Question

f:R->[0,+,~]f:R->[0,+,~],f(x)=|x| is one one and onto or not​

Answer 1

f:R→R, given by f(x)=ex.

one−one:

Let x1 and x2 be any two elements in the domain (R), such that f(x1)=f(x2)

Calculate f(x1):

⇒  f(x1)=ex1

Calculate f(x2):

⇒  f(x2)=ex2

Now, f(x1)=f(x2)

⇒  ex1=ex2

⇒  x1=x2

∴  f is one-one function.

onto:

We know that range of ex is (0,∞)=R+

⇒  Co-domain =R

Both are not same.

∴  f is not onto function.

If the co-domain is replaced by R+, then the co-domain and range become the same and in that case, f is onto and hence, it is a bijection.

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