Question

If $f\left(x\right)$ = $\frac{9^x}{9^x+3}$,find the value of $f\left(\frac{1}{50}\right)$ + $f\left(\frac{2}{50}\right)$ + ...... + $f\left(\frac{48}{50}\right)$ + $f\left(\frac{49}{50}\right)$

• A. 24
• B. 24.5
• C. 25
• D. 25.5

Answer 1

The correct option is B

24.5

$f\left(x\right)$ = $\frac{9^x}{9^x+3}$
$f(1-x)$ = $\frac{9^{1-x}}{9^{1-x}+3}=\frac{\frac{9}{9^x}}{[\left(\frac{9}{9^x}\right)+3]}=\frac{3}{3+9^x}$
f(x) + f(1 - x) = 1
So, $f\left(\frac{1}{50}\right)$ = 1 - $f(1-\frac{1}{50})$ = 1 - $f\left(\frac{49}{50}\right)$ $\to$ $f\left(\frac{1}{50}\right)$ + $f\left(\frac{49}{50}\right)$ = 1
Similarly $f\left(\frac{2}{50}\right)$ + $f\left(\frac{48}{50}\right)$ = 1
So, the given expression becomes:1 + 1 + ..... 24 times + $f\left(\frac{25}{50}\right)$ = 24 + $f\left(\frac{1}{2}\right)$
We have f(x) + f(1 - x) = 1.
Put x = $\frac{1}{2}$, 2$f\left(\frac{1}{2}\right)$ = 1 $\to$ $f\left(\frac{1}{2}\right)$ = $\frac{1}{2}$
so, the answer is 24.5. Hence option (b)