JEE Main 26 Feb 2021 Shift 1 Solved Paper

Given, |f(x)−f(y)|≤|x−y|2
⇒‌‌‌

|f(x)−f(y)|

|x−y|

≤|x−y|
Now, taking the limit,

lim

x→y

|‌

f(x)−f(y)

x−y

|≤

lim

x→y

|x−y|
⇒‌‌|f′(y)|≤0‌‌ [using the definition of f′(y) ]
⇒‌‌f′(y)=0
[since, modulus value can never be less than 0 ]
On integrating it, we get
f(y)=c‌ (constant) ‌
Given, f(0)=1 gives C=1
∴‌‌f(y)=1‌∀y∈R
From given options, f(x)>0‌∀x∈R is satisfied only.
Hence, answer will be option (d).