TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Let x and y be any two elements in the domain (Z), such that
f(x+y)=f(x)+f(y)...(i)
Differentiating above expression w.r.t ' y, keeping x constant, we get
f′(x+y)=f(y)
Let y=0⇒f′(x+0)=f′(0)
and Assume f′(0)=K
∴‌‌f′(x)=K
Integrating on both sides, we get
Now putting x=y=0 in Eq. (i), we get
Let x=0, from Eq. (ii)
f(0)=K×0+C or 0=0+C or C=0
∴‌‌f(x)=kx
Case (i) If k>0, then f(x) is strictly increasing.
Case (ii) If k<0, then f(x) is strictly decreasing
So, function is injective
Also f(x) where every element in the codomain is a valid output of the function i.e., range is equal to codomain.
So, function is surjective also.
Therefore, there are two bijective functions.