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Range of the function f defined by (x)...

Range of the function f defined by
`(x) =[(1)/(sin{x})]` (where `[.]` and `{x}` denotes
greatest integer and fractional part of x respectively) is

A

I,the set of integers

B

N,the set of natural number

C

W, the set of whole number

D

`{2,3,4,…..}`

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The correct Answer is:
To find the range of the function \( f(x) = \left\lfloor \frac{1}{\sin \{x\}} \right\rfloor \), where \( \{x\} \) denotes the fractional part of \( x \), we can follow these steps: ### Step 1: Understand the fractional part of \( x \) The fractional part of \( x \), denoted as \( \{x\} \), is defined as: \[ \{x\} = x - \lfloor x \rfloor \] This means \( \{x\} \) takes values in the interval \( [0, 1) \). ### Step 2: Analyze the sine function The sine function, \( \sin \{x\} \), will take values in the interval \( [0, \sin(1)] \) for \( \{x\} \) in \( [0, 1) \). The sine function is continuous and increasing in this interval, reaching its maximum at \( \sin(1) \). ### Step 3: Determine the range of \( \frac{1}{\sin \{x\}} \) Since \( \sin \{x\} \) approaches \( 0 \) as \( \{x\} \) approaches \( 0 \), \( \frac{1}{\sin \{x\}} \) will approach \( +\infty \). The minimum value occurs when \( \{x\} \) is at its maximum, which is \( \sin(1) \). Therefore, the minimum value of \( \frac{1}{\sin \{x\}} \) is: \[ \frac{1}{\sin(1)} \] ### Step 4: Analyze the greatest integer function The function \( f(x) = \left\lfloor \frac{1}{\sin \{x\}} \right\rfloor \) will take integer values. Since \( \frac{1}{\sin \{x\}} \) can take any value from \( \frac{1}{\sin(1)} \) to \( +\infty \), we need to find the smallest integer greater than or equal to \( \frac{1}{\sin(1)} \). ### Step 5: Conclusion on the range The range of \( f(x) \) will be all natural numbers \( n \) such that: \[ n \geq \left\lceil \frac{1}{\sin(1)} \right\rceil \] Since \( \sin(1) \) is a positive number less than \( 1 \), \( \frac{1}{\sin(1)} \) is greater than \( 1 \). Thus, the range of \( f(x) \) is: \[ \text{Range of } f(x) = \{ n \in \mathbb{N} : n \geq \left\lceil \frac{1}{\sin(1)} \right\rceil \} \]

To find the range of the function \( f(x) = \left\lfloor \frac{1}{\sin \{x\}} \right\rfloor \), where \( \{x\} \) denotes the fractional part of \( x \), we can follow these steps: ### Step 1: Understand the fractional part of \( x \) The fractional part of \( x \), denoted as \( \{x\} \), is defined as: \[ \{x\} = x - \lfloor x \rfloor \] This means \( \{x\} \) takes values in the interval \( [0, 1) \). ...
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