Question

Let f : \(\R → \R\) be a function such that
20(x - y) ≤ f(x) - f(y) ≤ 20(x - y) + 2(x - y)²
for all x, y ∈ \(\R\) and f(0) = 2. Then f(101) equals __________

Answer 1

Correct Answer: 2022

To solve for \( f(101) \) based on the inequality constraint \( 20(x-y) \leq f(x) - f(y) \leq 20(x-y) + 2(x-y)^2 \), we begin by setting \( y = 0 \). This gives us:
\( 20x \leq f(x) - f(0) \leq 20x + 2x^2 \).
Since \( f(0) = 2 \), it follows that:
\( 20x + 2 \leq f(x) \leq 20x + 2x^2 + 2 \).
Substituting \( x = 101 \), we calculate the bounds:
Lower bound: \( f(101) \geq 20 \times 101 + 2 = 2022 \).
Upper bound: \( f(101) \leq 20 \times 101 + 2 \times 101^2 + 2 = 2022 + 2 \times 10201 = 22424 \).
The range [2022, 22424] contains one number within the expected min-max range given, which is 2022.
Hence, \( f(101) = 2022 \)