Let f be a differentiable function from R to... | Definite Integration

JEE MAIN 2019

09-01-19 S2

Question

Let $f$ be a differentiable function from $R$ to $R$ such that $|f(x)-f(y)| \leqslant 2|x-y|^{\frac{3}{2}}$, for all $x, y \in R$. If $f(0)=1$ then $\int_0^1 f^2(x) d x$

Select the correct option:

$0$

$\frac{1}{2}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & \because f: R \rightarrow R \\ & \text { and }|f(x)-f(y)| \leq 2 \cdot|x-y|^{3 / 2} \\ & \Rightarrow\left|\frac{f(x)-f(y)}{x-y}\right| \leq 2 \sqrt{x-y} \\ & \Rightarrow \ln _{x \rightarrow y}\left|\frac{f(x)-f(y)}{x-y}\right| \leq \ln _{x \rightarrow y} 2 \sqrt{x-y} \\ & \Rightarrow\left|f^{\prime}(x)\right|=0 \\ & \therefore f(x) \text { is a constant function. }\end{aligned}$

Question Tags

JEE Main

Mathematics

Easy

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