$$f\left( x \right) = x + \sqrt {{x^2}} $$ is a function from $$R \to R.$$ Then $$f\left( x \right)$$ is :
A.
injective
B.
surjective
C.
bijective
D.
none of these
Answer :
none of these
Solution :
$$f\left( x \right) = x + \left| x \right|.$$ Clearly, $$f\left( { - 1} \right) = f\left( { - 2} \right) = ...... = 0$$
So, $$f$$ is many-one.
Also, $$f\left( x \right) \geqslant 0$$ for all $$x.$$ So, it is not 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: