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Q.
The number of bijective functions $f: Z \rightarrow Z$ such that $f(x +y)=f(x)+f(y) \forall x, y \in Z$, is
TS EAMCET 2020
Solution:
Let $x$ and $y$ be any two elements in the domain $(Z)$, such that
$f(x +y)=f(x)+f(y) \ldots$ (i)
Differentiating above expression w.r.t ' $y$, keeping $x$ constant, we get
$f'(x+ y)=f(y)$
Let $y=0 \Rightarrow f'(x+0)=f'(0)$
and Assume $f'(0)=K$
$\therefore f'(x)=K$
Integrating on both sides, we get
$f(x)=K x+C\{C \rightarrow$ integration constant $\} \ldots$ (ii)
Now putting $x=y=0$ in Eq. (i), we get
$f(0+0)=f(0)+f(0) \Rightarrow f(0)=2 f(0) \text { or } f(0)=0$
Let $x=0$, from Eq. (ii)
$f(0)=K \times 0+C \text { or } 0=0+C \text { or } C=0$
$\therefore f(x)=k x$
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.