JEE Advance - Mathematics (2009 - Paper 1 Offline - No. 9)
Let $$f$$ be a non-negative function defined on the interval $$[0,1]$$.
If $$\int\limits_0^x {\sqrt {1 - {{(f'(t))}^2}dt} = \int\limits_0^x {f(t)dt,0 \le x \le 1} } $$, and $$f(0) = 0$$, then
If $$\int\limits_0^x {\sqrt {1 - {{(f'(t))}^2}dt} = \int\limits_0^x {f(t)dt,0 \le x \le 1} } $$, and $$f(0) = 0$$, then
$$f\left( {{1 \over 2}} \right) < {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) > {1 \over 3}$$
$$f\left( {{1 \over 2}} \right) > {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) > {1 \over 3}$$
$$f\left( {{1 \over 2}} \right) < {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) < {1 \over 3}$$
$$f\left( {{1 \over 2}} \right) > {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) < {1 \over 3}$$
Explanation
We have, $$f' = \pm \sqrt {1 - {f^2}} $$
$$ \Rightarrow f(x) = \sin x$$ or $$f(x) = - \sin x$$ (not possible)
$$ \Rightarrow f(x) = \sin x$$
Also, $$x > \sin x\forall x > 0$$.
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