Let $$f$$ be a non-negative function in $$[0, 1]$$ and twice differentiable in $$(0, 1)$$. If
$$\int_0^x \sqrt{1 - (f'(t))^2} dt = \int_0^x f(t) dt$$, $$0 \leq x \leq 1$$ and $$f(0) = 0$$, then $$\lim_{x \to 0} \frac{1}{x^2} \int_0^x f(t) dt$$:

We are told that the non-negative function $$f$$ is twice differentiable on $$(0,1)$$, that $$f(0)=0$$ and that for every $$x$$ with $$0\le x\le 1$$

$$\int_{0}^{x}\sqrt{\,1-(f'(t))^{2}\,}\;dt=\int_{0}^{x}f(t)\;dt.$$

We first recall the Fundamental Theorem of Calculus, which states that if $$G(x)=\int_{0}^{x}g(t)\,dt$$ with $$g$$ continuous, then $$G'(x)=g(x).$$ Applying this rule to both integrals and differentiating each side with respect to $$x$$ we obtain

$$\sqrt{\,1-(f'(x))^{2}\,}=f(x)\qquad(0<x<1).$$

Because both sides are non-negative, we can safely square the equality. Squaring gives

$$1-(f'(x))^{2}=f(x)^{2},$$

which we rearrange to

$$(f'(x))^{2}=1-f(x)^{2}.$$

Taking square-roots,

$$f'(x)=\pm\sqrt{\,1-f(x)^{2}\,}.$$

The sign is decided by the initial information. We have $$f(0)=0$$ and $$f$$ is non-negative near the origin, so for tiny positive $$x$$ the function must be increasing. Hence $$f'(0)>0,$$ and continuity of the derivative forces the positive sign:

$$f'(x)=\sqrt{\,1-f(x)^{2}\,}\qquad(0\le x\le 1).$$

Setting $$x=0$$ in the earlier squared equation gives

$$1-(f'(0))^{2}=f(0)^{2}=0\;\;\Longrightarrow\;\;(f'(0))^{2}=1\;\;\Longrightarrow\;\;f'(0)=1.$$

Now differentiate the identity $$f(x)^{2}+(f'(x))^{2}=1$$ with respect to $$x$$. Using the Chain Rule,

$$2f(x)\,f'(x)+2f'(x)\,f''(x)=0.$$

Dividing by the non-zero factor $$2f'(x)$$ (recall $$f'(x)>0$$), we obtain the linear second-order differential equation

$$f''(x)+f(x)=0.$$

The general solution of $$y''+y=0$$ is well known: $$y(x)=A\sin x+B\cos x.$$ We now impose the initial conditions.

Using $$f(0)=0$$, we get $$B\cos 0 =0 \Longrightarrow B=0.$$

Using $$f'(0)=1$$, we have $$f'(x)=A\cos x,$$ so $$f'(0)=A\cos 0=A=1.$$

Therefore

$$f(x)=\sin x\qquad(0\le x\le 1).$$

The problem asks for the limit

$$\lim_{x\to 0}\frac{1}{x^{2}}\int_{0}^{x}f(t)\,dt.$$

Substituting $$f(t)=\sin t,$$ we first evaluate the inside integral:

$$\int_{0}^{x}\sin t\,dt=\bigl[-\cos t\bigr]_{0}^{x}=1-\cos x.$$

Thus the required expression becomes

$$\frac{1-\cos x}{x^{2}}.$$

To find its limit as $$x\to 0,$$ we recall the standard Maclaurin expansion

$$\cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\cdots.$$

Hence

$$1-\cos x=\frac{x^{2}}{2}-\frac{x^{4}}{24}+\cdots,$$

and dividing by $$x^{2}$$ gives

$$\frac{1-\cos x}{x^{2}}=\frac{1}{2}-\frac{x^{2}}{24}+\cdots.$$

As $$x\to 0,$$ the higher-order terms vanish, leaving the limit

$$\lim_{x\to 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2}.$$

Hence, the correct answer is Option D.