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Question:

Let f be a non-negative function defined on the interval [0,1]. If ∫0x√1−(f'(t))2dt = ∫0xf(t)dt, 0≤x≤1, and f(0)=0, then

f(1/2)<1/2 and f(1/3)<1/3

f(1/2)<1/2 and f(1/3)>1/3

f(1/2)>1/2 and f(1/3)>1/3

f(1/2)>1/2 and f(1/3)<1/3

Solution:

∫0x√1−(f'(t))2dt = ∫0xf(t)dt
Differentiating both sides using Leibnitz rule,
f(x) = √1−(f'(x))2 ⇒ f(x) = sinx
And we know , x > sinx ∀ x > 0.
⇒ f(x) < x
Hence option 'A' is correct choice.