Let $ f: R \to [\frac{1}{2} , 1]$ and $f(x+2) = \frac{1}{2} +\sqrt{f(x) -f(x)^2}$ Problem : 
Let $ f: R \to [\frac{1}{2} , 1]$ and $f(x+2) = \frac{1}{2} +\sqrt{f(x) -f(x)^2}$ 
Then which of the following is always true 
$(a) f(2) = f(7)$ 
$(b) f(4) = f(10) $ 
$(c) f(2) =f(4) $
Please suggest how to proceed in this problem , will be of great help please guide thanks 
 A: We first note that the function that takes us from $f(x)$ to $f(x+2)$, namely
$$g(x)=\frac 12+\sqrt{x-x^2}$$
is the upper-right quadrant of an circle, and if its domain is $\left[\frac 12,1\right]$ then so is its range. We can see that $g(x)$ is a decreasing function, a bijection from $\left[\frac 12,1\right]$ to itself. (These statements are easily confirmed if you know conic sections, or are also confirmed by graphing.)
The important point is that if $f(x)$ is defined, so is $f(x+2)$. That is also the only restriction on $f$, so we can choose any values at all for a domain $[a,a+2)$ and then the values of $f$ are then defined in a sort-of-periodic way for all real  numbers.
Choice (a) cannot always be true, because we can define $f(2)$ and $f(3)$ any way we want, then $f(3)$ alone determines $f(3+2)$ and $f(3+2+2)$. So there is no relationship between $f(2)$ and $f(7)$.
We can check (c) with the formula:
$$f(4)=f(2+2)=\frac 12+\sqrt{f(2)-f(2)^2}$$
If we let $f(2)=1$ then $f(4)=\frac 12+\sqrt{1-1^2}=\frac 12$, so (c) is certainly not always true.
Now we'll set $f(4)=\frac 12$, then we get
$$f(6)=f(4+2)=\frac 12+\sqrt{\frac 12-\left(\frac 12\right)^2}=1$$
$$f(8)=f(6+2)=\frac 12+\sqrt{1-1^2}=\frac 12$$
$$f(10)=f(8+2)=\frac 12+\sqrt{\frac 12-\left(\frac 12\right)^2}=1$$
So here $f(4)\ne f(10)$, and (b) is not always true.

Interestingly, if we do the calculations we will see that $g(g(x))=x$. We can also see that from the graph of $g(x)$, which is symmetric with the line $y=x$ and is therefore its own inverse. So we can say that
$$f(x+4)=f(x+2+2)=g(f(x+2))=g(g(f(x)))=f(x)$$
So we do get that $f(2)=f(6)=f(10)$ and $f(4)=f(8)$. But your question does not ask about those.
