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JEE MainMathematicsToughContinuity and differentiability

Let f be a differentiable function from R to R such that
∣f(x)−f(y)∣≤2∣x−y∣3/2|f(x) - f(y)| \leq 2|x - y|^{3/2}, for all x, y ∈\in R. If f(0) = 1
then ∫01\mspace2muf2\int_{0}^{1}\mspace{2mu} f^{2}(x)dx is equal to

Show answer and solution
∣f(x)−f(y)∣≤2∣x−y∣3/2|f(x) - f(y)| \leq 2|x - y|^{3/2}
∣f(x)−f(y)x−y∣≤2∣x−y∣1/2\left| \frac{f(x) - f(y)}{x - y} \right| \leq 2|x - y|^{1/2}
lim⁡y→x\mspace2mu∣f(x)−f(y)x−y∣≤lim⁡y→x\mspace2mu2∣x−y∣1/2\lim_{y \rightarrow x}\mspace{2mu}\left| \frac{f(x) - f(y)}{x - y} \right| \leq \lim_{y \rightarrow x}\mspace{2mu} 2|x - y|^{1/2}
⇒ ∣f′(x)∣≤0\Rightarrow \ \left| f^{'}(x) \right| \leq 0 ⇒\Rightarrow
f′(x)=0f^{'}(x) = 0 ⇒f(x)=c as f(0)=1⇒\Rightarrow f(x) = c\ as\ f(0) = 1 \Rightarrow
f(x)=1f(x) = 1
⇒ ∫01\mspace2muf2(x)dx=1\Rightarrow \ \int_{0}^{1}\mspace{2mu} f^{2}(x)dx = 1

Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx

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