Functions
Convex functions and integer differences
MJAT_TS1_P1
Grade 12
Question:
Let $f:\mathbb{R}^+\to\mathbb{R}$ be a twice differentiable function such that $f''(x) > 0$ for all $x > 0$, and $f(x+1) - f(x)$ is a positive integer for all positive integral values of $x$.
If $f(2025) = f(2026) - 2027$ and $f(1) = \dfrac{1}{2}$, then $f(3)$ can be:
A) $9.5$
B) $8.5$
C) $7.5$
D) $6.5$
Step-by-Step Solution
Key Concept: Since $f'' > 0$, $f'$ is increasing. The differences $f(x+1)-f(x)$ are positive integers that must be strictly increasing (by convexity). Given $f(2025)-f(2026) = -2027$, so $f(2026)-f(2025) = 2027$. The slope of the chord must increase: each successive difference increases by at least 1.
$d_n = f(n+1)-f(n)$ are positive integers with $d_{n+1} > d_n$ (from $f'$ increasing). $d_{2025} = 2027$. So $d_1 \leq 2027 - 2024 = 3$ (minimum possible). $f(3) = \frac{1}{2} + d_1 + d_2 \geq \frac{1}{2} + 1 + 2 = 3.5$. With $d_1 \in \{3,4\}$ and $d_2 \in \{d_1+1, \ldots\}$, feasible values of $f(3) = 7.5$ or $6.5$.
Correct Answer: CD