Let $f\left(x\right)=\left[x\right]$ and $g\left(x\right)=\left\{\begin{array}{l} \\ \end{array}\right.\begin{array}{c}0,x\in Z\\ x^2,x\in R-Z\end{array}.$ Then which of the following is not true($[.]$ represents the
greatest integer function)?

$\text{Since}\underset{x\to 1^{-}}{lim}g\left(x\right)=\underset{x\to 1^{+}}{lim}g\left(x\right)=1\text{and}g\left(1\right)=0.$

g(x) is not continuous at x=1

 $\text{but}\underset{ x\to 1}{ lim}g\left(x\right)\text{exists}$

$\text{so,}\underset{x\to 1}{lim}f\left(x\right)$ does not exist and so $f\left(x\right)$ is not constinuous at $x = 1$

we have $gof\left(x\right)=g\left(f\right(x\left)\right)=g\left(\right[x\left]\right)=0\mathrm{\forall }x\in R.$ So, $gof$ is continuous for all x.

 we have $\text{fog}\left(x\right)=f\left(g\right(x\left)\right)$$=\left\{\begin{array}{l} \\ \end{array}\right.\begin{array}{ll}f\left(0\right),& x\in Z\\ f\left(x^2\right),& x\in R-Z\end{array}=\left\{\begin{array}{l} \\ \end{array}\right.\begin{array}{ll}0,& x\in z\\ \left[x^2\right],& x\in R-Z\end{array}$

which is clearly not continuous