Question

. Express the function f: A →R. f(x) = x2
– 1. where A = { -9, 0, 4 , 9) as a set of ordered
pairs.

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+x²

⇒x1/2 −x2

=0⇒(x1−x² )(x +x² )=0

⇒x 1 =±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.