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AP EAMCET · Maths · Functions

Let \(f: \mathbf{R} \rightarrow \mathbf{R}\) be a continuous function such that for any two real numbers \(x\) and \(y\), \(|f(x)-f(y) \leq 10| x-.\left.y\right|^{201}\), then

  1. A \(f(2019)=f(2020)+1\)
  2. B \(f(2019)+f(2022)=2 f(2021)\)
  3. C \(f(2019)=f(2020)+8\)
  4. D \(f(2019)=f(2020)+2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(f(2019)+f(2022)=2 f(2021)\)

Step-by-step Solution

Detailed explanation

Given, \(f: \mathbf{R} \rightarrow \mathbf{R}\) is continuous \[ \begin{aligned} & |f(x)-f(y)| \leq 10|x-y|^{201} \\ \Rightarrow & \frac{|f(x)-f(y)|}{|x-y|} \leq 10|x-y|^{200} \end{aligned} \] Put, \(y=x+h\) and taking limit \(h \rightarrow 0\) on both sides…