JEE Mains · Maths · STD 12 - 1. relation and function
If \(f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in R\) then \(f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\ldots \ldots . .+f\left(\frac{2022}{2023}\right)\) is equal to
- A \(2011\)
- B \(1010\)
- C \(2010\)
- D \(1011\)
Answer & Solution
Correct Answer
(D) \(1011\)
Step-by-step Solution
Detailed explanation
\(f(x)=\frac{4^x}{4^x+2}\) \(f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2}\) \(=\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)}\) \(=\frac{4^x}{4^x+2}+\frac{2}{2+4^x}\) \(=1\) \(\Rightarrow f(x)+f(1-x)=1\)…
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