Let $f$ be a non-negative differentiable function on $[0,1]$. Let $f$ be a non-negative differentiable function on $[0,1]$ such that $\int_{0}^{x} \sqrt {1-\{f'(t)\}^2}\ dt = \int_{0}^{x} f(t)\ dt$, $0 \le x \le 1$ and $f(0)=0$. Then which of the following is true?
$(1)$ $f(\frac {1} {2}) < \frac {1} {2}$ and $f(\frac {1} {3}) > \frac {1} {3}$.
$(2)$ $f(\frac {1} {2}) > \frac {1} {2}$ and $f(\frac {1} {3}) > \frac {1} {3}$.
$(3)$ $f(\frac {1} {2}) < \frac {1} {2}$ and $f(\frac {1} {3}) < \frac {1} {3}$.
$(4)$ $f(\frac {1} {2}) > \frac {1} {2}$ and $f(\frac {1} {3}) < \frac {1} {3}$.
I have first differentiated the above given expression and obtained $\{f(t)\}^2 + \{f'(t)\}^2=1$ for all $t \in [0,1].$ Clearly from this expression it follows that $|f(t)| \le 1$ and $|f'(t)| \le 1$ for all $t \in [0,1].$ Since $f$ is non-negative on $[0,1]$ so we have $0 \le f(t) \le 1$ for all $t \in [0,1].$ Now let us construct a function $g$ on $[0,1]$ defined by $g(t)=f(t)-t$ , $t \in [0,1]$.Then clearly $g$ is differentiable on $[0,1]$ since $f$ is so. Now $g'(t)=f'(t)-1$ , for all $ t \in [0,1]$. Since $|f'(t)| \le 1$ for all $t \in [0,1]$ so we have $-2 \le g'(t) \le 0$ for all $t \in [0,1].$ This shows that the function $g$ is monotonic decreasing on $[0,1].$ So $g(\frac {1} {2}) \le g(0)=0$ and $g(\frac {1} {3}) \le g(0)=0$ since it is given that $f(0)=0.$ So we have $f(\frac {1} {2}) \le \frac {1} {2}$ and $f(\frac {1} {3}) \le \frac {1} {3}.$ 
Now from the above fact it is clear that $(1), (2)$ and $(4)$ are all false. Hence I think $(3)$ is the correct option. But I have confused about the strict inequality in option $(3)$ which I have failed to found. Please help me in this regard and also please mention the reason for holding strict inequality.
Thank you in advance.
 A: Lauds to Arnab Chatterjee for a sweet analysis which I think has advantages over what I offer here since it may be more generalizable to similar problems.
Having said that, here's a quick and easy way of solving this.  Note that 
$f(x) = \sin x.  \tag{1}$
If we differentiate the given equation
$\displaystyle \int_{0}^{x} \sqrt {1-\{f'(t)\}^2}\ dt = \displaystyle\int_{0}^{x} f(t)\ dt \tag{2}$
with respect to $x$, we obtain, just as did our OP,
$ \sqrt {1-\{f'(x)\}^2}\ dt = f(x); \tag{3}$
from which, as was pointed out,
$(f(x))^2 + (f'(x))^2 = 1. \tag{4}$
Now any time I see a function whose square added to the square of its derivative yields $1$, a bell goes off in my head and I think, "sines and cosines"!  Indeed, we may re-arrange (4) to read
$f'(x) = \sqrt{1 - (f(x))^2}; \tag{5}$
the function $\sqrt{1 - f^2}$ is differentiable in on $[0, 1)$ hence locally Lipschitz there; it follows there is a unique solution $f(x)$ to (5) on $[0, 1)$ such that $f(0) = 0$, so (1) yields the only possibility.  We can then extend $f(x)$ to $[0, 1]$ by continuity.
Since
$\sin x < x \tag{6}$
for $x > 0$, the only possible correct answer is (3).
Using the theory of ODEs here is perhaps overkill, though it does solve the problem; but this is one reason I like Arnab Chatterjee's method.
