Let `f : R to R` be defined by
`f(x) = ((x^(6) +1)x(x+1)+x^(6)+1)/(x^(2) + x+1)`, Then f is

Options

A. one-to-one but not onto
B. onto but not one-to-one
C. both one-to-one and onto
D. neither one-to-one nor onto

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{(x^6 + 1)x(x + 1) + (x^6 + 1)}{x^2 + x + 1} \).

### Step 1: Simplify the Function
We start by simplifying the function. The numerator can be rewritten as:
\[ f(x) = \frac{(x^6 + 1)(x(x + 1) + 1)}{x^2 + x + 1} \]
This simplifies to:
\[ f(x) = \frac{(x^6 + 1)(x^2 + x + 1)}{x^2 + x + 1} \]
Since \( x^2 + x + 1 \) is not zero for any real \( x \) (its discriminant is negative), we can cancel it out:
\[ f(x) = x^6 + 1 \]

### Step 2: Determine the Nature of the Function
Now, we need to determine whether the function \( f(x) = x^6 + 1 \) is one-one (injective) or onto (surjective).

#### One-One Function
A function is one-one if it is either always increasing or always decreasing. We find the derivative:
\[ f'(x) = 6x^5 \]
- For \( x > 0 \), \( f'(x) > 0 \) (increasing).
- For \( x < 0 \), \( f'(x) < 0 \) (decreasing).
- At \( x = 0 \), \( f'(0) = 0 \).

Since the function is not consistently increasing or decreasing across its entire domain, it is not one-one.

#### Onto Function
To check if the function is onto, we need to find the range of \( f(x) \):
- The minimum value of \( f(x) = x^6 + 1 \) occurs at \( x = 0 \), which gives \( f(0) = 1 \).
- As \( x \) approaches \( \pm \infty \), \( f(x) \) approaches \( \infty \).

Thus, the range of \( f(x) \) is \( [1, \infty) \).

Since the codomain is \( \mathbb{R} \) (which includes all real numbers), and the range \( [1, \infty) \) does not cover all real numbers, \( f(x) \) is not onto.

### Conclusion
The function \( f(x) = x^6 + 1 \) is neither one-one nor onto.

### Final Answer
The function \( f \) is many-one.

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