The function `f(x) =cot x` is discontinuous on set

Options

A. `{x=npi:ninZ }`
B. `{x = 2npi : n in Z }`
C. ` {x =(2n+1)(pi)/(2),n in Z}`
D. `{x=(npi)/(2), n in Z}`

Explanation

To determine where the function \( f(x) = \cot x \) is discontinuous, we can follow these steps:

### Step 1: Understand the function
The function \( f(x) = \cot x \) can be expressed in terms of sine and cosine:
\[ f(x) = \cot x = \frac{\cos x}{\sin x} \]

### Step 2: Identify points of discontinuity
The cotangent function is discontinuous wherever the denominator is zero. Therefore, we need to find where \( \sin x = 0 \).

### Step 3: Solve for \( \sin x = 0 \)
The sine function is equal to zero at integer multiples of \( \pi \):
\[ \sin x = 0 \quad \Rightarrow \quad x = n\pi, \quad n \in \mathbb{Z} \]
where \( n \) is any integer.

### Step 4: Conclusion about discontinuity
Thus, the function \( f(x) = \cot x \) is discontinuous at all points where \( x = n\pi \), where \( n \) is an integer.

### Final Answer
The function \( f(x) = \cot x \) is discontinuous on the set:
\[ \{ x = n\pi \mid n \in \mathbb{Z} \} \]

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