Consider functions \(f,g:\mathbb{Z}\rightarrow \mathbb{R}\), \(f\left(x\right)=ax+m\), \(g\left(x\right)=bx+n\) where \(a,b,m\), and \(n\) are real constants. Let \(A\) and \(B\) ranges of \(f\) and \(g\) respectively. Therefore, which of the following statement are correct:

  • Let \(A\ =\ B\), therebefore, \(a=b\) and \(m=n\);
  • Let \(A=\mathbb{Z}\), so \(a=\ 1\);
  • Let \(a,b,m,n\in Z\), with \(a=b\) and \(m\ =\ n\), so \(A\ =\ B\);

I.

\begin{equation} f:\mathbb{Z}\rightarrow \mathbb{R}:f\left(x\right)=ax+m \end{equation}

\begin{equation} g:\mathbb{Z}\rightarrow \mathbb{R}:g\left(x\right)=bx+n \end{equation}

Consider

\begin{equation} f\left(x\right)=3x+1 \end{equation}

\begin{equation} g\left(x\right)=3x+4 \end{equation}

Evaluation \(-2\ \le\ \ x\ \le2\). Hence, we will obtain the ranges for $f\left(x\right)=3x+1$

\begin{equation} A: \left \lbrace -5,-2,1,4,7 \right \rbrace \end{equation}

And for $g\left(x\right)=3x+4$

\begin{equation} B∶ \left \lbrace -2,1,4,7,10 \right \rbrace \end{equation}

It’s easy to see that $A\ \neq B$, so is False.

II.

$f:\mathbb{Z} \rightarrow \mathbb{R}$. $y=f\left(x\right)=ax+m$ and $a\neq1\ \land a\ \in \mathbb{Z}$. We should get the following range

\begin{equation} A: \left \lbrace \ldots,m-2a,m-a,m,a+m,2a+m,\ldots \right \rbrace \end{equation}

So, there not all the integers’ values. If $a\ \geq2$, so $y\ -\ x\ =\ a\geq2\ \rightarrow\ y\ \geq x\ +\ 2$. Hence, this statement is True.

III.

$\mathbb{Z} \rightarrow \mathbb{R}$. $f(x) = ax+m$.
$A: \left \lbrace \ldots,m-2a,m-a,m,a+m,2a+m,\ldots \right \rbrace$
$g: \mathbb{Z} \rightarrow \mathbb{R}$. $g(x) = ax-m$
$B: \left \lbrace \ldots,-m-2a,-m-a,-m,a-m,2a-m,\ldots \right \rbrace$

Hence, $A \ne B$. This statement is False.