Question

If $f\left(x\right)=1-f(1-x)$, then the value of $f\left(\frac{1}{999}\right)+f\left(\frac{2}{999}\right)+....+f\left(\frac{998}{999}\right)$ is

• A. $998$
• B. $599$
• C. $499$
• D. $Noneofthese$

Answer 1

The correct option is C $499$

$f\left(x\right)=1-f(1-x)$

$\Rightarrow f\left(x\right)+f(1-x)=1$

Therefore,

$f\left(\frac{1}{999}\right)+f(1-\frac{1}{999})=1$

$f\left(\frac{1}{999}\right)+f\left(\frac{998}{999}\right)=1$

Similarly,

$f\left(\frac{2}{999}\right)+f\left(\frac{997}{999}\right)=1$

and so on.....

Therefore,

$f\left(\frac{1}{999}\right)+f\left(\frac{2}{999}\right)+f\left(\frac{2}{999}\right)+.....+f\left(\frac{998}{999}\right)$

$=[f\left(\frac{1}{999}\right)+f\left(\frac{998}{999}\right)]+[f\left(\frac{2}{999}\right)+f\left(\frac{997}{999}\right)]+.....[f\left(\frac{499}{999}\right)+f\left(\frac{500}{999}\right)]$

$=1+1+1+.....\text{499 terms}$

$=499$