Let f be any function defined on $\mathbf{R}$ and let it satisfy the condition : \[ |f(\mathrm{x})-f(\mathrm{y})| \leq\left|(\mathrm{x}-\mathrm{y})^{2}\right|, \forall(\mathrm{x}, \mathrm{y}) \in \mathrm{R} \] If $f(0)=1$, then :

(A) $f(\mathrm{x})>0, \forall \mathrm{x} \in \mathbf{R}$

(B) $f(\mathrm{x})=0, \forall \mathrm{x} \in \mathbf{R}$

(C) $f(x)$ can take any value in $\mathbf{R}$

(D) $f(\mathrm{x})<0, \forall \mathrm{x} \in \mathbf{R}$

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Continuity and Differentiability

Let f be any function defined on

R

and let it satisfy the condition :

∣ f (x) - f (y) ∣ \leq (x - y)^{2}, \forall (x, y) \in R

f (0) = 1

, then :

Answer: (A)

For

x \neq = y

the given condition can be divided by

∣ x - y ∣

\frac{f ( x ) - f ( y )}{x - y} \leq ∣ x - y ∣

Letting

y \to x

, the right-hand side tends to

0

, so the difference quotient tends to

0

; thus

f

is differentiable everywhere with

f^{'} (x) = 0

.
A function whose derivative vanishes on all of

R

is constant, and

f (0) = 1

makes that constant

1

, so

f (x) = 1

for every

x

.
Therefore

f (x) > 0

for all

x \in R

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option