Question

Let $f\left(x\right)$ be a function such that $f\left(x\right).f\left(y\right)=f(x+y),f\left(0\right)=1,f\left(1\right)=4$. If $2g\left(x\right)=f\left(x\right)(1-g(x\left)\right)$, then:

• A. $g\left(x\right)+g(1-x)=0$
• B. $g\left(x\right)=1-g(1-x)$
• C. $\sum _{k=1}^{9}g\left(\frac{k}{10}\right)=\frac{9}{2}$
• D. $\sum _{k=1}^{18}g\left(\frac{k}{19}\right)=9$

Answer 1

Answer: (B)

$g\left(x\right)=1-g(1-x)$

Given $2g\left(x\right)=f\left(x\right)(1-g(x\left)\right)\Rightarrow g\left(x\right)=\frac{f\left(x\right)}{2+f\left(x\right)}$

$g(1-x)=\frac{f(1-x)}{2+f(1-x)}$

$g\left(x\right)+g(1-x)=\frac{f\left(x\right)}{2+f\left(x\right)}+\frac{f(1-x)}{2+f(1-x)}=\frac{2f\left(x\right)+f\left(x\right)f(1-x)+2f(1-x)+f\left(x\right)f(1-x)}{4+f\left(x\right)f(1-x)+2f\left(x\right)+2f(1-x)}$

$=\frac{2f\left(x\right)+f(x+1-x)+2f(1-x)+f(x+1-x)}{4+f(x+1-x)+2f\left(x\right)+2f(1-x)}=\frac{2f\left(x\right)+2f\left(1\right)+2f(1-x)}{4+f\left(1\right)+2f\left(x\right)+2f(1-x)}$

$=\frac{2f\left(x\right)+8+2f(1-x)}{8+2f\left(x\right)+2f(1-x)}=1$

$g\left(x\right)+g(1-x)=1\Rightarrow g\left(x\right)=1-g(1-x)$