Search any question & find its solution
Question:
Answered & Verified by Expert
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 $f(1)$ equal
Options:
Solution:
2085 Upvotes
Verified Answer
The correct answer is:
$0$
$\begin{aligned} & \lim _{x \rightarrow y}\left|\frac{f(x)-f(y)}{x-y}\right| \leq \lim _{x \rightarrow y}|x-y| \text { or }\left|f^{\prime}(x)\right| \leq 0 \\ & \quad \Rightarrow f^{\prime}(x)=0 \Rightarrow f(x) \text { is constant, As } f(0)=0 \\ & \quad \therefore f(1)=0 .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.