IIT-JEE 2009 · Paper 1 · Mathematics Q23

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MathematicsSingle correct+3 / -1Question 23 of 60

Let $f$ be a non-negative function defined on the interval $[0,1]$ . If
$\int_0^x\sqrt{1-(f'(t))^2}\,dt=\int_0^xf(t)\,dt,\qquad0\le x\le1,$
and $f(0)=0$ , then

f\left(\frac12\right)<\frac12

and

f\left(\frac13\right)>\frac13

f\left(\frac12\right)>\frac12

and

f\left(\frac13\right)>\frac13

f\left(\frac12\right)<\frac12

and

f\left(\frac13\right)<\frac13

f\left(\frac12\right)>\frac12

and

f\left(\frac13\right)<\frac13

Answer and worked solution

Official answer: C

Differentiate. $\sqrt{1-f'(x)^2}=f(x)$ , so $f'^2+f^2=1$ with $f(0)=0$ ; this gives $f(x)=\sin x$ (the non-negative solution, valid on [0, 1]).

Compare. $\sin x<x$ for $x>0$ : $\sin\frac12<\frac12$ and $\sin\frac13<\frac13$ .

(C)

Topics:Variable separable Fundamental theorem of calculus

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IIT-JEE 2009 · Paper 1, Mathematics question 23