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 o 0} {1 \over {{x^2}}}\int_0^x {f(t)dt} :

Question

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 	o 0} {1 \over {{x^2}}}\int_0^x {f(t)dt} :

Accepted Answer

Answer: D. equals {1 \over 2} \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 	o 0} {{\int\limits_0^x {\sin t\,dt} } \over {{x^2}}}\left( {{0 \over 0}} ight) = {1 \over 2}

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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