Question

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

• A. 1
• B. 48
• C. -48
• D. -1

Answer 1

Answer: (B)

For $\frac{1}{97}\le x\le \frac{96}{97}$,
$1<4^x<4$
$\Rightarrow \frac{1}{3}<\frac{4^x}{4^x+2}<\frac{4}{6}$
$\Rightarrow 0.33<\frac{4^x}{4^x+2}<0.66$
$\Rightarrow 0.33\times 96<\frac{4^x}{4^x+2}\times 96<0.66\times 96$
$=31.68<\frac{4^x}{4^x+2}\times 96<63.36$