Question

Cheak following function is onto or not?

f :[-1, 1] R defined by
f(x)= x/2​

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+x¹

=1+x²

⇒x¹−x²

=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.