If `f(x)=(2x(sinx+tanx))/(2[(x+2pi)/(pi)]-3)` then it is (where [.] denotes the greatest integer function)

Options

A. odd
B. Even
C. many one and onto
D. one-one

Explanation

To solve the problem, we need to analyze the function \( f(x) = \frac{2x(\sin x + \tan x)}{2\left(\frac{x + 2\pi}{\pi}\right) - 3} \) and determine its properties: whether it is an even function, odd function, many-one function, or onto function.

### Step 1: Simplify the Function
The function can be rewritten as:
\[ f(x) = \frac{2x(\sin x + \tan x)}{2\left(\frac{x + 2\pi}{\pi}\right) - 3} \]
This simplifies to:
\[ f(x) = \frac{2x(\sin x + \tan x)}{\frac{2x + 4\pi}{\pi} - 3} \]
\[ = \frac{2x(\sin x + \tan x)}{\frac{2x + 4\pi - 3\pi}{\pi}} = \frac{2x(\sin x + \tan x)}{\frac{2x + \pi}{\pi}} = \frac{2\pi x(\sin x + \tan x)}{2x + \pi} \]

### Step 2: Determine if the Function is Even or Odd
To check if \( f(x) \) is even or odd, we need to evaluate \( f(-x) \):
\[ f(-x) = \frac{2\pi(-x)(\sin(-x) + \tan(-x)}{2(-x) + \pi} \]
Using the properties of sine and tangent:
\[ \sin(-x) = -\sin(x), \quad \tan(-x) = -\tan(x) \]
Thus,
\[ f(-x) = \frac{2\pi(-x)(-\sin(x) - \tan(x))}{-2x + \pi} = \frac{2\pi x(\sin(x) + \tan(x))}{\pi - 2x} \]

Now, we compare \( f(-x) \) with \( f(x) \):
\[ f(-x) \neq f(x) \quad \text{and} \quad f(-x) \neq -f(x) \]
This indicates that \( f(x) \) is neither even nor odd.

### Step 3: Determine if the Function is Many-One or Onto
To check if \( f(x) \) is a many-one function, we need to see if there are different values of \( x \) that yield the same \( f(x) \).

Since \( \sin(x) \) and \( \tan(x) \) are periodic functions, \( f(x) \) will also be periodic. Therefore, there will be multiple \( x \) values that yield the same output, confirming that \( f(x) \) is a many-one function.

### Step 4: Determine if the Function is Onto
The range of \( \sin(x) \) and \( \tan(x) \) is \( \mathbb{R} \). Since \( f(x) \) is constructed from these functions, and given that it can take on all real values due to the periodic nature of sine and tangent, we can conclude that \( f(x) \) is onto.

### Conclusion
Based on the analysis:
- \( f(x) \) is neither even nor odd.
- \( f(x) \) is a many-one function.
- \( f(x) \) is onto.

Thus, the final answer is that \( f(x) \) is a many-one function and onto function.