JEE Main 31-Aug-2021 Shift 1 Solved Paper

Section: Mathematics

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Question : 65 of 90

Marks: +1, -0

Let f be a non-negative function in [0,1] and twice differentiable in (0,1). If ∫0x√1−(f′(t))2dt=∫0xf(t)dt,0≤x≤1 and f(0)=0, then

lim

x→0

1

x2

‌∫0xf(t)dt

[31 Aug 2021 Shift 1]

equals 0
equals 1
does not exist
equals
1
2

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