Calculate $$f(x_1):$$
$$\Rightarrow$$ $$f(x_1)=2x_1$$
Now, calculate $$f(x_2):$$
$$\Rightarrow$$ $$f(x_2)=2x_2$$
Now, $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$2x_1=2x_2$$
$$\Rightarrow$$ $$x_1=x_2$$
Hence, if $$f(x_1)=f(x_2),$$ $$x_1=x_2$$ the function $$f$$ is $$one-one$$.
Now, $$f(x)=2x$$
Let $$f(x)=y,$$ such that $$y\in N$$
$$\Rightarrow$$ $$2x=y$$
$$\Rightarrow$$ $$x=\dfrac{y}{2}$$
If $$y=1$$
$$x=\dfrac{1}{2}=0.5,$$ which is not possible as $$x\in N$$
Hence, $$f$$ is $$not$$ onto.
So, the function $$f:N\rightarrow N, $$ given by $$f(x)=2x$$, is one-one but not onto.
$$(ii)$$ Let the function $$f:N\rightarrow N$$, given by $$f(1)=f(2)=1$$
Here, $$f(x)=f(1)=1$$ and
$$\Rightarrow$$ $$f(x)=f(2)=1$$
Since, different elements $$1,2$$ have same image $$1,$$
$$\therefore$$ $$f$$ is not one-one.
Let $$f(x)=y,$$ such that $$y\in N$$
Here, $$y$$ is a natural number and for every $$y$$, there is a value of $$x$$ which is natural number.
Hence $$f$$ is onto.
So, the function $$f:N\rightarrow N$$, given by $$f(1)=f(2)=1$$ is not one-one but onto.
$$(iii)$$ Let function $$f:R\rightarrow R,$$ given by $$f(x)=x^2$$
Calculate $$f(x_1):$$
$$\Rightarrow$$ $$f(x_1)=(x_1)^2$$
Calculate $$f(x_2):$$
$$\Rightarrow$$ $$f(x_2)=(x_2)^2$$
Now, $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$(x_1)^2=(x_2)^2$$
$$\Rightarrow$$ $$x_1=x_2$$ or $$x_1=-x_2$$
Since, $$x_1$$ does not have unique image, it is $$not$$ one-one.
Now, $$f(x)=x^2$$
Let $$f(x)=y,$$ such that $$y\in R$$
$$\Rightarrow$$ $$x^2=y$$
$$\Rightarrow$$ $$x-\pm\sqrt{y}$$
Since, $$y$$ is real number, then it can be negative also.
Putting $$y=-5$$
$$x=\pm\sqrt{-5}$$
Which is not possible as the root of a negative number is not real.
Hence, $$x$$ is not real, so $$f$$ is not onto.
$$\therefore$$ Function $$f:R\rightarrow R,$$ given by $$f(x)=x^2$$ is neither one-one nor onto.