QJEEM21Mafc89: JEE Main 2021 Mathematics - Class 12 JEEM21Mafc89 Solution: Let f be a non-negative function in [0, 1] and twice…

Question

Verified Solution • 2025–26 Session moderate

Let f be a non-negative function in [0, 1] and twice differentiable in (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

\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt}

:

Accepted Answer

\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } ,\,0 \le x \le 1

differentiating both the sides

\sqrt {1 - {{(f'(x))}^2}} = f(x)

\Rightarrow 1 - {(f'(x))^2} = {f^2}(x)

{{f'(x)} \over {\sqrt {1 - {f^2}(x)} }} = 1

{\sin ^{ - 1}}f(x) = x + C

\because

f(0) = 0 \Rightarrow C = 0 \Rightarrow f(x) = \sin x

Now,

\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^x {\sin t\,dt} } \over {{x^2}}}\left( {{0 \over 0}} \right) = {1 \over 2}

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