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Definite Integration question

2021 · 31 Aug · Shift 1 · Q23

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PYQ Library
JEE Main
Mathematics
Definite Integration
2021 · 31 Aug · Shift 1 · Q23

Definite Integration question

2021 · 31 Aug · Shift 1 · Q23

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1

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}

:

A
equals 0
B
equals 1
C
does not exist
D
equals ${1 \over 2}$

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