Practicing Success
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 |
1 2 0 -1 |
0 |
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 |