Question

If $f\left(x\right)=\frac{4^x}{4^x+2}\mathrm{for}\mathrm{all}x\in R.$
Which of the following is/are true?

• 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 $f\left(x\right)+f(1-x)=1$

Given that,
$f\left(x\right)=\frac{4^x}{4^x+2}\mathrm{for}\mathrm{all}x\in Rf\left(1-x\right)=\frac{4^{1-x}}{4^{1-x}+2}=\frac{\frac{4}{4^x}}{\frac{4}{4^x}+2}=\frac{4}{4+2\cdot 4^x}\Rightarrow f\left(1-x\right)=\frac{4}{2\left(2+4^x\right)}=\frac{2}{4^x+2}f\left(x\right)+f\left(1-x\right)=\frac{4^x}{4^x+2}+\frac{2}{4^x+2}=\frac{4^x+2}{4^x+2}=1$