We are given that a function $f(x)$ is continuous in the interval $(0,2$) and $f(0) = f(2) =-1$, $f(1) =1$
Consider a function $g(y)$ such that $g(y) = f(y) -f(y+1)$
From the given values we have,
$g(0) = f(0) – f(1) = -1-1 = -2 $
$g(1) = f(1) – f(2) = 1-(-1) =2 $
Notice how $g(0)$ and $g(1)$ have opposite signs, from a simple corollary of the Intermediate Value Theorem:
If a continuous function defined on an interval is sometimes positive and sometimes negative it must be 0 at some point.
Thus there will exist atleast one root in the interval $(0,1)$ such that
$g(y) = 0$
$ f(y) -f(y+1) =0$
$ f(y) = f(y+1)$
Thus, option A is true.
For every $y$ in the interval $(0,1)$, $f(y) = f(2-y)$ need not be true. Since, we are given that $f(0) = -1$ and $f(1) =1$ the curve will be an increasing one ( need not be strictly monotonically increasing, but it will be increasing as the function needs to reach till $y = 1$ at $x=1$.). So, its not necessary that every y follows the condition in option B hence, it can be eliminated.
Just by knowing three random points of a curve we cannot determine the maxima or minima of an unknown function, be it any interval. Thus, option C is eliminated
Let us consider again that $g(y) = f(y) + f(2-y)$
We have,
$g(0) = f(0) + f(2) = -2$
$g(1) = f(1) +f(1) = 2$
With a similar argument as above we can prove $f(y) = -f(2-y)$
Correct options: $A,D$