Question

The number of bijective functions $f:Z\to Z$ such that $f(x+y)=f(x)+f(y)\forall x,y\in Z$, is

• A. two
• B. four
• C. zero
• D. infinitely many

Answer 1

Answer: (A)

Let $x$ and $y$ be any two elements in the domain $(Z)$, such that
$f(x+y)=f(x)+f(y)\dots$ (i)
Differentiating above expression w.r.t ' $y$, keeping $x$ constant, we get
$f^{\prime }(x+y)=f(y)$
Let $y=0\Rightarrow f^{\prime }(x+0)=f^{\prime }(0)$
and Assume $f^{\prime }(0)=K$
$\therefore f^{\prime }(x)=K$
Integrating on both sides, we get
$f(x)=Kx+C\{C\to$ integration constant $\}\dots$ (ii)
Now putting $x=y=0$ in Eq. (i), we get
$f(0+0)=f(0)+f(0)\Rightarrow f(0)=2f(0)\text{ or }f(0)=0$
Let $x=0$, from Eq. (ii)
$f(0)=K\times 0+C\text{ or }0=0+C\text{ or }C=0$
$\therefore f(x)=kx$
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.