Question

If $f\left(x\right)=\frac{4^x}{4^x+2}$, then the value of $f\left(x\right)+f(1-x)$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $ca{n}^{\prime }tbedetermined$

Answer 1

Answer: (C)

$1$

Given $f\left(x\right)=\frac{4^x}{4^x+2}$

$\therefore f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}$

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

$\Rightarrow f(1-x)=\frac{4}{4+4^x\cdot 2}=\frac{2}{2+4^x}$

Therefore,

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

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