Question

Let $f\left(x\right)=\frac{4^x}{4^x+2}$, then $f\left(x\right)+f\left(1-x\right)$ is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is B

$1$

Explanation of correct answer :

Finding the value of $f\left(x\right)+f\left(1-x\right)$ :

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

We need $f\left(1-x\right)$. To find $f\left(1-x\right)$, replace $x$ with $(1-x)$ in $f\left(x\right)$.

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

$$\begin{gathered} =\frac{4^1\times 4^{-x}}{4^1\times 4^{-x}+2} \\ =\frac{{\displaystyle \frac{4}{4^x}}}{{\displaystyle \frac{4}{4^x}}+2} \\ =\frac{4}{4+2\times 4^x} \\ =\frac{2}{2+4^x}(\text{Taking}2\text{common form both numerator and denominator}) \end{gathered}$$

Now finding, $f\left(x\right)+f\left(1-x\right)$

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

$$\begin{gathered} =\frac{4^x+2}{4^x+2} \\ =1 \end{gathered}$$

Hence, the value of $f\left(x\right)+f\left(1-x\right)$ is $1$.

Thus, the correct answer is Option (B).