Explanation

(c)
Given, $f(x) = x^{3} + 5x + 1$
Now, $f^{'}(x) = 3x^{2} + 5 > 0,\ \forall x \in R$
$\therefore f(x)$ is strictly increasing function.
$\therefore f(x)$ is one-one function.
Clearly, $f(x)$ is a continous function and also increasing on $R$,
$\lim_{x \rightarrow - \infty}{f(x) = - \infty\ and\ \lim_{x \rightarrow \infty}{= \infty}}$
$\therefore f(x)$ takes every value between $- \infty$ and $\infty$
Thus, $f(x)$is onto function.