CUET Preparation Today
The function $f(x) = \cot x$ is discontinuous on the set |
$\{x = n\pi : n \in \mathbb{Z}\}$ $\{x = 2n\pi : n \in \mathbb{Z}\}$ $\left\{x = (2n + 1)\frac{\pi}{2} : n \in \mathbb{Z}\right\}$ $\left\{x = \frac{n\pi}{2} : n \in \mathbb{Z}\right\}$ |
$\{x = n\pi : n \in \mathbb{Z}\}$ |
The correct answer is Option (1) → $\{x = n\pi : n \in \mathbb{Z}\}$ ## We know that, $f(x) = \cot x$ is continuous in $R - \{x = n\pi \in \mathbb{Z}\}$. Since, $f(x) = \cot x = \frac{\cos x}{\sin x} \quad [∵\sin x = 0 \text{ at } x = n\pi, n \in \mathbb{Z}]$ Hence, $f(x) = \cot x$ is discontinuous on the set $\{x = n\pi : n \in \mathbb{Z}\}$. |
