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Integral of sqrt(1 - f'(t)^2) equals the integral of f, with f(0) = 0: find f

The same question, asked 2 times: 2021 Paper 22*, 9 Apr 2024. Every wording below, oldest first.

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  1. Question 1

    2021 Paper 22

    65
    Let ff be a non-negative function in [0,1]\lbrack 0,1\rbrack and twice differentiable in (0,1)(0,1). If ∫0x1−(f′(t))2dt=∫0xf(t)dt\int_{0}^{x} \sqrt{1 -\left( f^{'}(t) \right)^{2}}dt =\int_{0}^{x} f(t)dt, 0≤x≤10 \leq x\leq 1 and f(0)=0f(0) = 0, then lim⁡x→01x2∫0xf(t)dt\lim_{x\rightarrow 0} \frac{1}{x^{2}}\int_{0}^{x} f(t)dt:

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  2. Question 2

    9 Apr 2024

    153
    Let ∫0x1−(y′(t))2dt=∫0xy(t)dt,0≤x≤3,y≥0\int_{0}^{x} \sqrt{1 -\left( y^{'}(t) \right)^{2}}dt =\int_{0}^{x} y(t)dt,0 \leq x \leq 3,y \geq 0, y(0)=0y(0) = 0. Then at x=2,y"+y+1x = 2,y" + y + 1 is equal to:

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