Question

If $f\left(x\right)$ and $g_a\left(x\right)$ are real valued function defined as $g_a\left(x\right)=\frac{a^x}{a^x+\sqrt{a}},$ where $a>0$ and $f\left(x\right)=g_9\left(x\right)$, then the value of $\left[\sum _{r=1}^{1995}f\left(\frac{r}{1996}\right)\right]$ is
(where [.] denotes greatest integer function)

• A. $995$
• B. $996$
• C. $997$
• D. $998$

Answer 1

The correct option is C $997$

Given :
$g_a\left(x\right)=\frac{a^x}{a^x+\sqrt{a}},$ where $a>0$ and $f\left(x\right)=g_9\left(x\right)$
Then
$f\left(x\right)=\frac{9^x}{9^x+3}\Rightarrow f(1-x)=\frac{9^{1-x}}{9^{1-x}+3}\Rightarrow f(1-x)=\frac{3}{3+9^x}\Rightarrow f\left(x\right)+f(1-x)=1$

Now,
$\sum _{r=1}^{1995}f\left(\frac{r}{1996}\right)=f\left(\frac{1}{1996}\right)+f\left(\frac{2}{1996}\right)+\dots +f\left(\frac{998}{1996}\right)+\dots +f\left(\frac{1994}{1996}\right)+f\left(\frac{1995}{1996}\right)=997+f\left(\frac{1}{2}\right)=997+\frac{1}{2}=997.5\therefore \left[\sum _{r=1}^{1995}f\left(\frac{r}{1996}\right)\right]=997$