Question Id : 12713 |
Context : JEE Main 2021 (31 August Morning Shift)
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} $ :
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