Sets, Relations & Functions
Functional Symmetry / Sum
nta_pyq_2025_apr
Grade 11
Question:
If $f(x) = \dfrac{2^x}{2^x + \sqrt{2}}$, $x \in \mathbb{R}$, then $\displaystyle\sum_{k=1}^{81} f\!\left(\frac{k}{82}\right)$ is equal to:
$81\sqrt{2}$
41
82
$\frac{81}{2}$
Step-by-Step Solution
Key Concept: Show $f(x)+f(1-x)=1$ by direct computation. Pair terms $f(k/82)+f(1-k/82)=f(k/82)+f((82-k)/82)=1$ for $k=1,\ldots,40$, and compute $f(41/82)$ separately.
$f(x)+f(1-x) = \frac{2^x}{2^x+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}} = 1$. Pairing: $\sum_{k=1}^{81}f(k/82) = 40\times1 + f(41/82) = 40+\frac{1}{2} = \frac{81}{2}$.
Correct Answer: $\frac{81}{2}$