Question

If $f\left( x \right) = \dfrac{{{9^x}}}{{{9^x} + 3}}$ then what will be the value of$f\left( {\dfrac{1}{{1996}}} \right) + f\left( {\dfrac{2}{{1996}}} \right) + \cdots \cdots \cdots + f\left( {\dfrac{{1995}}{{1996}}} \right)$
A. $997$
B.$997.5$
C.$998$
D.$998.5$

Answer 1

In this question, we need to find the value for$f\left( {x - 1} \right)$, and then we will be adding both $f\left( x \right)$ and $f\left( {x - 1} \right)$ together, and then we will be solving step by step by using simple mathematics formulae and tactics. At the end we will substitute the values of$f\left( x \right) + f\left( {x - 1} \right)$. After that we have to substitute this value in the question asked.

Complete step by step solution:
As given in the question,
$f\left( x \right) = \dfrac{{{9^x}}}{{{9^x} + 3}}$
We will first find the value for $f\left( {x - 1} \right)$,
$f\left( {1 - x} \right) = \dfrac{{{9^{1 - x}}}}{{{9^{1 - x}} + 3}}$
Now separating the terms according to the property of indices
$f\left( {1 - x} \right) = \dfrac{{9 \times {9^{ - x}}}}{{9 \times {9^{ - x}} + 3}}$
Dividing the numerator and denominator by ${9^{ - x}}$ , we get,
$\,f\left( {1 - x} \right) = \dfrac{9}{{9 + 3 \times {9^x}}}$
Dividing the numerator and denominator by $3$, we get,
$f\left( {1 - x} \right) = \dfrac{3}{{3 + {9^x}}}$
Now, adding both the equations i.e. $f\left( x \right) + f\left( {1 - x} \right)$ , we get,
$f\left( x \right) + f\left( {1 - x} \right) = \dfrac{{{9^x}}}{{{9^x} + 3}} + \dfrac{3}{{3 + {9^x}}}$
As the denominator of both the term is same , addition of numerator will take place directly,
$f\left( x \right) + f\left( {1 - x} \right) = \dfrac{{{9^x} + 3}}{{3 + {9^x}}}$
Hence, we get,
$f\left( x \right) + f\left( {1 - x} \right) = 1$
Since, there are a total of $1996$ terms, numbers of pairs will be $998$.
Now putting $x = \left( {\dfrac{1}{{1996}}} \right),\left( {\dfrac{2}{{1996}}} \right) \cdots \cdots \cdots \left( {\dfrac{{998}}{{1996}}} \right)$ in the above equation, we get,

$$f\left( {\dfrac{1}{{1996}}} \right) + \left( {\dfrac{{1995}}{{1996}}} \right) = 1 \\\ f\left( {\dfrac{2}{{1996}}} \right) + f\left( {\dfrac{{1994}}{{1996}}} \right) = 1 \\\ \cdots \cdots \cdots \\\ \cdots \cdots \cdots \\\ f\left( {\dfrac{{998}}{{1996}}} \right) + f\left( {\dfrac{{998}}{{1996}}} \right) = 1 \\\$$

From the above equation, we can conclude that,
$f\left( {\dfrac{{998}}{{1996}}} \right) = \dfrac{1}{2}$
Adding all the above terms together, we get,
$f\left( {\dfrac{1}{{1996}}} \right) + f\left( {\dfrac{2}{{1996}}} \right) \cdots \cdots \cdots f\left( {\dfrac{{1995}}{{1996}}} \right) = \left( {1 + 1 + \cdots \cdots upto997} \right) + \dfrac{1}{2}$
After substituting the values, we get,
$f\left( {\dfrac{1}{{1996}}} \right) + f\left( {\dfrac{2}{{1996}}} \right) \cdots \cdots \cdots f\left( {\dfrac{{1995}}{{1996}}} \right) = 997 + \dfrac{1}{2}$
We know that,$\dfrac{1}{2} = 0.5$
$f\left( {\dfrac{1}{{1996}}} \right) + f\left( {\dfrac{2}{{1996}}} \right) \cdots \cdots \cdots f\left( {\dfrac{{1995}}{{1996}}} \right) = 997 + 0.5$
Therefore, we conclude that,
$f\left( {\dfrac{1}{{1996}}} \right) + f\left( {\dfrac{2}{{1996}}} \right) \cdots \cdots \cdots f\left( {\dfrac{{1995}}{{1996}}} \right) = 997.5$
Hence, option b is correct.

Note: Students should understand and solve the question very carefully. Students should know all the formulae and domains. Students should be able to apply the formula as per the problem. Students should use the sign convention properly while solving the problem. Students should not get confused between the number of terms and number of pairs formed.