Question

If $|f(x_{1})-f(x_{2})|\le(x_{1}-x_{2})^{2},x_{1},x_{2}\in\mathbb{R}$ and if $f(0)=2026$, then $f(2025)$ is equal to

• A. 2025
• B. 4051
• C. 4050
• D. 2026
• E. 1

Hint:

Math Tip: Any condition of the form $|f(x) - f(y)| \le K|x - y|^\alpha$ where $\alpha>1$ mathematically guarantees that the function $f(x)$ is a flat, horizontal constant line. Thus, $f(\text{anything}) = f(\text{initial value})$.

Answer 1

The Correct Option is D

Concept:
Calculus - Definition of Derivative using Inequalities.
Step 1: Rearrange the given inequality.
Given: $|f(x_1) - f(x_2)| \le (x_1 - x_2)^2$.
Divide both sides by $|x_1 - x_2|$, assuming $x_1 \neq x_2$: $$ \frac{|f(x_1) - f(x_2)|}{|x_1 - x_2|} \le |x_1 - x_2| $$ $$ \left| \frac{f(x_1) - f(x_2)}{x_1 - x_2} \right| \le |x_1 - x_2| $$
Step 2: Apply the limit to form the derivative definition.
Let $x_1 = x + h$ and $x_2 = x$. Substitute these into the inequality: $$ \left| \frac{f(x+h) - f(x)}{h} \right| \le |h| $$ Take the limit as $h \rightarrow 0$ on both sides: $$ \lim_{h\rightarrow0} \left| \frac{f(x+h) - f(x)}{h} \right| \le \lim_{h\rightarrow0} |h| $$
Step 3: Evaluate the limits.
The expression inside the absolute value on the left is the formal definition of the derivative $f'(x)$: $$ |f'(x)| \le 0 $$
Step 4: Determine the nature of the function $f(x)$.
Since the absolute value of any real quantity cannot be strictly less than $0$, it must be exactly $0$: $$ |f'(x)| = 0 \implies f'(x) = 0 $$ If the derivative of a function is zero everywhere ($f'(x) = 0$ for all $x \in \mathbb{R}$), then the function must be a constant. Let $f(x) = c$.
Step 5: Find the value of the constant.
We are given the initial condition $f(0) = 2026$.
Since $f(x)$ is a constant function, it evaluates to 2026 for any input: $$ f(x) = 2026 $$
Step 6: Evaluate the target value.
Substitute $x = 2025$ into our constant function: $$ f(2025) = 2026 $$