Question

The function $f:R\to R,f\left(x\right)=x^2$ is
(a) injective but not surjective
(b) surjective but not injective
(c) injective as well as surjective
(d) neither injective nor surjective

Answer 1

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y). Then,

$$\begin{gathered} x^2=y^2 \\ \Rightarrow x=\pm y \end{gathered}$$

So, f is not one-one.

Surjectivity:
$$\begin{gathered} \mathrm{As}f\left(-1\right)={\left(-1\right)}^2=1 \\ \mathrm{and}f\left(1\right)=1^2=1, \\ f\left(-1\right)=f\left(1\right) \end{gathered}$$
So, both -1 and 1 have the same images.
$\Rightarrow$f is not onto.
So, the answer is (d).

x2+x+1=y2+y+1(x2−y2)+(x−y)=0(x+y)(x−y)+(x−y)=0(x−y)(x+y+1)=0x−y=0 (x+y+1) cannot be zero because x and y are natural numbers)x=y