Question

If $f(x)=\frac{4^x}{4^x+2},$ 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

Answer: (B)

Since, $f(x)=\frac{4^x}{4^x+2}$
$\therefore$ $f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}$
$=\frac{4}{4+{2.4}^x}=\frac{2}{2+4^x}$
$\Rightarrow$ $f(x)+f(1-x)=1$
On putting $x=\frac{1}{97},\frac{2}{97},.....,\frac{48}{97},$
we get $f\left(\frac{1}{97}\right)+f\left(\frac{2}{97}\right)+....+f\left(\frac{96}{97}\right)=48$