Let f be any function defined on R and let it satisfy the condition : |f(x) - f(y)| ≤ |(x - y)²|,∀ (x,y) ∈ R
If f(0) = 1, then :

Solution

Maths sir · 4 steps

Divide by |x - y| and let y → x: the derivative is squeezed to zero everywhere.

1. Divide both sides. For x ≠ y, divide by |x - y|. This gives |(f(x) - f(y))/(x - y)| ≤ |x - y|.

   |f'(x)| = lim(y → x)|(f(x) - f(y))/(x - y)| ≤ lim(y → x)|x - y| = 0

2. Derivative is zero. The difference quotient is squeezed between 0 and |x - y|, which goes to 0. So the limit exists and f'(x) = 0 for every real x, which means f is a constant function.
3. Fix the constant. Since f(0) = 1, the constant is 1. So f(x) = 1 for all x ∈ ℝ.
4. So. f(x) = 1 > 0 for every x, which is option (c).

The idea to keep. Whenever |f(x) - f(y)| is bounded by |x - y|ⁿ with n > 1, divide by |x - y| and take the limit. You get f' = 0, so f is constant, and one given value fixes it.

Answer from the NTA final answer key. Solution by PrepWiser.