If f is a real valued differentiable function satisfying $|f(x)-f(y)| \leq(x-y)^2, x, y \in R$ and f(0) = 0, then f(1) is equals
Answer & explanation
Correct answer: option 3
We have,
$|f(x)-f(y)| \leq(x-y)^2$ for all x, y ∈ R
$\Rightarrow |f(x)-f(y)| \leq|x-y|^2$ for all x, y ∈ R
$\Rightarrow \left|\frac{f(x)-f(y)}{x-y}\right| \leq|x-y|$ for all x, y ∈ R
$\Rightarrow \lim\limits_{x \rightarrow y}\left|\frac{f(x)-f(y)}{x-y}\right| \leq \lim\limits_{x \rightarrow y}|x-y|$ for all y ∈ R
⇒ |f'(y)| ≤ 0 for all y ∈ R
⇒ f'(y) = 0 for all y ∈ R
⇒ f(y) = Constant (= k say) for all y ∈ R
But, f(0) = 0
∴ k = 0
Hence, f(y) = 0 for all y ∈ R. So, f(1) = 0