Question

If $f\left( x \right)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}$ then $f\left( \dfrac{1}{2012} \right)+f\left( \dfrac{2}{2012} \right)+...+f\left( \dfrac{2011}{2012} \right)$ is equal to
$\begin{aligned} & a)1005 \\\ & b)1005.5 \\\ & c)1006 \\\ & d)1006.5 \\\ \end{aligned}$

Answer 1

To solve this equation we will first find the value of $f\left( 1-x \right)$ . Now we will add and find the value of $f\left( x \right)+f\left( 1-x \right)$ . Now we will use the obtained value to simplify the above series and hence find the value of the sum given.

Complete step by step solution:
Now let us consider the given function $f\left( x \right)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}$
Now we know that the series in RHS has inputs of the form x and 1 – x.
Hence first let us find the value of $f\left( 1-x \right)$ .
Now substituting 1 – x in place of x we get,
$\Rightarrow f\left( 1-x \right)=\dfrac{{{9}^{1-x}}}{{{9}^{1-x}}+3}$
Now we know that ${{a}^{m-n}}=\dfrac{{{a}^{m}}}{{{a}^{n}}}$ Hence using this we get,
$\Rightarrow f\left( 1-x \right)=\dfrac{\dfrac{9}{{{9}^{x}}}}{\dfrac{9}{{{9}^{x}}}+3}$
Now taking LCM in the we get,
$\Rightarrow f\left( 1-x \right)=\dfrac{\dfrac{9}{{{9}^{x}}}}{\dfrac{9+{{3.9}^{x}}}{{{9}^{x}}}}=\dfrac{9}{9+{{3.9}^{x}}}$
Now let us calculate the value of $f\left( x \right)+f\left( 1-x \right)$
$\Rightarrow f\left( x \right)+f\left( 1-x \right)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}+\dfrac{9}{9+{{3.9}^{x}}}$
Taking LCM in the equation we get,
$\Rightarrow f\left( x \right)+f\left( 1-x \right)=\dfrac{{{9}^{x}}\left( 9+{{3.9}^{x}} \right)+9\left( {{9}^{x}}+3 \right)}{\left( {{9}^{x}}+3 \right)\left( 9+{{3.9}^{x}} \right)}$
Now taking 3 common in the first bracket we get,
$\Rightarrow f\left( x \right)+f\left( 1-x \right)=\dfrac{{{9}^{x}}3\left( 3+{{9}^{x}} \right)+9\left( {{9}^{x}}+3 \right)}{\left( {{9}^{x}}+3 \right)\left( 9+{{3.9}^{x}} \right)}$
Now using the distributive property we get,
$\Rightarrow f\left( x \right)+f\left( 1-x \right)=\dfrac{\left( {{9}^{x}}+3 \right)\left( 9+{{3.9}^{x}} \right)}{\left( {{9}^{x}}+3 \right)\left( 9+{{3.9}^{x}} \right)}=1$
Hence we get, $f\left( x \right)+f\left( 1-x \right)=1$ .
Now let us combine all the terms of type f(x) and f(1 – x). hence we can say that
$f\left( \dfrac{1}{2012} \right)+f\left( \dfrac{2011}{2012} \right)=1$ , $f\left( \dfrac{2}{2012} \right)+f\left( \dfrac{2010}{2012} \right)=1$ and going so on we till $f\left( \dfrac{1005}{1996} \right)+f\left( \dfrac{1007}{1996} \right)=1$
Now we also have $f\left( \dfrac{1006}{2012} \right)+f\left( \dfrac{1006}{2012} \right)=1$
Hence we can say $f\left( \dfrac{1006}{2012} \right)=\dfrac{1}{2}$
Now substituting these values in the series we get,
$\begin{aligned} & \Rightarrow f\left( \dfrac{1}{2012} \right)+f\left( \dfrac{2}{2012} \right)+...+f\left( \dfrac{2011}{2012} \right)=\left( 1+1+1+...1005times \right)+\dfrac{1}{2} \\\ & \Rightarrow f\left( \dfrac{1}{2012} \right)+f\left( \dfrac{2}{2012} \right)+...+f\left( \dfrac{2011}{2012} \right)=1005.5 \\\ \end{aligned}$

So, the correct answer is “Option b”.

Note: Now note that in the series $f\left( \dfrac{1}{2012} \right)+f\left( \dfrac{2}{2012} \right)+...+f\left( \dfrac{2011}{2012} \right)$ we have 2011 terms. Now since we are pairing the terms up and there are an odd number of terms we can say that 1 term will remain unpaired. Hence we calculate $\dfrac{2011}{2}=1005.5$ . Hence we understand the last term that will be paired is 1005 and 1007. And hence 1006 will remain unpaired. Now we can easily calculate the value of 1006 as 2012 – 1006 = 1006 and is in the for x and 1 – x.