Let f be a non-negative function in 0, 1 and twice differentiable in (0, 1). If ∫0x1−(f1(t))2dt=∫0xf(t)dt, 0≤x≤1 and f(0)=0, then limx→01x2∫0xf(t)dt

∫0x 1-f1(t)2 dt=∫0x f(t) dtdifferentiate w.r.t 'x'1-f1(x)2=f(x)1-f1(x)2=f2(x)1-f2(x)=f1(x)⇒ f1(x)1-f2(x)=1 ⇒ ∫ f1(x)1-f2(x) dx=∫ 1 dx ⇒ sin-1 (f(x))=x+C x=0 ⇒ sin-1(f(x))=0+C 0=C ∴ sin-1(f(x))=x f(x)=sin x ltx→0 1x2 ∫0x f(t) dt =ltx→0 ∫0x sin t dtx2 =ltx→0 sin x2x=12

Let f be a non-negative function in 0, 1 and twice differentiable in (0, 1). If ∫0x1−(f1(t))2dt=∫0xf(t)dt, 0≤x≤1 and f(0)=0, then limx→01x2∫0xf(t)dt

∫0x 1-f1(t)2 dt=∫0x f(t) dtdifferentiate w.r.t 'x'1-f1(x)2=f(x)1-f1(x)2=f2(x)1-f2(x)=f1(x)⇒ f1(x)1-f2(x)=1 ⇒ ∫ f1(x)1-f2(x) dx=∫ 1 dx ⇒ sin-1 (f(x))=x+C x=0 ⇒ sin-1(f(x))=0+C 0=C ∴ sin-1(f(x))=x f(x)=sin x ltx→0 1x2 ∫0x f(t) dt =ltx→0 ∫0x sin t dtx2 =ltx→0 sin x2x=12

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Let f be a non-negative function in $[0, 1]$ and twice differentiable in (0, 1). If $\int_{0}^{x} \sqrt{1 - {(f^{1} (t))}^{2}} d t = \int_{0}^{x} f (t) d t, 0 \leq x \leq 1$ and $f (0) = 0,$ then $\underset{x \to 0}{l i m} \frac{1}{x^{2}}$ $\int_{0}^{x} f (t) d t$

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equals 0

equals 1

equals $\frac{1}{2}$

does not exist

answer is A.

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Detailed Solution

$\int_{0}^{x} \sqrt{1 - {(f^{1} (t))}^{2}} dt = \int_{0}^{x} f (t) dt$

differentiate w.r.t 'x'

$\begin{array}{rcl} \sqrt{1 - {(f^{1} (x))}^{2}} & = & f (x) \\ 1 - {(f^{1} (x))}^{2} & = & f^{2} (x) \\ \sqrt{1 - f^{2} (x)} & = & f^{1} (x) \end{array}$

$\Rightarrow \frac{f^{1} (x)}{\sqrt{1 - f^{2} (x)} = 1} \Rightarrow \int \frac{f^{1} (x)}{\sqrt{1 - f^{2} (x)}} dx = \int 1 dx \Rightarrow \sin^{- 1} (f (x)) = x + C x = 0 \Rightarrow \sin^{- 1} (f (x)) = 0 + C 0 = C ∴ \sin^{- 1} (f (x)) = x f (x) = \sin x ()$

$\begin{array}{rcl} \underset{x \to 0}{lt} \frac{1}{x^{2}} \int_{0}^{x} f (t) dt \\ = & \underset{x \to 0}{lt} \frac{\int_{0}^{x} \sin t dt}{x^{2}} \\ = & \underset{x \to 0}{lt} \frac{\sin x}{2 x} = \frac{1}{2} \end{array}$