Question

Let $f$ be a non-negative function defined on the interval [0,1].if $\int_0^x \sqrt{1-(f'(t))^2}dt = \int_0^x f(t) dt , 0 \le x \le 1$ and $f (0) = 0$, then

• A. $f\bigg(\frac{1}{2}\bigg) < \frac{1}{2} \, and \, f \bigg(\frac{1}{3}\bigg) > \frac{1}{3}$
• B. $f\bigg(\frac{1}{2}\bigg) > \frac{1}{2} \, and \, f \bigg(\frac{1}{3}\bigg) > \frac{1}{3}$
• C. $f\bigg(\frac{1}{2}\bigg) < \frac{1}{2} \, and \, f \bigg(\frac{1}{3}\bigg) < \frac{1}{3}$
• D. $f\bigg(\frac{1}{2}\bigg) > \frac{1}{2} \, and \, f \bigg(\frac{1}{3}\bigg) < \frac{1}{3}$

Answer 1

The Correct Option is C

Given $\int_0^x \sqrt{1-(f'(t))^2}dt = \int_0^x f(t) dt , 0 \le x \le 1$
Differentiating both sides w.r.t. x by using Leibnitz's
rule, we get
$\sqrt{1- \{f'(x) \}^2} =f(x) \, \, \Rightarrow \, \, f'(x) = \pm \sqrt {1- \{f(x)\}^2}$
$\Rightarrow \, \, \, \, \, \, \, \int \frac{f'(x)}{\sqrt{1-\{(x)\}^2}}dx =\pm \int dx \, \Rightarrow \, \, sin^{-1}\{f(x)\} =\pm x+c$
Put $ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, x=0 \, \Rightarrow \, \, sin^(-1) \{f(0)\}=c$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, c=sin^{-1} (0)=0 \, \, \, \, \, \, \, \, \, \, \, \, [\because f(0)=0]$
$\therefore \, \, \, \, \, f(x)=\pm sin x $
but $ \, \, \, \, f(x) \ge 0, \forall \, x\in \, [0,1]$
$\therefore \, \, f(x) =sin x$
As we know that,
sin x < x , $\forall $ x > 0
$\therefore \, \, \, \, \, \, \, \, \, sin\bigg(\frac{1}{2}\bigg)