If `f:R->R` and `f(x)=sin(pi{x})/(x^4+3x^2+7)`, where `{}` is a fractional part of x , then

Options

A. f is injective
B. f is not one-one and non-constant
C. f is a surjective
D. f is a zero function

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{\sin(\pi \{x\})}{x^4 + 3x^2 + 7} \), where \( \{x\} \) denotes the fractional part of \( x \).

### Step 1: Understanding the Function
The fractional part of \( x \), denoted as \( \{x\} \), is defined as \( x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). Therefore, for any integer \( n \), \( \{n\} = 0 \).

### Step 2: Evaluating Specific Values
Let's evaluate \( f(3) \) and \( f(1) \):
- For \( f(3) \):
\[ f(3) = \frac{\sin(\pi \{3\})}{3^4 + 3 \cdot 3^2 + 7} = \frac{\sin(\pi \cdot 0)}{81 + 27 + 7} = \frac{0}{115} = 0 \]

- For \( f(1) \):
\[ f(1) = \frac{\sin(\pi \{1\})}{1^4 + 3 \cdot 1^2 + 7} = \frac{\sin(\pi \cdot 0)}{1 + 3 + 7} = \frac{0}{11} = 0 \]

### Step 3: Conclusion on Injectivity
Since \( f(3) = 0 \) and \( f(1) = 0 \) but \( 3 \neq 1 \), the function \( f \) is not injective (one-to-one).

### Step 4: Checking for Constant Function
Now, let's check if \( f \) is a constant function by evaluating \( f(1.1) \):
\[ f(1.1) = \frac{\sin(\pi \{1.1\})}{(1.1)^4 + 3(1.1)^2 + 7} \]
Calculating \( \{1.1\} = 1.1 - 1 = 0.1 \):
\[ f(1.1) = \frac{\sin(0.1\pi)}{(1.1)^4 + 3(1.1)^2 + 7} \]
The numerator \( \sin(0.1\pi) \) is not zero, and the denominator is a positive value, so \( f(1.1) \neq 0 \). Thus, \( f \) is not a constant function.

### Step 5: Checking for Surjectivity
To check if \( f \) is surjective (onto), we analyze the range of \( f \):
- The numerator \( \sin(\pi \{x\}) \) varies between -1 and 1.
- The denominator \( x^4 + 3x^2 + 7 \) is always positive for all real \( x \).

Thus, \( f(x) \) will always yield values between \(-1\) and \(1\), but it cannot cover all real numbers \( \mathbb{R} \). Therefore, \( f \) is not surjective.

### Final Conclusion
Based on the evaluations:
- \( f \) is not injective.
- \( f \) is not constant.
- \( f \) is not surjective.

Thus, the correct option is that \( f \) is neither injective nor surjective, and it is not a constant function.