First slide

Functions (XII)

Question

Let $f (x) = \frac{9^{x}}{9^{x} + 3}$ , then

Moderate

A

$f (x) + f (1 - x) = 1$
B

$f (x) + f (1 - x) = - 1$
C

$f (\frac{1}{1996}) + f (\frac{2}{1996}) + f (\frac{3}{1996}) + \dots + f (\frac{1995}{1996}) = 998$
D

$f (\frac{1}{1996}) + f (\frac{2}{1996}) + f (\frac{3}{1996}) + \dots + f (\frac{1995}{1996}) = 997 \frac{1}{2}$

Solution

$f (x) = \frac{9^{x}}{9^{x} + 3} - - - - (1)$
and $f (1 - x) = \frac{9^{1 - x}}{9^{1 - x} + 3}$
$\begin{aligned} \Rightarrow f (1 - x) = \frac{\frac{9}{9^{x}}}{\frac{9}{y^{x}} + 3} = \frac{9}{9 + 3 {.9}^{x}} \\ f (1 - x) = \frac{9}{3 (3 + 9^{x})} - - - - (2) \end{aligned}$
Adding (1) and (2), we get
$\begin{aligned} f (x) + f (1 - x) & = \frac{9^{x}}{9^{x} + 3} + \frac{9}{3 (3 + 9^{x})} \\ = \frac{3 \cdot 9^{x} + 9}{3 (9^{x} + 3)} = \frac{3 (9^{x} + 3)}{3 (9^{x} + 3)} \\ ∴ f (x) + f (1 - x) & = 1 \end{aligned}$
Now, putting $x = \frac{1}{1996}, \frac{2}{1996}, \frac{3}{1996}, \dots \frac{998}{1996},$ , in (3), we get $f (\frac{1}{1996}) + f (\frac{1995}{1996}) = 1$
$\begin{matrix} f (\frac{1}{1996}) + f (\frac{2}{1996}) + \dots + f (\frac{1995}{1996}) \\ = (1 + 1 + 1 + \dots + 997) + \frac{1}{2} \\ = 997 + \frac{1}{2} = 997 \frac{1}{2} \end{matrix}$
$\dots \dots \dots \dots \dots \dots \dots \dots$
$\begin{aligned} \Rightarrow f (\frac{997}{1996}) + f (\frac{999}{1996}) = 1 \\ \Rightarrow f (\frac{998}{1996}) + f (\frac{998}{1996}) = 1 or f (\frac{998}{1996}) = \frac{1}{2} \end{aligned}$
Adding all the above expressions, we get
$\begin{aligned} f (\frac{1}{1996}) + f (\frac{2}{1996}) + \dots + f (\frac{1995}{1996}) \\ = (1 + 1 + 1 + \dots + 997) + \frac{1}{2} \\ = 997 + \frac{1}{2} = 997 \frac{1}{2} \end{aligned}$

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