Question

If $f:R\to R$ be given by $f\left(x\right)=\frac{4^x}{4^x+2}$ for all $xϵR$. Then,

• A. $f\left(x\right)=f(1-x)$
• B. $f\left(x\right)+f(1-x)=0$
• C. $f\left(x\right)+f(1-x)=1$
• D. $f\left(x\right)+f(x-1)=1$

Answer 1

The correct option is C

$f\left(x\right)+f(1-x)=1$

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