Question

If $f\left(x\right)=\frac{9^x}{9^x+3}$, then $f\left(\frac{1}{2012}\right)+f\left(\frac{2}{2012}\right)+....+f\left(\frac{2011}{2012}\right)$ is equal to

• A. $1005$
• B. $1005.5$
• C. $1006$
• D. $1006.5$

Answer 1

The correct option is B $1005.5$

Given that $f\left(x\right)=\frac{9^x}{9^x+3}$

$\Rightarrow f(1-x)=\frac{9^{(1-x)}}{9^{(1-x)}+3}=\frac{3}{3+9^x}$

So we get $f\left(x\right)+f(1-x)=\frac{9^x}{9^x+3}+\frac{3}{3+9^x}=1$

The given equation can be written as

$y=f\left(\frac{1}{2012}\right)+f\left(\frac{2}{2012}\right)+.......+f\left(\frac{2011}{2012}\right)$

$=f\left(\frac{1}{2012}\right)+f\left(\frac{2}{2012}\right)+.....+f\left(\frac{1006}{2012}\right)+....+f(1-\frac{2}{2012})+f(1-\frac{1}{2012})$

$\Rightarrow y=f\left(\frac{1}{2012}\right)+f(1-\frac{1}{2012})+f\left(\frac{2}{2012}\right)+f(1-\frac{2}{2012}).....+f\left(\frac{1006}{2012}\right)$

$\Rightarrow y=1+1+1+......\left(1005terms\right)+f\left(\frac{1}{2}\right)=1005+0.5=1005.5$