Question

Let f: R=> R be defined by fix) = x'. Then f is
not an injection because f(-1) = 1 = f(1).​

Answer 1

Given, function f:R→R such that f(x)=1+x2, 

Let A and B be two sets of real numbers.

Let x1,x2∈A such that f(x1)=f(x2).

⇒1+x12=1+x22⇒x12−x22=0⇒(x1−x2)(x1+x2)=0

⇒x1=±x2. Thus f(x1)=f(x2) does not imply that x1=x2.

For instance, f(1)=f(−1)=2, i.e. , two elements (1, -1) of A have the same image in B. So, f is many-one function.

Now, y=1+x2⇒x=y−1⇒elements < y have no pre-image in A (for instance an element -2 in the codomain has no pre-image in the domain A). So, f is not onto.

Hence, f is neither one-one onto. So, it is not bijective.