Question

If $f\left(x\right)=\frac{4^x}{4^x+2}$, then $f\left(\frac{1}{97}\right)+f\left(\frac{2}{97}\right)+.......+f\left(\frac{96}{97}\right)$ is equal to:

• A. $1$
• B. $48$
• C. $-48$
• D. $-1$

Answer 1

The correct option is B

$48$

Explanation for the correct option.

Step 1: Find the value of $f\left(x\right)+f(1-x)$.

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

$\begin{array}{rcl}f\left(1-x\right)& =& \frac{4^{1-x}}{4^{1-x}+2}\\ & =& \frac{{\displaystyle \frac{4}{4^x}}}{{\displaystyle \frac{4}{4^x}}+2}\\ & =& \frac{4}{4+2\left(4^x\right)}\\ & =& \frac{2}{4^x+2}\end{array}$

Now,

$\begin{array}{rcl}f\left(x\right)+f(1-x)& =& \frac{4^x}{4^x+2}+\frac{2}{4^x+2}\\ & =& 1\end{array}$

Step 2: Find the value of $f\left(\frac{1}{97}\right)+f\left(\frac{2}{97}\right)+.......+f\left(\frac{96}{97}\right)$.

$\begin{array}{rcl}f\left(\frac{1}{97}\right)+f\left(\frac{96}{97}\right)& =& f\left(\frac{1}{97}\right)+f\left(1-\frac{1}{97}\right)\\ & =& 1\left[byf\left(x\right)+f\left(1-x\right)\right]\end{array}$

Similarly,

$\begin{array}{rcl}f\left(\frac{2}{97}\right)+f\left(\frac{95}{97}\right)& =& f\left(\frac{2}{97}\right)+f\left(1-\frac{2}{97}\right)\\ & =& 1\end{array}$

So, in $f\left(\frac{1}{97}\right)+f\left(\frac{2}{97}\right)+.......+f\left(\frac{96}{97}\right)$, there will be total $48$ such groups whose sum will be $1$.

Therefore,

$\begin{array}{rcl}f\left(\frac{1}{97}\right)+f\left(\frac{2}{97}\right)+.......+f\left(\frac{96}{97}\right)& =& 1\left(48\right)\\ & =& 48\end{array}$

Hence, option B is correct.