Sets, Relations & Functions
Symmetric Functional Equations
nta_pyq_2023_jan
Grade 11
Question:
If $f(x) = \dfrac{2^{2x}}{2^{2x}+2}$, $x \in \mathbb{R}$, then $f\!\left(\dfrac{1}{2023}\right) + f\!\left(\dfrac{2}{2023}\right) + \cdots + f\!\left(\dfrac{2022}{2023}\right)$ is equal to
Step-by-Step Solution
Key Concept: Use $f(x)+f(1-x)=1$; pair terms equidistant from beginning and end.
$f(x)+f(1-x)=1$. Pairing $f\!\left(\frac{k}{2023}\right)+f\!\left(1-\frac{k}{2023}\right)=1$ for $k=1,\ldots,1011$. Sum $= 1011$.
Correct Answer: 4