Define F(x) as the product of two real functions  $f_1\left(x\right)=x,x\in R,and{f}_2\left(x\right)=\left\{\begin{array}{l}\sin \frac{1}{x} \begin{array}{cc},& ifx\ne 0\end{array}\\ 0 \begin{array}{cc},& ifx=0\end{array} \end{array}\right.$    as follows:
$F\left(x\right)=\{\begin{array}{cc}{f}_1\left(x\right).f_2\left(x\right)& ifx\ne 0\\ 0,& ifx=0\end{array}$
Statement -1: F(x) is continuous on R.
Statement -2 : $f_1\left(x\right)and{f}_2\left(x\right)$ are continuous on R.
 

$F\left(x\right)=\{\begin{array}{cc}x\sin \left(\frac{1}{x}\right),& x\ne 0\\ 0& ,x=0\end{array}$

$\underset{x\to 0}{\lim }F\left(x\right)=\underset{x\to 0}{\lim }x\sin \left(\frac{1}{x}\right)=0$

Also F (0) = 0

$\Rightarrow \underset{x\to 0}{\lim }F\left(x\right)=F\left(0\right)$

$\Rightarrow F\left(x\right)$ is continuous at x=0

$\Rightarrow F\left(x\right)$ is continuous for all real numbers
Statement-1 is true
$f_1\left(x\right)=x$

$\Rightarrow$it is continuous on R

$f_2\left(x\right)=\{\begin{array}{cc}\sin \left(\frac{1}{x}\right)& ,x\ne 0\\ 0& ,x=0\end{array}$

$\underset{x\to 0}{\lim }\sin \frac{1}{x}$does not exist

$\Rightarrow$it is not continuous at x=0

$\Rightarrow f_2\left(x\right)$ is discontinuous on R 
Thus statement-2 is false.