Le `f: R to R` be a function defined by `f(x) = x^(2009) + 2009x + 2009`
Then f(x) is

Options

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

Answer

To determine the properties of the function \( f(x) = x^{2009} + 2009x + 2009 \), we need to analyze whether it is one-one (injective) and onto (surjective).

### Step 1: Check if the function is one-one

A function is one-one if it is either strictly increasing or strictly decreasing. We can check this by finding the derivative of the function.

**Derivative Calculation:**
\[ f'(x) = \frac{d}{dx}(x^{2009} + 2009x + 2009) = 2009x^{2008} + 2009 \]

### Step 2: Analyze the derivative

Now, we need to analyze the sign of \( f'(x) \).

1. The term \( 2009x^{2008} \) is always non-negative for all \( x \) since \( 2008 \) is an even power.
2. The constant term \( 2009 \) is positive.

Thus, we can conclude:
\[ f'(x) = 2009x^{2008} + 2009 > 0 \quad \text{for all } x \in \mathbb{R} \]

Since \( f'(x) > 0 \) for all \( x \), the function \( f(x) \) is strictly increasing.

### Step 3: Conclusion on one-one property

Since \( f(x) \) is strictly increasing, it is a one-one function.

### Step 4: Check if the function is onto

A function is onto if its range is equal to its codomain. The codomain of \( f(x) \) is \( \mathbb{R} \).

To determine the range, we note that \( f(x) \) is a polynomial function. The leading term \( x^{2009} \) dominates as \( x \to \infty \) and \( x \to -\infty \).

1. As \( x \to \infty \), \( f(x) \to \infty \).
2. As \( x \to -\infty \), since \( 2009 \) is odd, \( f(x) \to -\infty \).

Thus, the range of \( f(x) \) is all real numbers \( \mathbb{R} \).

### Step 5: Conclusion on onto property

Since the range of \( f(x) \) is \( \mathbb{R} \) and the codomain is also \( \mathbb{R} \), we conclude that \( f(x) \) is onto.

### Final Conclusion

The function \( f(x) = x^{2009} + 2009x + 2009 \) is both one-one and onto.

### Summary of Results
- **One-One:** Yes
- **Onto:** Yes

Thus, the correct option is that \( f(x) \) is both one-one and onto.

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