GATE 2025 DA Q.59: solution and answer

GATE 2025 · DAQ.59 · 2 marks · NAT

Let $f : R \to R$ be such that $∣ f (x) - f (y) ∣ \leq (x - y)^{2}$ for all $x, y \in R$ . Then $f (1) - f (0) =$ ______ (Answer in integer)

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

$f (1) - f (0) = 0$

Official answer: 0Source: Official answer key, GATE 2025 DA (IIT Roorkee)

Fix $x$ and put $y = x + h$ with $h \neq = 0$ . The condition gives $∣ f (x + h) - f (x) ∣ \leq h^{2}$ .
Divide by $∣ h ∣$ : $\frac{f ( x + h ) - f ( x )}{h} \leq ∣ h ∣$ .
Let $h \to 0$ . The right side tends to $0$ , so by the squeeze theorem the difference quotient tends to $0$ . Hence $f^{'} (x)$ exists and $f^{'} (x) = 0$ for every $x \in R$ .
A function with zero derivative everywhere on $R$ is constant (mean value theorem: $f (1) - f (0) = f^{'} (c) \cdot 1 = 0$ for some $c \in (0, 1)$ ). So $f (1) = f (0)$ .
Check without derivatives: split $[0, 1]$ into $n$ equal steps of length $\frac{1}{n}$ . The triangle inequality gives $∣ f (1) - f (0) ∣ \leq \sum_{k = 1}^{n} f (\frac{k}{n}) - f (\frac{k - 1}{n}) \leq n \cdot \frac{1}{n ^{2}} = \frac{1}{n}$ . This holds for every $n$ , and $\frac{1}{n} \to 0$ , so $∣ f (1) - f (0) ∣ = 0$ .
Hence $f (1) - f (0) = 0$ .

KEY IDEA

A bound of the form $∣ f (x) - f (y) ∣ \leq C ∣ x - y ∣^{p}$ with $p > 1$ forces $f$ to be constant, because the difference quotient is squeezed to $0$ and so the derivative is $0$ everywhere.

WATCH OUT FOR

Treating $f$ as an arbitrary function and guessing a value such as $1$ from the right-hand side $(1 - 0)^{2}$ . That bound only says $∣ f (1) - f (0) ∣ \leq 1$ . It becomes $0$ only after splitting the interval into many small steps, or by showing the derivative is $0$ .

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 6 Limits & continuity questions, with notes

GATE 2025 DA Q.59: solution and answer

Answer and solution

More Limits & continuity questions