Question

Let $f\left(x\right)=\frac{{2025}^x}{45+{2025}^x}$, then the value of $f\left(\frac{1}{2025}\right)+f\left(\frac{2}{2025}\right)+f\left(\frac{3}{2025}\right)+...+f\left(\frac{2024}{2025}\right)$ is equal to

• A. $1011$
• B. $1012$
• C. $1013$
• D. $2024$

Answer 1

Answer: (B)

$1012$

We know $(45{)}^2=2025$

Let $a=45$

$\to f\left(x\right)=\frac{a^{2x}}{a+a^{2x}}$

$Val=f\left(\frac{1}{2025}\right)+f\left(\frac{2}{2025}\right)+......+f\left(\frac{2024}{2025}\right)$

Now by seeing carefully we can observe that we are asked something of form $f\left(y\right)+f(1-y)$ where

y=$\frac{1}{2025},\frac{2}{2025},.......,\frac{1012}{2025}$

$f\left(y\right)+f(1-y)=\frac{a^{2y}}{a+a^{2y}}+\frac{a^{2(1-y)}}{a+a^{2(1-y)}}$

$\frac{a^{2y}}{a+a^{2y}}+\frac{a^2{a}^{-2y}}{a+a^2{a}^{-2y}}=\frac{a^{2y}}{a+a^{2y}}+\frac{a}{a^{2y}+a}=1$

$\therefore Val=1+1+\mathrm{1.....1012}$ times$=1012$