Question

if $f:R\to R$ is given by $f\left(x\right)=\frac{4^x}{4^x+2}$ for all $xϵR$ then $f\left(\frac{1}{1997}\right)+f\left(\frac{2}{1997}\right)+---+f\left(\frac{1996}{1997}\right)=$

• A. $997.5$
• B. $998$
• C. $998.5$
• D. $999$

Answer 1

The correct option is B $998$

Given that,

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

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

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

Now, $f\left(\frac{1}{1997}\right)+f\left(\frac{2}{1997}\right)+f\left(\frac{3}{1997}\right)+.......................+f\left(\frac{1996}{1997}\right)$

$=\left[f\right(\frac{1}{1997})+f(\frac{1996}{1997}\left)\right]+\left[f\right(\frac{2}{1997})+f(\frac{1995}{1997}\left)\right].......................+\left[f\right(\frac{998}{1997})+f(\frac{999}{1997}\left)\right]$ $\left(998pairs\right)$

$=\left[f\right(\frac{1}{1997})+f(1-\frac{1}{1997}\left)\right]+\left[f\right(\frac{2}{1997})+f(1-\frac{2}{1997}\left)\right].......................+\left[f\right(\frac{998}{1997})+f(1-\frac{998}{1997}\left)\right]$ $\left(998pairs\right)$

$=1+1+1+.............+1$

$=998$ (Ans)

Option B is correct.