WBJEE · Maths · Definite Integration
Let \(f\) be a non-negative function defined in \([0, \pi / 2], f^{\prime}\) exists and be continuous for all \(x\) and \(\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t\) and \(f(0)=0\). Then
- A \(f\left(\frac{1}{2}\right) < \frac{1}{2}\) and \(f\left(\frac{1}{3}\right) > \frac{1}{3}\)
- B \(f\left(\frac{1}{2}\right) > \frac{1}{2}\) and \(f\left(\frac{1}{3}\right) < \frac{1}{3}\)
- C \(f\left(\frac{4}{3}\right) < \frac{4}{3}\) and \(f\left(\frac{2}{3}\right) < \frac{2}{3}\)
- D \(\mathrm{f}\left(\frac{4}{3}\right) > \frac{4}{3}\) and \(\mathrm{f}\left(\frac{2}{3}\right) > \frac{2}{3}\)
Answer & Solution
Correct Answer
(C) \(f\left(\frac{4}{3}\right) < \frac{4}{3}\) and \(f\left(\frac{2}{3}\right) < \frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t\) Using Leibnitz Rule,…
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