Let $f : R \to R$ be a function defined by $f(x) = x^3 + x$, then check whether $f$ is a bijection or not.

$f$ is a bijective function.

The correct answer is Option (3) → $f$ is a bijective function. ##

We have $f : R \to R$ defined by $f(x) = x^3 + x$

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$

$\Rightarrow x_1^3 + x_1 = x_2^3 + x_2 \Rightarrow x_1^3 - x_2^3 + x_1 - x_2 = 0$

$\Rightarrow (x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2 + 1) = 0 \Rightarrow x_1 - x_2 = 0 \Rightarrow x_1 = x_2$

$[∵x_1^2 + x_1x_2 + x_2^2 \ge 0 \text{ for all } x_1, x_2 \in R.$

$∴x_1^2 + x_1x_2 + x_2^2 + 1 \ge 1 \text{ for all } x_1, x_2 \in R]$

$\Rightarrow f$ is one-one 

Let $y$ be any arbitrary element of $R$, then there exist $x \in R$ such that $f(x) = y$.

$\Rightarrow x^3 + x = y \Rightarrow x^3 + x - y = 0$

Since, odd degree equation has atleast one real root.

Thus, for every value of $y$, the equation $x^3 + x - y = 0$ has a real root $\alpha$, such that $\alpha^3 + \alpha - y = 0$

$\Rightarrow f(\alpha) = y$

Thus, for every $y \in R, \exists \, \alpha \in R$ such that $f(\alpha) = y$

So, $f$ is onto function

Hence, $f : R \to R$ is a bijection.