Let `f((x+y)/2)=(f(x)+f(y))/2fora l lr e a lxa n dy` If `f^(prime)(0)` exists and equals `-1a n df(0)=1,t h e n f i n d f(2)dot`

Explanation

Suppose `A(x, f(x)) and B(y, f(y))` be any two points on the curve `y=f(x)`.
`" "`If `M` is the mid-point of AB, then the coordinates of `M` are `((x+y)/(2), (f(x)+f(y))/(2))`.
Now the foot of the perpendicular from the M on the x-axis is L. ML meets the curve `y=f(x)` at P.
According to the graph, the coordinates of P are `((x+y)/(2), f((x+y)/(2))) and PL gt ML`.
`rArr" "f((x+y)/(2)) gt (f(x)+ f(y))/(2)`
But given `f((x+y)/(2))= (f(x)+f(y))/(2)`, which is possible when `P to M`.
That is, P lies on AB. Hence `y=f(x)` must be a linear function.
Let `f(x)=ax+b" "rArr f(0)= 0+b=1 " "`(given )
and `f'(x)=a " "rArr f'(0)=a=-1" "`(given )
`therefore " "f(x)=-x+1`
`therefore" "f(2)=-2+1=-1`