Let `f(x)=((2sinx+sin2x)/(2cosx+sin2x)*(1-cosx)/(1-sinx)): x in R.`
Consider the following statements.
I. Domain of f is R.
II. Range of f is R.
III. Domain of f is `R-(4n-1)pi/2, n in I.`
IV. Domain of f is `R-(4n+1)pi/2, n in I.`
Which of the following is correct?

Options

A. (a)I and II
B. (b)II and III
C. (c)III and IV
D. (d)II, III and IV

Answer

To solve the problem, we need to analyze the function given and determine its domain and range. The function is defined as:

\[ f(x) = \frac{(2\sin x + \sin 2x)}{(2\cos x + \sin 2x)} \cdot \frac{(1 - \cos x)}{(1 - \sin x)} \]

### Step 1: Identify the Denominator
The function is defined when the denominator is not equal to zero. Thus, we need to find when:

\[ (2\cos x + \sin 2x)(1 - \sin x) \neq 0 \]

### Step 2: Analyze Each Factor
1. **For \( 1 - \sin x \neq 0 \)**:
- This implies \( \sin x \neq 1 \).
- The values of \( x \) where \( \sin x = 1 \) are given by:
\[ x = \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \]

2. **For \( 2\cos x + \sin 2x \neq 0 \)**:
- We can rewrite \( \sin 2x \) as \( 2\sin x \cos x \), so:
\[ 2\cos x + 2\sin x \cos x \neq 0 \]
- Factoring out \( 2\cos x \):
\[ 2\cos x(1 + \sin x) \neq 0 \]
- This gives us two conditions:
- \( \cos x \neq 0 \) implies \( x \neq \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \)
- \( 1 + \sin x \neq 0 \) implies \( \sin x \neq -1 \), which occurs at:
\[ x = \frac{3\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \]

### Step 3: Combine Conditions for Domain
The function is undefined at:
- \( x = \frac{\pi}{2} + 2n\pi \) (where \( \sin x = 1 \))
- \( x = \frac{3\pi}{2} + 2n\pi \) (where \( \sin x = -1 \))
- \( x = \frac{\pi}{2} + n\pi \) (where \( \cos x = 0 \))

Thus, the domain of \( f \) is:
\[ \mathbb{R} - \left\{ \frac{(4n-1)\pi}{2}, n \in \mathbb{Z} \right\} \]

### Step 4: Determine the Range
To find the range of \( f(x) \):
- The function \( f(x) \) can be simplified to:
\[ f(x) = 2\tan^2 x \]
- Since \( \tan x \) can take any real value, \( \tan^2 x \) will take values from \( 0 \) to \( +\infty \).
- Therefore, \( f(x) \) can take values from \( 0 \) to \( +\infty \).

### Conclusion
From our analysis:
- The domain of \( f \) is \( \mathbb{R} - \left\{ \frac{(4n-1)\pi}{2}, n \in \mathbb{Z} \right\} \).
- The range of \( f \) is \( [0, +\infty) \).

### Final Answer
- Statement I: False (Domain is not all of \( \mathbb{R} \))
- Statement II: False (Range is not all of \( \mathbb{R} \))
- Statement III: True (Domain is \( \mathbb{R} - \left\{ \frac{(4n-1)\pi}{2}, n \in \mathbb{Z} \right\} \))
- Statement IV: False (Domain is not \( \mathbb{R} - \left\{ \frac{(4n+1)\pi}{2}, n \in \mathbb{Z} \right\} \))

Thus, the correct statement is **III**.