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. None of these

Answer 1

Answer: (B)

$48$

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

$f\left(1-x\right)=\frac{4^{1-x}}{4^{1-x}+2}=\frac{\frac{4}{4^x}}{\frac{4}{4^x}+2}=\frac{4}{4+{2.4}^x}=\frac{2}{2+4^x}$

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

$\Rightarrow f\left(\frac{1}{97}\right)+f\left(\frac{2}{97}\right)+......+f\left(\frac{96}{97}\right)$

$=\left[f\left(\frac{1}{97}\right)+f\left(\frac{96}{97}\right)\right]+\left[f\left(\frac{2}{97}\right)+f\left(\frac{95}{97}\right)\right]+.....+\left[f\left(\frac{48}{97}\right)+f\left(\frac{49}{97}\right)\right]$

$=1+1+1+1+.......+1(\therefore 48ones)$

$=48$