)=\left(\frac{3}{4}\right)$ , then $f\left(\frac{9}{16}\right)$ =

• A. $\frac{3}{4}$
• B. $\frac{9}{16}$
• C. $\frac{\sqrt{3}}{2}$
• D. $0$

Answer 1

The correct option is A $\frac{3}{4}$

$f(x+y)=f\left(xy\right)-\left(i\right)f\left(\frac{3}{4}\right)=\left(\frac{3}{4}\right)f\left(\frac{9}{16}\right)=f(\frac{3}{4}\times \frac{3}{4})from\left(i\right)f(\frac{3}{4}\times \frac{3}{4})=f(\frac{3}{4}+\frac{3}{4})f(\frac{3}{4}\times 2)=f\left(\frac{3}{2}\right)\because f\left(\frac{3}{4}\right)=\frac{3}{4}constantfunc.\therefore f\left(\frac{3}{2}\right)=\frac{3}{4}$