Question
For real $$x,$$ let $$f\left( x \right) = {x^3} + 5x + 1,$$ then
A.
$$f$$ is onto $$R$$ but not one-one
B.
$$f$$ is one-one and onto $$R$$
C.
$$f$$ is neither one-one nor onto $$R$$
D.
$$f$$ is one-one but not onto $$R$$
Answer :
$$f$$ is one-one and onto $$R$$
Solution :
Given that $$f\left( x \right) = {x^3} + 5x + 1$$
$$\therefore f'\left( x \right) = 3{x^2} + 5 > 0,\forall x \in R$$
$$ \Rightarrow f\left( x \right)$$ is strictly increasing on $$R$$
$$ \Rightarrow f\left( x \right)$$ is one one
$$\therefore $$ Being a polynomial $$f\left( x \right)$$ is cont. and inc.
on $$R$$ with $$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = - \infty $$
and $$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \infty $$
$$\therefore $$ Range of $$f = \left( { - \infty ,\infty } \right) = R$$
Hence $$f$$ is onto also. So, $$f$$ is one one and onto $$R$$.