Step 1: Understanding the modulus function.
The modulus function $f(x) = |x|$ maps every real number $x$ to a non-negative real number. Its codomain is $\mathbb{R}^{+}$ (non-negative real numbers).
Step 2: Check one-one property.
If $f(a) = f(b)$, then $|a| = |b|$. This means either $a = b$ or $a = -b$.
Thus, different inputs (e.g., $2$ and $-2$) give the same output. Hence, $f(x)$ is **not one-one**, but **many-one**.
Step 3: Check onto property.
For every $y \in \mathbb{R}^{+}$, there exists an $x \in \mathbb{R}$ such that $f(x) = |x| = y$.
Thus, the function is **onto** $\mathbb{R}^{+}$.
Step 4: Conclusion.
The function is **many-one and onto**, so the correct answer is (B).
A relation \( R = \{(a, b) : a = b - 2, b \geq 6 \} \) is defined on the set \( \mathbb{N} \). Then the correct answer will be:
The principal value of the \( \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) \) will be: