Question

If \[f(x)=\dfrac{{{4}^{x}}}{{{4}^{x}}+2}\], then \[f\left( \dfrac{1}{97} \right)+f\left( \dfrac{2}{97} \right)+........f\left( \dfrac{96}{97} \right)\] is equal to

(1) 1
(2) 48
(3) -48
(4) -1

Answer 1

Hint: In this question, F(x) is given. From this, we First put x=1-x and we find out f(1-x). After solving f(1-x) we add both f(x) and f(1-x) values and after putting these equations equal to a and given series equal to b we are able to get our desired answer.

Complete step by step Solution:
We have given the equation \[f(x)=\dfrac{{{4}^{x}}}{{{4}^{x}}+2}\]
And we have to find out the value of \[f\left( \dfrac{1}{97} \right)+f\left( \dfrac{2}{97} \right)+........f\left( \dfrac{96}{97} \right)\]
To find out the value, first we put f (x) = f (1-x)
F (1-x) = $f\left( \dfrac{2}{97} \right)+f\left( \dfrac{95}{97} \right)$$\dfrac{{{4}^{1-x}}}{{{4}^{1-x}}+2}$
Solving the above equation, we get
$\dfrac{{{4}^{1-x}}}{{{4}^{1-x}}+2}$ = $\dfrac{\dfrac{4}{{{4}^{x}}}}{\dfrac{4}{{{4}^{x}}}+2}$
Which is equal to $\dfrac{4}{4+{{2.4}^{x}}}$
And $\dfrac{4}{4+{{2.4}^{x}}}$= $\dfrac{2}{2+{{4}^{x}}}$
Now we add both f(x) and f(1-x), we get
F(x) + F(1-x) = $\dfrac{{{4}^{x}}}{{{4}^{x}}+2}$ + $\dfrac{2}{2+{{4}^{x}}}$
Which is equal to $\dfrac{{{4}^{x}}+2}{{{4}^{x}}+2}$ = 1
Let value of f (x) =a
That is a = 1
Now \[f\left( \dfrac{1}{97} \right)+f\left( \dfrac{2}{97} \right)+........f\left( \dfrac{96}{97} \right)\]
And that value is equal to b
Now \[f\left( \dfrac{1}{97} \right)+f\left( \dfrac{2}{97} \right)+........f\left( \dfrac{96}{97} \right)\] =$f\left( \dfrac{1}{97} \right)+f\left( \dfrac{96}{97} \right)$
Which is equal to $f\left( \dfrac{1}{97} \right)+f\left( 1-\dfrac{1}{97} \right)$ = 1
Similarly $f\left( \dfrac{2}{97} \right)+f\left( \dfrac{95}{97} \right)$ = 1
So , b = 1
Total pairs = 48
a+ a + a + a + a + a + . . . . . . . . . . . . . . . . . . . . . . . + a = b
i.e. 1 + 1 + 1 + 1 + 1 + 1 + ……………………. +1 ( 48 one's )
48a = b ( a =1, b= 1)
 Hence the required sum = 48

Hence, the correct option is 2.

Note: Many students do mistakes while expanding any series. It is important to be very careful while adding or subtraction the powers. If the power has a negative sign then the value comes into the denominator. We cannot multiply the terms with power and without power so these values come with a multiply sign.