The function f: R→R, f(x)= |x| (R is the set of real numbers) is

The correct answer is Option (4) → neither injective nor surjective

Given: $f: \mathbb{R} \to \mathbb{R}, \ f(x) = |x|$

Check Injectivity:

If $f(a) = f(b)$, then $|a| = |b|$ ⇒ $a = b$ or $a = -b$

For example, $f(2) = |2| = 2$ and $f(-2) = |-2| = 2$ ⇒ $f(2) = f(-2)$ but $2 \ne -2$

So, $f(x)$ is not injective

Check Surjectivity:

Codomain is $\mathbb{R}$ but range of $f(x) = |x|$ is $[0, \infty)$

So negative numbers in $\mathbb{R}$ are not attained by $f(x)$

Therefore, $f$ is not surjective

Conclusion:

The function is neither injective nor surjective