JEE MAIN - Mathematics (2009 - No. 3)
For real x, let f(x) = x3 + 5x + 1, then
f is one-one but not onto R
f is onto R but not one-one
f is one-one and onto R
f is neither one-one nor onto R
Explanation
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 $$f(x)$$ 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 and onto $$R.$$
$$\therefore$$ $$\,\,\,\,\,$$ $$f'\left( x \right) = 3{x^2} + 5 > 0,\,\,\,\forall x \in R$$
$$ \Rightarrow f\left( x \right)\,\,$$ is strictly increasing on $$R$$
$$ \Rightarrow f\left( x \right)$$ is one one
$$\therefore$$ $$\,\,\,\,\,\,\,$$ Being a polynomial $$f(x)$$ 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 and onto $$R.$$
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