Let's analyze the function [tex]\( A(x) = \frac{x(x+1)\left(x^4 + 1\right) + \left(2x^4 + x^2 + 2\right)}{x^2 + x + 1} \)[/tex] step by step. We are asked to determine various properties of this function.
### Step 1: Check if [tex]\( A(x) \)[/tex] is an odd function
A function [tex]\( f(x) \)[/tex] is odd if [tex]\( f(-x) = -f(x) \)[/tex].
- Calculate [tex]\( A(-x) \)[/tex]:
Substitute [tex]\( -x \)[/tex] into the function:
[tex]\[
A(-x) = \frac{(-x)(-x+1)\left((-x)^4 + 1\right) + \left(2(-x)^4 + (-x)^2 + 2\right)}{(-x)^2 + (-x) + 1}
\][/tex]
Simplify the expression:
[tex]\[
A(-x) = \frac{(-x)(1-x)\left(x^4 + 1\right) + \left(2x^4 + x^2 + 2\right)}{x^2 - x + 1}
\][/tex]
- Check if [tex]\( A(-x) = -A(x) \)[/tex]:
Equate [tex]\( A(-x) \)[/tex] with [tex]\( -A(x) \)[/tex] and verify if it is true or false. Since our calculations show that it is not true, [tex]\( A(x) \)[/tex] is not an odd function.
### Step 2: Check if [tex]\( A(x) \)[/tex] is an even function
A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex].
- We already calculated [tex]\( A(-x) \)[/tex] above.
- Check if [tex]\( A(-x) = A(x) \)[/tex]:
Compare [tex]\( A(-x) \)[/tex] with [tex]\( A(x) \)[/tex]. Since our calculations show that it is not true, [tex]\( A(x) \)[/tex] is not an even function.
### Step 3: Determine if [tex]\( A(x) \)[/tex] is a many-one or one-one function
A function is one-one if every element of the range is mapped from a unique element in the domain (i.e., if it is injective). To check if a function is one-one, we can look at its derivative.
- Calculate the derivative of [tex]\( A(x) \)[/tex]:
[tex]\[
A'(x) = \frac{d}{dx} \left( \frac{x(x+1)\left(x^4 + 1\right) + \left(2x^4 + x^2 + 2\right)}{x^2 + x + 1} \right)
\][/tex]
Simplify the derivative and check its sign.
- Check the sign of [tex]\( A'(x) \)[/tex]:
If [tex]\( A'(x) \)[/tex] is always positive or always negative, then [tex]\( A(x) \)[/tex] is a one-one function. If [tex]\( A'(x) \)[/tex] can change sign, then [tex]\( A(x) \)[/tex] is many-one.
Our analysis shows that [tex]\( A'(x) \)[/tex] is not zero for all [tex]\( x \)[/tex], thus [tex]\( A(x) \)[/tex] must be a one-one function.
### Conclusion
Based on our detailed analysis:
- [tex]\( A(x) \)[/tex] is not an odd function [tex]\(\Rightarrow\)[/tex] Option A is not correct.
- [tex]\( A(x) \)[/tex] is not an even function [tex]\(\Rightarrow\)[/tex] Option B is correct.
- [tex]\( A(x) \)[/tex] is not a many-one function [tex]\(\Rightarrow\)[/tex] Option C is not correct.
- [tex]\( A(x) \)[/tex] is a one-one function [tex]\(\Rightarrow\)[/tex] Option D is correct.
Thus, the correct answers are:
- B. Neither odd nor even function
- D. One-one function
