Statement 1: The function,  $f:R\to R$ given by  $f\left(x\right)=2x$ is one-one and onto.
Statement 2 : The function  $f:R\to R$ defined by $f\left(x\right)=x^2$  is both one-one and onto. 

Let,  $f\left(x_1\right)=f\left(x_2\right)\Rightarrow 2x_1=2x_2\Rightarrow x_1=x_2$.
Thus, f is one-one
Let  $y\in R$ (co-domain) and  $f\left(x\right)=y\Rightarrow 2x=y$
$\Rightarrow x=\frac{1}{2}y\in \mathbf{R}$ , the codomain. That is, for every 
$y\in \mathbf{R}$  there exists a real number  $\frac{y}{2}\in R,$ such that  $f\left(\frac{y}{2}\right)=y$. Hence  $f$ is onto.
Clearly,  $f\left(2\right)=f(-2)$ but  $2\ne -2$. Thus,  $f$ is not one-one.
Now, $-2\in R$ (codomain), then $x^2=-2$ has no real solution  $\Rightarrow f$  is not onto.