Let $$R$$ be the set of real numbers. If $$f:R \to R$$ is a function defined by $$f\left( x \right) = {x^2},$$ then $$f$$ is:
A.
Injective but not surjective
B.
Surjective but not injective
C.
Bijective
D.
None of these.
Answer :
None of these.
Solution :
$$f\left( x \right) = {x^2}$$ is many one as $$f\left( 1 \right) = f\left( { - 1} \right) = 1$$
Also $$f$$ is into as $$- ve$$ real number have no pre-image.
$$\therefore F$$ is neither injective nor surjective.
Releted MCQ Question on Calculus >> Function
Releted Question 1
Let $$R$$ be the set of real numbers. If $$f:R \to R$$ is a function defined by $$f\left( x \right) = {x^2},$$ then $$f$$ is: