Question

# Consider the given function, $f\left( x \right)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}$ , then $f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{2}{1996} \right)+......$(a) 997(b) 997.5(c) 998(d) 998.5

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Hint: We are given that $f\left( x \right)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3},$ we first have to find the value of f (1 – x) and then check how the function behaves when we add. Then, once we get that f(x) + f(1 – x) = 1. Then we will solve to find the sum of $f\left( \dfrac{1}{1996} \right)+.......f\left( \dfrac{1995}{1996} \right).$

We are given that the function f is represented as
$f\left( x \right)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}$
We have to find $f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{2}{1996} \right)+......+f\left( \dfrac{1995}{1996} \right).$ First of all, we will find the value of f(1 – x). Now,
$f\left( 1-x \right)=\dfrac{{{9}^{1-x}}}{{{9}^{1-x}}+3}$
We can write ${{9}^{1-x}}$ as $\dfrac{9}{{{9}^{x}}}.$ Hence, we get
$\Rightarrow f\left( 1-x \right)=\dfrac{\dfrac{9}{{{9}^{x}}}}{\dfrac{9}{{{9}^{x}}}+3}$
Now simplifying the above equation, we get,
$\Rightarrow f\left( 1-x \right)=\dfrac{9}{9+{{3.9}^{x}}}$
Taking 3 common from numerator and denominator, we get,
$\Rightarrow f\left( 1-x \right)=\dfrac{3}{3}\left[ \dfrac{3}{3+{{9}^{x}}} \right]$
$\Rightarrow f\left( 1-x \right)=\dfrac{3}{3+{{9}^{x}}}$
Now, adding f(x) + f(1 – x), we get,
$f\left( x \right)+f\left( 1-x \right)=\dfrac{{{9}^{x}}}{3+{{9}^{x}}}+\dfrac{3}{3+{{9}^{x}}}$
$\Rightarrow f\left( x \right)+f\left( 1-x \right)=\dfrac{{{9}^{x}}+3}{3+{{9}^{x}}}$
$\Rightarrow f\left( x \right)+f\left( 1-x \right)=1$
We get the sum of f(x) and f(1 – x) as 1 for any value of x. So, we get the following terms.
For $x=\dfrac{1}{1996}$
$\Rightarrow 1-x$
$\Rightarrow 1-\dfrac{1}{1996}$
$\Rightarrow \dfrac{1995}{1996}$
Since f(x) + f(1 – x) = 1, we get,
$f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{1995}{1996} \right)=1$
Similarly, for $x=\dfrac{2}{1996},$
$\Rightarrow 1-x$
$\Rightarrow 1-\dfrac{2}{1996}$
$\Rightarrow \dfrac{1994}{1996}$
So,
$f\left( \dfrac{2}{1996} \right)+f\left( \dfrac{1994}{1996} \right)=1$
And in the same manner for other terms from 1 to 1995, we have $\dfrac{1995-1}{2}=997$ pair of f(x) + f(1 – x).
Leaving $f\left( \dfrac{998}{1996} \right)$ aside. So,
\begin{align} & f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{2}{1996} \right)+........+f\left( \dfrac{1994}{1996} \right)+f\left( \dfrac{1995}{1996} \right)= \\ & ++.......+f\left( \dfrac{998}{1996} \right) \\ \end{align}
$=+f\left( \dfrac{998}{1996} \right)$
$=997+f\left( \dfrac{998}{1996} \right).....\left( i \right)$
Now, $f\left( \dfrac{998}{1996} \right)=f\left( \dfrac{1}{2} \right)$
As $f\left( x \right)=\dfrac{{{9}^{x}}}{3+{{9}^{x}}}$ so,
$f\left( \dfrac{1}{2} \right)=\dfrac{{{9}^{\dfrac{1}{2}}}}{3+{{9}^{\dfrac{1}{2}}}}$
Since, ${{9}^{\dfrac{1}{2}}}={{\left( {{3}^{2}} \right)}^{\dfrac{1}{2}}}=3,$ we get,
$\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{3}{3+3}$
$\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{1}{2}=0.5.....\left( ii \right)$
Now using (ii) in (i), we get,
$f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{2}{1996} \right)+.......+f\left( \dfrac{1995}{1996} \right)=997+0.5$
$\Rightarrow f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{2}{1996} \right)+.......+f\left( \dfrac{1995}{1996} \right)=997.5$
So, the correct answer is “Option B”.

Note: While making a pair, we consider that from 1 to 1995, there are 1995 numbers which is odd, hence one item will be left out. When the denominator is the same for two fractions, the fractions are simply added up by adding their numerator value without solving for LCM.