Question

If $f(x)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}$ then \[f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{2}{1996} \right)+...f\left( \dfrac{1995}{1996} \right)\] is equal to
A. $997$
B. $997.5$
C. $998$
D. $998.5$

Answer 1

Hint:Hint: We need to find the value of \[f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{2}{1996} \right)+...f\left( \dfrac{1995}{1996} \right)\] . First we have to get a relation between $f(x)$ and $f(1-x)$ . Put the value of $x$ as $(1-x)$ in $f(x)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}$ . Now add $f(x)$ and $f(1-x)$ . Rearrange the equation \[f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{2}{1996} \right)+...f\left( \dfrac{1995}{1996} \right)\] in the form $f(x)+f(1-x)$ . There will be $997$ such terms plus $f\left( \dfrac{998}{1996} \right)$ . Simplify further and by substituting in $f(x)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}$ we will get the value of $f\left( \dfrac{998}{1996} \right)$ or $f\left( \dfrac{1}{2} \right)$ .

Complete step by step answer:
We need to find the value of \[f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{2}{1996} \right)+...f\left( \dfrac{1995}{1996} \right)\] given that $f(x)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}...(i)$ .
First let us find $f(1-x)$ . For this, replace $x$ by $1-x$ in equation $(i)$ .
Therefore, \[f(1-x)=\dfrac{{{9}^{1-x}}}{{{9}^{1-x}}+3}\] .
We know that ${{a}^{m+n}}={{a}^{m}}{{a}^{n}}$ . So the above equation can be written as
Therefore, \[f(1-x)=\dfrac{9\times {{9}^{-x}}}{\left( 9\times {{9}^{-x}} \right)+3}\] .
Now taking ${{9}^{-x}}$ common from denominator, we get
\[f(1-x)=\dfrac{9\times {{9}^{-x}}}{{{9}^{-x}}\left( 9+\dfrac{3}{{{9}^{-x}}} \right)}\]
Now cancelling ${{9}^{-x}}$ from numerator and denominator, we get
\[f(1-x)=\dfrac{9}{\left( 9+3\times {{9}^{x}} \right)}\]
Now, take $3$ common from the denominator terms. We get
\[f(1-x)=\dfrac{9}{3(3+{{9}^{x}})}\]
Cancelling $3$ from numerator and denominator, we get
\[f(1-x)=\dfrac{3}{(3+{{9}^{x}})}\]
Now let us find $f(x)+f(1-x)$ .
That is, $f(x)+f(1-x)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}+\dfrac{3}{(3+{{9}^{x}})}$
Adding, we get
$f(x)+f(1-x)=\dfrac{{{9}^{x}}+3}{{{9}^{x}}+3}$
Cancelling the numerator and denominator, we get
$f(x)+f(1-x)=1...(a)$
Now consider \[f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{2}{1996} \right)+...f\left( \dfrac{1995}{1996} \right)\] . The first and last term are of the form in equation $(a)$ .
Therefore, we can group it in the form \[f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{1995}{1996} \right)+f\left( \dfrac{2}{1996} \right)+f\left( \dfrac{1994}{1996} \right)+...\]
i.e. \[f\left( \dfrac{1}{1996} \right)+f\left( \dfrac{1995}{1996} \right)+f\left( \dfrac{2}{1996} \right)+f\left( \dfrac{1994}{1996} \right)+...\]
i.e. $1+1+....$
We need to find how many such pairs will be there. For that we take \[1995\] and divide it by $2$ .
So there will be $997$ such pairs and one pair will be left alone.
i.e. $1+1+....+1+f\left( \dfrac{998}{1996} \right)$ \[\]
$=997+f\left( \dfrac{998}{1996} \right)$
Dividing $\dfrac{998}{1996}$ , we get
$997+f\left( \dfrac{1}{2} \right)$
 To get the value of $f\left( \dfrac{1}{2} \right)$ substitute $x=\dfrac{1}{2}$ in equation $(i)$ .
i.e. $997+\dfrac{{{9}^{\dfrac{1}{2}}}}{{{9}^{\dfrac{1}{2}}}+3}$
Solving the root terms, we get
$997+\dfrac{3}{3+3}$
Further simplifying, we get
$997+\dfrac{3}{6}=997+\dfrac{1}{2}$
$=997+0.5$
$=997.5$
Hence, the correct option is B.

Note:
 $f(x)+f(1-x)=1$ is the base of this problem. The last step few steps can also be done in the following way:
We need to find how many such pairs will be there. For that, we take \[1995\] and divide it by $2$ .
So the answer will be $997.5$ .